Change of Basis

Representations vary with the bases. For instance, with respect to the bases $ℰ_{2}$ and

$B = ⟨ (\begin{matrix} 1 \\ 1 \end{matrix}), (\begin{matrix} 1 \\ - 1 \end{matrix}) ⟩$

${\vec{e}}_{1} \in ℝ^{2}$ has these different representations.

${Rep}_{ℰ_{2}} ({\vec{e}}_{1}) = (\begin{matrix} 1 \\ 0 \end{matrix}) {Rep}_{B} ({\vec{e}}_{1}) = (\begin{matrix} 1 / 2 \\ 1 / 2 \end{matrix})$

The same holds for maps: with respect to the basis pairs $ℰ_{2}, ℰ_{2}$ and $ℰ_{2}, B$ , the identity map has these representations.

${Rep}_{ℰ_{2}, ℰ_{2}} (id) = (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}) {Rep}_{ℰ_{2}, B} (id) = (\begin{matrix} 1 / 2 & 1 / 2 \\ 1 / 2 & - 1 / 2 \end{matrix})$

This section shows how to translate among the representations. That is, we will compute how the representations vary as the bases vary.

Changing Representations of Vectors

In converting ${Rep}_{B} (\vec{v})$ to ${Rep}_{D} (\vec{v})$ the underlying vector $\vec{v}$ doesn’t change. Thus, the translation between these two ways of expressing the vector is accomplished by the identity map on the space, described so that the domain space vectors are represented with respect to $B$ and the codomain space vectors are represented with respect to $D$ .

$\begin{matrix} V_{wrt B} \\ \begin{matrix} id & ↓ \end{matrix} \\ V_{wrt D} \end{matrix}$

(This diagram is vertical to fit with the ones in the next subsection.)

Definition 1.1 The change of basis matrix for bases $B, D \subset V$ is the representation of the identity map $id : V \to V$ with respect to those bases.

${Rep}_{B, D} (id) = (\begin{array}{ccccc} ⋮ & ⋮ \\ {Rep}_{D} ({\vec{β}}_{1}) & \dots & {Rep}_{D} ({\vec{β}}_{n}) \\ ⋮ & ⋮ \end{array})$

Remark 1.2 A better name would be ‘change of representation matrix’ but the above name is standard.

The next result supports the definition.

Lemma 1.3 To convert from the representation of a vector $\vec{v}$ with respect to $B$ to its representation with respect to $D$ use the change of basis matrix.

${Rep}_{B, D} (id) {Rep}_{B} (\vec{v}) = {Rep}_{D} (\vec{v})$

Conversely, if left-multiplication by a matrix changes bases $M \cdot {Rep}_{B} (\vec{v}) = {Rep}_{D} (\vec{v})$ then $M$ is a change of basis matrix.

Proof The first sentence holds because matrix-vector multiplication represents a map application and so ${Rep}_{B, D} (id) \cdot {Rep}_{B} (\vec{v}) = {Rep}_{D} (id (\vec{v})) = {Rep}_{D} (\vec{v})$ for each $\vec{v}$ . For the second sentence, with respect to $B, D$ the matrix $M$ represents a linear map whose action is to map each vector to itself, and is therefore the identity map.

QED

Example 1.4 With these bases for $ℝ^{2}$ ,

$B = ⟨ (\begin{matrix} 2 \\ 1 \end{matrix}), (\begin{matrix} 1 \\ 0 \end{matrix}) ⟩ D = ⟨ (\begin{matrix} - 1 \\ 1 \end{matrix}), (\begin{matrix} 1 \\ 1 \end{matrix}) ⟩$

because

${Rep}_{D} (id ((\begin{matrix} 2 \\ 1 \end{matrix}))) = {(\begin{matrix} - 1 / 2 \\ 3 / 2 \end{matrix})}_{D} {Rep}_{D} (id ((\begin{matrix} 1 \\ 0 \end{matrix}))) = {(\begin{matrix} - 1 / 2 \\ 1 / 2 \end{matrix})}_{D}$

the change of basis matrix is this.

${Rep}_{B, D} (i d) = (\begin{matrix} - 1 / 2 & - 1 / 2 \\ 3 / 2 & 1 / 2 \end{matrix})$

For instance, this is the representation of ${\vec{e}}_{2}$

${Rep}_{B} ((\begin{matrix} 0 \\ 1 \end{matrix})) = (\begin{matrix} 1 \\ - 2 \end{matrix})$

and the matrix does the conversion.

$(\begin{matrix} - 1 / 2 & - 1 / 2 \\ 3 / 2 & 1 / 2 \end{matrix}) (\begin{matrix} 1 \\ - 2 \end{matrix}) = (\begin{matrix} 1 / 2 \\ 1 / 2 \end{matrix})$

Checking that vector on the right is ${Rep}_{D} ({\vec{e}}_{2})$ is easy.

We finish this subsection by recognizing the change of basis matrices as a familiar set.

Lemma 1.5 A matrix changes bases if and only if it is nonsingular.

Proof For the ‘only if’ direction, if left-multiplication by a matrix changes bases then the matrix represents an invertible function, simply because we can invert the function by changing the bases back. Because it represents a function that is invertible, the matrix itself is invertible, and so is nonsingular.

For ‘if’ we will show that any nonsingular matrix $M$ performs a change of basis operation from any given starting basis $B$ (having $n$ vectors, where the matrix is $n \times n$ ) to some ending basis.

If the matrix is the identity $I$ then the statement is obvious. Otherwise because the matrix is nonsingular Corollary IV.3.23 says there are elementary reduction matrices such that $R_{r} \dots R_{1} \cdot M = I$ with $r \geq 1$ . Elementary matrices are invertible and their inverses are also elementary so multiplying both sides of that equation from the left by ${R_{r}}^{- 1}$ , then by ${R_{r - 1}}^{- 1}$ , etc., gives $M$ as a product of elementary matrices $M = {R_{1}}^{- 1} \dots {R_{r}}^{- 1}$ .

We will be done if we show that elementary matrices change a given basis to another basis, since then ${R_{r}}^{- 1}$ changes $B$ to some other basis $B_{r}$ and ${R_{r - 1}}^{- 1}$ changes $B_{r}$ to some $B_{r - 1}$ , etc. We will cover the three types of elementary matrices separately; recall the notation for the three.

$M_{i} (k) (\begin{matrix} c_{1} \\ ⋮ \\ c_{i} \\ ⋮ \\ c_{n} \end{matrix}) = (\begin{matrix} c_{1} \\ ⋮ \\ k c_{i} \\ ⋮ \\ c_{n} \end{matrix}) P_{i, j} (\begin{matrix} c_{1} \\ ⋮ \\ c_{i} \\ ⋮ \\ c_{j} \\ ⋮ \\ c_{n} \end{matrix}) = (\begin{matrix} c_{1} \\ ⋮ \\ c_{j} \\ ⋮ \\ c_{i} \\ ⋮ \\ c_{n} \end{matrix}) C_{i, j} (k) (\begin{matrix} c_{1} \\ ⋮ \\ c_{i} \\ ⋮ \\ c_{j} \\ ⋮ \\ c_{n} \end{matrix}) = (\begin{matrix} c_{1} \\ ⋮ \\ c_{i} \\ ⋮ \\ k c_{i} + c_{j} \\ ⋮ \\ c_{n} \end{matrix})$

Applying a row-multiplication matrix $M_{i} (k)$ changes a representation with respect to $⟨ {\vec{β}}_{1}, \dots, {\vec{β}}_{i}, \dots, {\vec{β}}_{n} ⟩$ to one with respect to $⟨ {\vec{β}}_{1}, \dots, (1 / k) {\vec{β}}_{i}, \dots, {\vec{β}}_{n} ⟩$ .

$\begin{matrix} \vec{v} = c_{1} \cdot {\vec{β}}_{1} + \dots + c_{i} \cdot {\vec{β}}_{i} + \dots + c_{n} \cdot {\vec{β}}_{n} \\ \mapsto c_{1} \cdot {\vec{β}}_{1} + \dots + k c_{i} \cdot (1 / k) {\vec{β}}_{i} + \dots + c_{n} \cdot {\vec{β}}_{n} = \vec{v} \end{matrix}$

The second one is a basis because the first is a basis and because of the $k \neq 0$ restriction in the definition of a row-multiplication matrix. Similarly, left-multiplication by a row-swap matrix $P_{i, j}$ changes a representation with respect to the basis $⟨ {\vec{β}}_{1}, \dots, {\vec{β}}_{i}, \dots, {\vec{β}}_{j}, \dots, {\vec{β}}_{n} ⟩$ into one with respect to this basis $⟨ {\vec{β}}_{1}, \dots, {\vec{β}}_{j}, \dots, {\vec{β}}_{i}, \dots, {\vec{β}}_{n} ⟩$ .

$\begin{matrix} \vec{v} = c_{1} \cdot {\vec{β}}_{1} + \dots + c_{i} \cdot {\vec{β}}_{i} + \dots + c_{j} {\vec{β}}_{j} + \dots + c_{n} \cdot {\vec{β}}_{n} \\ \mapsto c_{1} \cdot {\vec{β}}_{1} + \dots + c_{j} \cdot {\vec{β}}_{j} + \dots + c_{i} \cdot {\vec{β}}_{i} + \dots + c_{n} \cdot {\vec{β}}_{n} = \vec{v} \end{matrix}$

And, a representation with respect to $⟨ {\vec{β}}_{1}, \dots, {\vec{β}}_{i}, \dots, {\vec{β}}_{j}, \dots, {\vec{β}}_{n} ⟩$ changes via left-multiplication by a row-combination matrix $C_{i, j} (k)$ into a representation with respect to $⟨ {\vec{β}}_{1}, \dots, {\vec{β}}_{i} - k {\vec{β}}_{j}, \dots, {\vec{β}}_{j}, \dots, {\vec{β}}_{n} ⟩$

$\begin{matrix} \vec{v} = c_{1} \cdot {\vec{β}}_{1} + \dots + c_{i} \cdot {\vec{β}}_{i} + c_{j} {\vec{β}}_{j} + \dots + c_{n} \cdot {\vec{β}}_{n} \\ \mapsto c_{1} \cdot {\vec{β}}_{1} + \dots + c_{i} \cdot ({\vec{β}}_{i} - k {\vec{β}}_{j}) + \dots + (k c_{i} + c_{j}) \cdot {\vec{β}}_{j} + \dots + c_{n} \cdot {\vec{β}}_{n} = \vec{v} \end{matrix}$

(the definition of $C_{i, j} (k)$ specifies that $i \neq j$ and $k \neq 0$ ).

QED

Corollary 1.6 A matrix is nonsingular if and only if it represents the identity map with respect to some pair of bases.

Exercises

Exercise 1.7 Worked answer

Recommended. In $ℝ^{2}$ , where

$D = ⟨ (\begin{matrix} 2 \\ 1 \end{matrix}), (\begin{matrix} - 2 \\ 4 \end{matrix}) ⟩$

find the change of basis matrices from $D$ to $ℰ_{2}$ and from $ℰ_{2}$ to $D$ . Multiply the two.

Back to Exercise 1.7

Answer. For the matrix to change bases from $D$ to $ℰ_{2}$ we need that ${Rep}_{ℰ_{2}} (id ({\vec{δ}}_{1})) = {Rep}_{ℰ_{2}} ({\vec{δ}}_{1})$ and that ${Rep}_{ℰ_{2}} (id ({\vec{δ}}_{2})) = {Rep}_{ℰ_{2}} ({\vec{δ}}_{2})$ . Of course, the representation of a vector in $ℝ^{2}$ with respect to the standard basis is easy.

${Rep}_{ℰ_{2}} ({\vec{δ}}_{1}) = (\begin{matrix} 2 \\ 1 \end{matrix}) {Rep}_{ℰ_{2}} ({\vec{δ}}_{2}) = (\begin{matrix} - 2 \\ 4 \end{matrix})$

Concatenating those two together to make the columns of the change of basis matrix gives this.

${Rep}_{D, ℰ_{2}} (id) = (\begin{matrix} 2 & - 2 \\ 1 & 4 \end{matrix})$

For the change of basis matrix in the other direction we can calculate ${Rep}_{D} (id ({\vec{e}}_{1})) = {Rep}_{D} ({\vec{e}}_{1})$ and ${Rep}_{D} (id ({\vec{e}}_{2})) = {Rep}_{D} ({\vec{e}}_{2})$ (this job is routine) or we can take the inverse of the above matrix. Because of the formula for the inverse of a $2 \times 2$ matrix, this is easy.

${Rep}_{ℰ_{2}, D} (id) = \frac{1}{10} \cdot (\begin{matrix} 4 & 2 \\ - 1 & 2 \end{matrix}) = (\begin{matrix} 4 / 10 & 2 / 10 \\ - 1 / 10 & 2 / 10 \end{matrix})$
Exercise 1.8 Worked answer

Which of these matrices could be used to change bases?
1. $(\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix})$
2. $(\begin{matrix} 0 & - 1 \\ 1 & - 1 \end{matrix})$
3. $(\begin{matrix} 2 & 3 & - 1 \\ 0 & 1 & 0 \end{matrix})$
4. $(\begin{matrix} 2 & 3 & - 1 \\ 0 & 1 & 0 \\ 4 & 7 & - 2 \end{matrix})$
5. $(\begin{matrix} 0 & 2 & 0 \\ 0 & 0 & 6 \\ 1 & 0 & 0 \end{matrix})$
Back to Exercise 1.8

Answer. If the matrix is nonsingular then it can be a change of basis matrix. For all of these matrices we can check that by eye.
1. This is nonsingular since the second row is not a multiple of the first.
2. This is nonsingular.
3. This matrix is singular, since to be nonsingular a matrix must be square.
4. This matrix is singular since twice the first row plus the second row equals the third row.
5. Nonsingular.
Exercise 1.9 Worked answer

Recommended. Find the change of basis matrix for $B, D \subseteq ℝ^{2}$ .
1. $B = ℰ_{2}$ , $D = ⟨ {\vec{e}}_{2}, {\vec{e}}_{1} ⟩$
2. $B = ℰ_{2}$ , $D = ⟨ (\begin{matrix} 1 \\ 2 \end{matrix}), (\begin{matrix} 1 \\ 4 \end{matrix}) ⟩$
3. $B = ⟨ (\begin{matrix} 1 \\ 2 \end{matrix}), (\begin{matrix} 1 \\ 4 \end{matrix}) ⟩$ , $D = ℰ_{2}$
4. $B = ⟨ (\begin{matrix} - 1 \\ 1 \end{matrix}), (\begin{matrix} 2 \\ 2 \end{matrix}) ⟩$ , $D = ⟨ (\begin{matrix} 0 \\ 4 \end{matrix}), (\begin{matrix} 1 \\ 3 \end{matrix}) ⟩$
Back to Exercise 1.9

Answer. Concatenate ${Rep}_{D} (id ({\vec{β}}_{1})) = {Rep}_{D} ({\vec{β}}_{1})$ and ${Rep}_{D} (id ({\vec{β}}_{2})) = {Rep}_{D} ({\vec{β}}_{2})$ to make the change of basis matrix ${Rep}_{B, D} (id)$ .
1. $(\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix})$
2. $(\begin{matrix} 2 & - 1 / 2 \\ - 1 & 1 / 2 \end{matrix})$
3. $(\begin{matrix} 1 & 1 \\ 2 & 4 \end{matrix})$
4. $(\begin{matrix} 1 & - 1 \\ - 1 & 2 \end{matrix})$
Exercise 1.10 Worked answer

Recommended. Find the change of basis matrix for each $B, D \subseteq 𝒫_{2}$ .
1. $B = ⟨ 1, x, x^{2} ⟩, D = ⟨ x^{2}, 1, x ⟩$
2. $B = ⟨ 1, x, x^{2} ⟩, D = ⟨ 1, 1 + x, 1 + x + x^{2} ⟩$
3. $B = ⟨ 2, 2 x, x^{2} ⟩, D = ⟨ 1 + x^{2}, 1 - x^{2}, x + x^{2} ⟩$
Back to Exercise 1.10

Answer. The vectors ${Rep}_{D} (id ({\vec{β}}_{1})) = {Rep}_{D} ({\vec{β}}_{1})$ , ${Rep}_{D} (id ({\vec{β}}_{2})) = {Rep}_{D} ({\vec{β}}_{2})$ , and ${Rep}_{D} (id ({\vec{β}}_{3})) = {Rep}_{D} ({\vec{β}}_{3})$ make the change of basis matrix ${Rep}_{B, D} (id)$ .
1. $(\begin{matrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix})$
2. $(\begin{matrix} 1 & - 1 & 0 \\ 0 & 1 & - 1 \\ 0 & 0 & 1 \end{matrix})$
3. $(\begin{matrix} 1 & - 1 & 1 / 2 \\ 1 & 1 & - 1 / 2 \\ 0 & 2 & 0 \end{matrix})$
E.g., for the first column of the first matrix, $1 = 0 \cdot x^{2} + 1 \cdot 1 + 0 \cdot x$ .
Exercise 1.11 Worked answer

For the bases in Exercise 1.9, find the change of basis matrix in the other direction, from $D$ to $B$ .
Back to Exercise 1.11

Answer. One way to go is to find ${Rep}_{B} ({\vec{δ}}_{1})$ and ${Rep}_{B} ({\vec{δ}}_{2})$ , and then concatenate them into the columns of the desired change of basis matrix. Another way is to find the inverse of the matrices that answer Exercise 1.9.
1. $(\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix})$
2. $(\begin{matrix} 1 & 1 \\ 2 & 4 \end{matrix})$
3. $(\begin{matrix} 2 & - 1 / 2 \\ - 1 & 1 / 2 \end{matrix})$
4. $(\begin{matrix} 2 & 1 \\ 1 & 1 \end{matrix})$
Exercise 1.12 Worked answer

Recommended. Decide if each changes bases on $ℝ^{2}$ . To what basis is $ℰ_{2}$ changed?
1. $(\begin{matrix} 5 & 0 \\ 0 & 4 \end{matrix})$
2. $(\begin{matrix} 2 & 1 \\ 3 & 1 \end{matrix})$
3. $(\begin{matrix} - 1 & 4 \\ 2 & - 8 \end{matrix})$
4. $(\begin{matrix} 1 & - 1 \\ 1 & 1 \end{matrix})$
Back to Exercise 1.12

Answer. A matrix changes bases if and only if it is nonsingular.
1. This matrix is nonsingular and so changes bases. Finding to what basis $ℰ_{2}$ is changed means finding $D$ such that
  
  ${Rep}_{ℰ_{2}, D} (id) = (\begin{matrix} 5 & 0 \\ 0 & 4 \end{matrix})$
  
  and by the definition of how a matrix represents a linear map, we have this.
  
  ${Rep}_{D} (id ({\vec{e}}_{1})) = {Rep}_{D} ({\vec{e}}_{1}) = (\begin{matrix} 5 \\ 0 \end{matrix}) {Rep}_{D} (id ({\vec{e}}_{2})) = {Rep}_{D} ({\vec{e}}_{2}) = (\begin{matrix} 0 \\ 4 \end{matrix})$
  
  Where
  
  $D = ⟨ (\begin{matrix} x_{1} \\ y_{1} \end{matrix}), (\begin{matrix} x_{2} \\ y_{2} \end{matrix}) ⟩$
  
  we can either solve the system
  
  $(\begin{matrix} 1 \\ 0 \end{matrix}) = 5 (\begin{matrix} x_{1} \\ y_{1} \end{matrix}) + 0 (\begin{matrix} x_{2} \\ y_{1} \end{matrix}) (\begin{matrix} 0 \\ 1 \end{matrix}) = 0 (\begin{matrix} x_{1} \\ y_{1} \end{matrix}) + 4 (\begin{matrix} x_{2} \\ y_{1} \end{matrix})$
  
  or else just spot the answer (thinking of the proof of Lemma 1.5).
  
  $D = ⟨ (\begin{matrix} 1 / 5 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 1 / 4 \end{matrix}) ⟩$
2. Yes, this matrix is nonsingular and so changes bases. To calculate $D$ , we proceed as above with
  
  $D = ⟨ (\begin{matrix} x_{1} \\ y_{1} \end{matrix}), (\begin{matrix} x_{2} \\ y_{2} \end{matrix}) ⟩$
  
  to solve
  
  $(\begin{matrix} 1 \\ 0 \end{matrix}) = 2 (\begin{matrix} x_{1} \\ y_{1} \end{matrix}) + 3 (\begin{matrix} x_{2} \\ y_{1} \end{matrix}) and (\begin{matrix} 0 \\ 1 \end{matrix}) = 1 (\begin{matrix} x_{1} \\ y_{1} \end{matrix}) + 1 (\begin{matrix} x_{2} \\ y_{1} \end{matrix})$
  
  and get this.
  
  $D = ⟨ (\begin{matrix} - 1 \\ 3 \end{matrix}), (\begin{matrix} 1 \\ - 2 \end{matrix}) ⟩$
3. No, this matrix does not change bases because it is singular.
4. Yes, this matrix changes bases because it is nonsingular. The calculation of the changed-to basis is as above.
  
  $D = ⟨ (\begin{matrix} 1 / 2 \\ - 1 / 2 \end{matrix}), (\begin{matrix} 1 / 2 \\ 1 / 2 \end{matrix}) ⟩$
Exercise 1.13 Worked answer

For each space find the matrix changing a vector representation with respect to $B$ to one with respect to $D$ .
1. $V = ℝ^{3}$ , $B = ℰ_{3}$ , $D = ⟨ (\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}), (\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}), (\begin{matrix} 0 \\ 1 \\ - 1 \end{matrix}) ⟩$
2. $V = ℝ^{3}$ , $B = ⟨ (\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}), (\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}), (\begin{matrix} 0 \\ 1 \\ - 1 \end{matrix}) ⟩$ , $D = ℰ_{3}$
3. $V = 𝒫_{2}$ , $B = ⟨ x^{2}, x^{2} + x, x^{2} + x + 1 ⟩$ , $D = ⟨ 2, - x, x^{2} ⟩$
Back to Exercise 1.13

Answer.
1. Start by computing the effect of the identity function on each element of the starting basis $B$ . Obviously this is the effect.
  
  $(\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) \overset{id}{⟼} (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}) \overset{id}{⟼} (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}) (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}) \overset{id}{⟼} (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix})$
  
  Now represent the three outputs with respect to the ending basis.
  
  ${Rep}_{D} ((\begin{matrix} 1 \\ 0 \\ 0 \end{matrix})) = (\begin{matrix} - 2 / 3 \\ 5 / 3 \\ - 1 / 3 \end{matrix}) {Rep}_{D} ((\begin{matrix} 0 \\ 1 \\ 0 \end{matrix})) = (\begin{matrix} 1 / 3 \\ - 1 / 3 \\ 2 / 3 \end{matrix}) {Rep}_{D} ((\begin{matrix} 0 \\ 0 \\ 1 \end{matrix})) = (\begin{matrix} 1 / 3 \\ - 1 / 3 \\ - 1 / 3 \end{matrix})$
  
  Concatenate them into a basis.
  
  ${Rep}_{B, D} (id) = (\begin{matrix} - 2 / 3 & 1 / 3 & 1 / 3 \\ 5 / 3 & - 1 / 3 & - 1 / 3 \\ - 1 / 3 & 2 / 3 & - 1 / 3 \end{matrix})$
2. One way to find this is to take the inverse of the prior matrix, since it converts bases in the other direction. Alternatively, we can compute these three
  
  ${Rep}_{ℰ_{3}} ((\begin{matrix} 1 \\ 2 \\ 3 \end{matrix})) = (\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}) {Rep}_{ℰ_{3}} ((\begin{matrix} 1 \\ 1 \\ 1 \end{matrix})) = (\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}) {Rep}_{ℰ_{3}} ((\begin{matrix} 0 \\ 1 \\ - 1 \end{matrix})) = (\begin{matrix} 0 \\ 1 \\ - 1 \end{matrix})$
  
  and put them in a matrix.
  
  ${Rep}_{B, D} (id) = (\begin{matrix} 1 & 1 & 0 \\ 2 & 1 & 1 \\ 3 & 1 & - 1 \end{matrix})$
3. Representing $id (x^{2})$ , $id (x^{2} + x)$ , and $id (x^{2} + x + 1)$ with respect to the ending basis gives this.
  
  ${Rep}_{D} (x^{2}) = (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}) {Rep}_{D} (x^{2} + x) = (\begin{matrix} 0 \\ - 1 \\ 1 \end{matrix}) {Rep}_{D} (x^{2} + x + 1) = (\begin{matrix} 1 / 2 \\ - 1 \\ 1 \end{matrix})$
  
  Put them together.
  
  ${Rep}_{B, D} (id) = (\begin{matrix} 0 & 0 & 1 / 2 \\ 0 & - 1 & - 1 \\ 1 & 1 & 1 \end{matrix})$
Exercise 1.14 Worked answer

Find bases such that this matrix represents the identity map with respect to those bases.

$(\begin{matrix} 3 & 1 & 4 \\ 2 & - 1 & 1 \\ 0 & 0 & 4 \end{matrix})$

Back to Exercise 1.14

Answer. This question has many different solutions. One way to proceed is to make up any basis $B$ for any space, and then compute the appropriate $D$ (necessarily for the same space, of course). Another, easier, way to proceed is to fix the codomain as $ℝ^{3}$ and the codomain basis as $ℰ_{3}$ . This way (recall that the representation of any vector with respect to the standard basis is just the vector itself), we have this.

$B = ⟨ (\begin{matrix} 3 \\ 2 \\ 0 \end{matrix}), (\begin{matrix} 1 \\ - 1 \\ 0 \end{matrix}), (\begin{matrix} 4 \\ 1 \\ 4 \end{matrix}) ⟩ D = ℰ_{3}$
Exercise 1.15 Worked answer

Consider the vector space of real-valued functions with basis $⟨ \sin (x), \cos (x) ⟩$ . Show that $⟨ 2 \sin (x) + \cos (x), 3 \cos (x) ⟩$ is also a basis for this space. Find the change of basis matrix in each direction.

Back to Exercise 1.15

Answer. Checking that $B = ⟨ 2 \sin (x) + \cos (x), 3 \cos (x) ⟩$ is a basis is routine. Call the natural basis $D$ . To compute the change of basis matrix ${Rep}_{B, D} (id)$ we must find ${Rep}_{D} (2 \sin (x) + \cos (x))$ and ${Rep}_{D} (3 \cos (x))$ , that is, we need $x_{1}, y_{1}, x_{2}, y_{2}$ such that these equations hold.

$\begin{aligned} x_{1} \cdot \sin (x) + y_{1} \cdot \cos (x) & = 2 \sin (x) + \cos (x) \\ x_{2} \cdot \sin (x) + y_{2} \cdot \cos (x) & = 3 \cos (x) \end{aligned}$

Obviously this is the answer.

${Rep}_{B, D} (id) = (\begin{matrix} 2 & 0 \\ 1 & 3 \end{matrix})$

For the change of basis matrix in the other direction we could look for ${Rep}_{B} (\sin (x))$ and ${Rep}_{B} (\cos (x))$ by solving these.

$\begin{aligned} w_{1} \cdot (2 \sin (x) + \cos (x)) + z_{1} \cdot (3 \cos (x)) & = \sin (x) \\ w_{2} \cdot (2 \sin (x) + \cos (x)) + z_{2} \cdot (3 \cos (x)) & = \cos (x) \end{aligned}$

An easier method is to find the inverse of the matrix found above.

${Rep}_{D, B} (id) = {(\begin{matrix} 2 & 0 \\ 1 & 3 \end{matrix})}^{- 1} = \frac{1}{6} \cdot (\begin{matrix} 3 & 0 \\ - 1 & 2 \end{matrix}) = (\begin{matrix} 1 / 2 & 0 \\ - 1 / 6 & 1 / 3 \end{matrix})$
Exercise 1.16 Worked answer

Where does this matrix

$(\begin{matrix} \cos (2 θ) & \sin (2 θ) \\ \sin (2 θ) & - \cos (2 θ) \end{matrix})$

send the standard basis for $ℝ^{2}$ ? Any other bases? Hint. Consider the inverse.

Back to Exercise 1.16

Answer. We start by taking the inverse of the matrix, that is, by deciding what is the inverse to the map of interest.

$\begin{aligned} {Rep}_{D, ℰ_{2}} (id) {Rep}_{D, ℰ_{2}} (id)^{- 1} & = \frac{1}{- \cos^{2} (2 θ) - \sin^{2} (2 θ)} \cdot (\begin{array}{c} - \cos (2 θ) & - \sin (2 θ) \\ - \sin (2 θ) & \cos (2 θ) \end{array}) \\ = (\begin{array}{c} \cos (2 θ) & \sin (2 θ) \\ \sin (2 θ) & - \cos (2 θ) \end{array}) \end{aligned}$

This is more tractable than the representation the other way because this matrix is the concatenation of these two column vectors

${Rep}_{ℰ_{2}} ({\vec{δ}}_{1}) = (\begin{matrix} \cos (2 θ) \\ \sin (2 θ) \end{matrix}) {Rep}_{ℰ_{2}} ({\vec{δ}}_{2}) = (\begin{matrix} \sin (2 θ) \\ - \cos (2 θ) \end{matrix})$

and representations with respect to $ℰ_{2}$ are transparent.

${\vec{δ}}_{1} = (\begin{matrix} \cos (2 θ) \\ \sin (2 θ) \end{matrix}) {\vec{δ}}_{2} = (\begin{matrix} \sin (2 θ) \\ - \cos (2 θ) \end{matrix})$

This pictures the action of the map that transforms $D$ to $ℰ_{2}$ (it is, again, the inverse of the map that is the answer to this question). The line lies at an angle $θ$ to the $x$ axis.

This map reflects vectors over that line. Since reflections are self-inverse, the answer to the question is: the original map reflects about the line through the origin with angle of elevation $θ$ . (Of course, it does this to any basis.)
Exercise 1.17 Worked answer

Recommended. What is the change of basis matrix with respect to $B, B$ ?

Back to Exercise 1.17

Answer. The appropriately-sized identity matrix.
Exercise 1.18 Worked answer

Prove that a matrix changes bases if and only if it is invertible.

Back to Exercise 1.18

Answer. Each is true if and only if the matrix is nonsingular.
Exercise 1.19 Worked answer

Finish the proof of Lemma 1.5.

Back to Exercise 1.19

Answer. What remains is to show that left multiplication by a reduction matrix represents a change from another basis to $B = ⟨ {\vec{β}}_{1}, \dots, {\vec{β}}_{n} ⟩$ .

Application of a row-multiplication matrix $M_{i} (k)$ translates a representation with respect to the basis $⟨ {\vec{β}}_{1}, \dots, k {\vec{β}}_{i}, \dots, {\vec{β}}_{n} ⟩$ to one with respect to $B$ , as here.

$\vec{v} = c_{1} \cdot {\vec{β}}_{1} + \dots + c_{i} \cdot (k {\vec{β}}_{i}) + \dots + c_{n} \cdot {\vec{β}}_{n} \mapsto c_{1} \cdot {\vec{β}}_{1} + \dots + (k c_{i}) \cdot {\vec{β}}_{i} + \dots + c_{n} \cdot {\vec{β}}_{n} = \vec{v}$

Apply a row-swap matrix $P_{i, j}$ to translates a representation with respect to the basis $⟨ {\vec{β}}_{1}, \dots, {\vec{β}}_{j}, \dots, {\vec{β}}_{i}, \dots, {\vec{β}}_{n} ⟩$ to one with respect to $⟨ {\vec{β}}_{1}, \dots, {\vec{β}}_{i}, \dots, {\vec{β}}_{j}, \dots, {\vec{β}}_{n} ⟩$ . Finally, applying a row-combination matrix $C_{i, j} (k)$ changes a representation with respect to $⟨ {\vec{β}}_{1}, \dots, {\vec{β}}_{i} + k {\vec{β}}_{j}, \dots, {\vec{β}}_{j}, \dots, {\vec{β}}_{n} ⟩$ to one with respect to $B$ .

$\begin{matrix} \vec{v} = c_{1} \cdot {\vec{β}}_{1} + \dots + c_{i} \cdot ({\vec{β}}_{i} + k {\vec{β}}_{j}) + \dots + c_{j} {\vec{β}}_{j} + \dots + c_{n} \cdot {\vec{β}}_{n} \\ \mapsto c_{1} \cdot {\vec{β}}_{1} + \dots + c_{i} \cdot {\vec{β}}_{i} + \dots + (k c_{i} + c_{j}) \cdot {\vec{β}}_{j} + \dots + c_{n} \cdot {\vec{β}}_{n} = \vec{v} \end{matrix}$

(As in the part of the proof in the body of this subsection, the various conditions on the row operations, e.g., that the scalar $k$ is nonzero, assure that these are all bases.)
Exercise 1.20 Worked answer

Recommended. Let $H$ be an $n \times n$ nonsingular matrix. What basis of $ℝ^{n}$ does $H$ change to the standard basis?

Back to Exercise 1.20

Answer. Taking $H$ as a change of basis matrix $H = {Rep}_{B, ℰ_{n}} (id)$ , its columns are

$(\begin{matrix} h_{1, i} \\ ⋮ \\ h_{n, i} \end{matrix}) = {Rep}_{ℰ_{n}} (id ({\vec{β}}_{i})) = {Rep}_{ℰ_{n}} ({\vec{β}}_{i})$

and, because representations with respect to the standard basis are transparent, we have this.

$(\begin{matrix} h_{1, i} \\ ⋮ \\ h_{n, i} \end{matrix}) = {\vec{β}}_{i}$

That is, the basis is the one composed of the columns of $H$ .
Exercise 1.21 Worked answer

Recommended.
1. In $𝒫_{3}$ with basis $B = ⟨ 1 + x, 1 - x, x^{2} + x^{3}, x^{2} - x^{3} ⟩$ we have this representation.
  
  ${Rep}_{B} (1 - x + 3 x^{2} - x^{3}) = {(\begin{matrix} 0 \\ 1 \\ 1 \\ 2 \end{matrix})}_{B}$
  
  Find a basis $D$ giving this different representation for the same polynomial.
  
  ${Rep}_{D} (1 - x + 3 x^{2} - x^{3}) = {(\begin{matrix} 1 \\ 0 \\ 2 \\ 0 \end{matrix})}_{D}$
2. State and prove that we can change any nonzero vector representation to any other.
Hint. The proof of Lemma 1.5 is constructive—it not only says the bases change, it shows how they change.
Back to Exercise 1.21

Answer.
1. We can change the starting vector representation to the ending one through a sequence of row operations. The proof tells us what how the bases change. We start by swapping the first and second rows of the representation with respect to $B$ to get a representation with respect to a new basis $B_{1}$ .
  
  ${Rep}_{B_{1}} (1 - x + 3 x^{2} - x^{3}) = {(\begin{matrix} 1 \\ 0 \\ 1 \\ 2 \end{matrix})}_{B_{1}} B_{1} = ⟨ 1 - x, 1 + x, x^{2} + x^{3}, x^{2} - x^{3} ⟩$
  
  We next add $- 2$ times the third row of the vector representation to the fourth row.
  
  ${Rep}_{B_{3}} (1 - x + 3 x^{2} - x^{3}) = {(\begin{matrix} 1 \\ 0 \\ 1 \\ 0 \end{matrix})}_{B_{2}} B_{2} = ⟨ 1 - x, 1 + x, 3 x^{2} - x^{3}, x^{2} - x^{3} ⟩$
  
  (The third element of $B_{2}$ is the third element of $B_{1}$ minus $- 2$ times the fourth element of $B_{1}$ .) Now we can finish by doubling the third row.
  
  ${Rep}_{D} (1 - x + 3 x^{2} - x^{3}) = {(\begin{matrix} 1 \\ 0 \\ 2 \\ 0 \end{matrix})}_{D} D = ⟨ 1 - x, 1 + x, (3 x^{2} - x^{3}) / 2, x^{2} - x^{3} ⟩$
2. Here are three different approaches to stating such a result. The first is the assertion: where $V$ is a vector space with basis $B$ and $\vec{v} \in V$ is nonzero, for any nonzero column vector $\vec{z}$ (whose number of components equals the dimension of $V$ ) there is a change of basis matrix $M$ such that $M \cdot {Rep}_{B} (\vec{v}) = \vec{z}$ . The second possible statement: for any ( $n$ -dimensional) vector space $V$ and any nonzero vector $\vec{v} \in V$ , where ${\vec{z}}_{1}, {\vec{z}}_{2} \in ℝ^{n}$ are nonzero, there are bases $B, D \subset V$ such that ${Rep}_{B} (\vec{v}) = {\vec{z}}_{1}$ and ${Rep}_{D} (\vec{v}) = {\vec{z}}_{2}$ . The third is: for any nonzero $\vec{v}$ member of any vector space (of dimension $n$ ) and any nonzero column vector (with $n$ components) there is a basis such that $\vec{v}$ is represented with respect to that basis by that column vector.
  
  The first and second statements follow easily from the third. The first follows because the third statement gives a basis $D$ such that ${Rep}_{D} (\vec{v}) = \vec{z}$ and then ${Rep}_{B, D} (id)$ is the desired $M$ . The second follows from the third because it is just a doubled application of it.
  
  A way to prove the third is as in the answer to the first part of this question. Here is a sketch. Represent $\vec{v}$ with respect to any basis $B$ with a column vector ${\vec{z}}_{1}$ . This column vector must have a nonzero component because $\vec{v}$ is a nonzero vector. Use that component in a sequence of row operations to convert ${\vec{z}}_{1}$ to $\vec{z}$ . (We could fill out this sketch as an induction argument on the dimension of $V$ .)
Exercise 1.22 Worked answer

Let $V, W$ be vector spaces, and let $B, \hat{B}$ be bases for $V$ and $D, \hat{D}$ be bases for $W$ . Where $h : V \to W$ is linear, find a formula relating ${Rep}_{B, D} (h)$ to ${Rep}_{\hat{B}, \hat{D}} (h)$ .

Back to Exercise 1.22

Answer. This is the topic of the next subsection.
Exercise 1.23 Worked answer

Recommended. Show that the columns of an $n \times n$ change of basis matrix form a basis for $ℝ^{n}$ . Do all bases appear in that way: can the vectors from any $ℝ^{n}$ basis make the columns of a change of basis matrix?

Back to Exercise 1.23

Answer. A change of basis matrix is nonsingular and thus has rank equal to the number of its columns. Therefore its set of columns is a linearly independent subset of size $n$ in $ℝ^{n}$ and it is thus a basis. The answer to the second half is also ‘yes’; all implications in the prior sentence reverse (that is, all of the ‘if …then …’ parts of the prior sentence convert to ‘if and only if’ parts).
Exercise 1.24 Worked answer

Recommended. Find a matrix having this effect.

$(\begin{matrix} 1 \\ 3 \end{matrix}) \mapsto (\begin{matrix} 4 \\ - 1 \end{matrix})$

That is, find a $M$ that left-multiplies the starting vector to yield the ending vector. Is there a matrix having these two effects?
1. $(\begin{matrix} 1 \\ 3 \end{matrix}) \mapsto (\begin{matrix} 1 \\ 1 \end{matrix}) (\begin{matrix} 2 \\ - 1 \end{matrix}) \mapsto (\begin{matrix} - 1 \\ - 1 \end{matrix})$
2. $(\begin{matrix} 1 \\ 3 \end{matrix}) \mapsto (\begin{matrix} 1 \\ 1 \end{matrix}) (\begin{matrix} 2 \\ 6 \end{matrix}) \mapsto (\begin{matrix} - 1 \\ - 1 \end{matrix})$
Give a necessary and sufficient condition for there to be a matrix such that ${\vec{v}}_{1} \mapsto {\vec{w}}_{1}$ and ${\vec{v}}_{2} \mapsto {\vec{w}}_{2}$ .
Back to Exercise 1.24

Answer. In response to the first half of the question, there are infinitely many such matrices. One of them represents with respect to $ℰ_{2}$ the transformation of $ℝ^{2}$ with this action.

$(\begin{matrix} 1 \\ 0 \end{matrix}) \mapsto (\begin{matrix} 4 \\ 0 \end{matrix}) (\begin{matrix} 0 \\ 1 \end{matrix}) \mapsto (\begin{matrix} 0 \\ - 1 / 3 \end{matrix})$

The problem of specifying two distinct input/output pairs is a bit trickier. The fact that matrices have a linear action precludes some possibilities.
1. Yes, there is such a matrix. These conditions
  
  $(\begin{matrix} a & b \\ c & d \end{matrix}) (\begin{matrix} 1 \\ 3 \end{matrix}) = (\begin{matrix} 1 \\ 1 \end{matrix}) (\begin{matrix} a & b \\ c & d \end{matrix}) (\begin{matrix} 2 \\ - 1 \end{matrix}) = (\begin{matrix} - 1 \\ - 1 \end{matrix})$
  
  can be solved
  
  $\begin{array}{rcrcrcrcr} a & + & 3 b & = & 1 \\ c & + & 3 d & = & 1 \\ 2 a & - & b & = & - 1 \\ 2 c & - & d & = & - 1 \end{array}$
  
  to give this matrix.
  
  $(\begin{matrix} - 2 / 7 & 3 / 7 \\ - 2 / 7 & 3 / 7 \end{matrix})$
2. No, because
  
  $2 \cdot (\begin{matrix} 1 \\ 3 \end{matrix}) = (\begin{matrix} 2 \\ 6 \end{matrix}) but 2 \cdot (\begin{matrix} 1 \\ 1 \end{matrix}) \neq (\begin{matrix} - 1 \\ - 1 \end{matrix})$
  
  no linear action can produce this effect.
3. A sufficient condition is that ${{\vec{v}}_{1}, {\vec{v}}_{2}}$ be linearly independent, but that’s not a necessary condition. A necessary and sufficient condition is that any linear dependences among the starting vectors appear also among the ending vectors. That is,
  
  $c_{1} {\vec{v}}_{1} + c_{2} {\vec{v}}_{2} = \vec{0} implies c_{1} {\vec{w}}_{1} + c_{2} {\vec{w}}_{2} = \vec{0} .$
  
  The proof of this condition is routine.

Changing Map Representations

The first subsection shows how to convert the representation of a vector with respect to one basis to the representation of that same vector with respect to another basis. We next convert the representation of a map with respect to one pair of bases to the representation with respect to a different pair—we convert from ${Rep}_{B, D} (h)$ to ${Rep}_{\hat{B}, \hat{D}} (h)$ . Here is the arrow diagram.

$\begin{matrix} V_{wrt B} & \to_{H}^{h} & W_{wrt D} \\ \begin{matrix} id & ↓ \end{matrix} & \begin{matrix} id & ↓ \end{matrix} \\ V_{wrt \hat{B}} & \to_{\hat{H}}^{h} & W_{wrt \hat{D}} \end{matrix}$

To move from the lower-left to the lower-right we can either go straight over, or else up to $V_{B}$ then over to $W_{D}$ and then down. So we can calculate $\hat{H} = {Rep}_{\hat{B}, \hat{D}} (h)$ either by directly using $\hat{B}$ and $\hat{D}$ , or else by first changing bases with ${Rep}_{\hat{B}, B} (id)$ then multiplying by $H = {Rep}_{B, D} (h)$ and then changing bases with ${Rep}_{D, \hat{D}} (id)$ .

Theorem 2.1 To convert from the matrix $H$ representing a map $h$ with respect to $B, D$ to the matrix $\hat{H}$ representing it with respect to $\hat{B}, \hat{D}$ use this formula.

$\begin{matrix} \hat{H} = {Rep}_{D, \hat{D}} (id) \cdot H \cdot {Rep}_{\hat{B}, B} (id) & (*) \end{matrix}$

Proof This is evident from the diagram.

QED

Example 2.2 The matrix

$T = (\begin{matrix} \cos (π / 6) & - \sin (π / 6) \\ \sin (π / 6) & \cos (π / 6) \end{matrix}) = (\begin{matrix} \sqrt{3} / 2 & - 1 / 2 \\ 1 / 2 & \sqrt{3} / 2 \end{matrix})$

represents, with respect to $ℰ_{2}, ℰ_{2}$ , the transformation $t : ℝ^{2} \to ℝ^{2}$ that rotates vectors through the counterclockwise angle of $π / 6$ radians.

Counterclockwise rotation through pi/6. The vector (1,3) in the left axes maps to ((minus 3 plus square root of 3)/2, (1 plus 3 square root of 3)/2) in the right axes. The mapping arrow is labelled t subscript pi/6.

We can translate $T$ to a representation with respect to these

$\hat{B} = ⟨ (\begin{matrix} 1 \\ 1 \end{matrix}) (\begin{matrix} 0 \\ 2 \end{matrix}) ⟩ \hat{D} = ⟨ (\begin{matrix} - 1 \\ 0 \end{matrix}) (\begin{matrix} 2 \\ 3 \end{matrix}) ⟩$

by using the arrow diagram above.

$\begin{matrix} ℝ_{wrt ℰ_{2}}^{2} & \to_{T}^{t} & ℝ_{wrt ℰ_{2}}^{2} \\ \begin{matrix} id & ↓ \end{matrix} & \begin{matrix} id & ↓ \end{matrix} \\ ℝ_{wrt \hat{B}}^{2} & \to_{\hat{T}}^{t} & ℝ_{wrt \hat{D}}^{2} \end{matrix}$

The picture illustrates that we can compute $\hat{T}$ either directly by going along the square’s bottom, or as in formula ( $*$ ) by going up on the left, then across the top, and then down on the right, with $\hat{T} = {Rep}_{ℰ_{2}, \hat{D}} (id) \cdot T \cdot {Rep}_{\hat{B}, ℰ_{2}} (id)$ . (Note again that the matrix multiplication reads right to left, as the three functions are composed and function composition reads right to left.)

Find the matrix for the left-hand side, the matrix ${Rep}_{\hat{B}, ℰ_{2}} (id)$ , in the usual way: find the effect of the identity matrix on the starting basis $\hat{B}$ — which is no effect at all—and then represent those basis elements with respect to the ending basis $ℰ_{2}$ .

${Rep}_{\hat{B}, ℰ_{2}} (id) = (\begin{matrix} 1 & 0 \\ 1 & 2 \end{matrix})$

This calculation is easy when the ending basis is the standard one.

There are two ways to compute the matrix for going down the square’s right side, ${Rep}_{ℰ_{2}, \hat{D}} (id)$ . We could calculate it directly as we did for the other change of basis matrix. Or, we could instead calculate it as the inverse of the matrix for going up ${Rep}_{\hat{D}, ℰ_{2}} (id)$ . That matrix is easy to find and we have a formula for the $2 \times 2$ inverse, so that’s what is in the equation below.

$\begin{aligned} {Rep}_{\hat{B}, \hat{D}} (t) & = {(\begin{array}{c} - 1 & 2 \\ 0 & 3 \end{array})}^{- 1} (\begin{array}{c} \sqrt{3} / 2 & - 1 / 2 \\ 1 / 2 & \sqrt{3} / 2 \end{array}) (\begin{array}{c} 1 & 0 \\ 1 & 2 \end{array}) \\ = (\begin{array}{c} (5 - \sqrt{3}) / 6 & (3 + 2 \sqrt{3}) / 3 \\ (1 + \sqrt{3}) / 6 & \sqrt{3} / 3 \end{array}) \end{aligned}$

The matrix is messier but the map that it represents is the same. For instance, to replicate the effect of $t$ in the picture, start with $\hat{B}$ ,

${Rep}_{\hat{B}} ((\begin{matrix} 1 \\ 3 \end{matrix})) = {(\begin{matrix} 1 \\ 1 \end{matrix})}_{\hat{B}}$

apply $\hat{T}$ ,

${(\begin{matrix} (5 - \sqrt{3}) / 6 & (3 + 2 \sqrt{3}) / 3 \\ (1 + \sqrt{3}) / 6 & \sqrt{3} / 3 \end{matrix})}_{\hat{B}, \hat{D}} {(\begin{matrix} 1 \\ 1 \end{matrix})}_{\hat{B}} = {(\begin{matrix} (11 + 3 \sqrt{3}) / 6 \\ (1 + 3 \sqrt{3}) / 6 \end{matrix})}_{\hat{D}}$

and check it against $\hat{D}$ .

$\frac{11 + 3 \sqrt{3}}{6} \cdot (\begin{matrix} - 1 \\ 0 \end{matrix}) + \frac{1 + 3 \sqrt{3}}{6} \cdot (\begin{matrix} 2 \\ 3 \end{matrix}) = (\begin{matrix} (- 3 + \sqrt{3}) / 2 \\ (1 + 3 \sqrt{3}) / 2 \end{matrix})$

Example 2.3 Changing bases can make the matrix simpler. On $ℝ^{3}$ the map

$(\begin{matrix} x \\ y \\ z \end{matrix}) \overset{t}{⟼} (\begin{matrix} y + z \\ x + z \\ x + y \end{matrix})$

is represented with respect to the standard basis in this way.

${Rep}_{ℰ_{3}, ℰ_{3}} (t) = (\begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix})$

Representing it with respect to

$B = ⟨ (\begin{matrix} 1 \\ - 1 \\ 0 \end{matrix}), (\begin{matrix} 1 \\ 1 \\ - 2 \end{matrix}), (\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}) ⟩$

gives a matrix that is diagonal.

${Rep}_{B, B} (t) = (\begin{matrix} - 1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & 2 \end{matrix})$

Naturally we usually prefer representations that are easier to understand. We say that a map or matrix has been diagonalized when we find a basis $B$ such that the representation is diagonal with respect to $B, B$ , that is, with respect to the same starting basis as ending basis. Chapter Five finds which maps and matrices are diagonalizable.

The rest of this subsection develops the easier case of finding two bases $B, D$ such that a representation is simple. Recall that the prior subsection shows that a matrix is a change of basis matrix if and only if it is nonsingular.

Definition 2.4 Same-sized matrices $H$ and $\hat{H}$ are matrix equivalent if there are nonsingular matrices $P$ and $Q$ such that $\hat{H} = P H Q$ .

Corollary 2.5 Matrix equivalent matrices represent the same map, with respect to appropriate pairs of bases.

Proof This is immediate from equation ( $*$ ) above.

QED

Exercise 2.24 checks that matrix equivalence is an equivalence relation. Thus it partitions the set of matrices into matrix equivalence classes.

All matrices:

The collection of matrices is partitioned into matrix-equivalence classes. Points H and H hat lie within the same region; dots indicate further classes.

H

matrix equivalent

\hat{H}

We can get insight into the classes by comparing matrix equivalence with row equivalence (remember that matrices are row equivalent when they can be reduced to each other by row operations). In $\hat{H} = P H Q$ , the matrices $P$ and $Q$ are nonsingular and thus each is a product of elementary reduction matrices by Lemma IV.4.7. Left-multiplication by the reduction matrices making up $P$ performs row operations. Right-multiplication by the reduction matrices making up $Q$ performs column operations. Hence, matrix equivalence is a generalization of row equivalence—two matrices are row equivalent if one can be converted to the other by a sequence of row reduction steps, while two matrices are matrix equivalent if one can be converted to the other by a sequence of row reduction steps followed by a sequence of column reduction steps.

Consequently, if matrices are row equivalent then they are also matrix equivalent since we can take $Q$ to be the identity matrix. The converse, however, does not hold: two matrices can be matrix equivalent but not row equivalent.

Example 2.6 These two are matrix equivalent

$(\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}) (\begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix})$

because the second reduces to the first by the column operation of taking $- 1$ times the first column and adding to the second. They are not row equivalent because they have different reduced echelon forms (both are already in reduced form).

We close this section by giving a set of representatives for the matrix equivalence classes.

Theorem 2.7 Any $m \times n$ matrix of rank $k$ is matrix equivalent to the $m \times n$ matrix that is all zeros except that the first $k$ diagonal entries are ones.

$(\begin{matrix} 1 & 0 & \dots & 0 & 0 & \dots & 0 \\ 0 & 1 & \dots & 0 & 0 & \dots & 0 \\ ⋮ \\ 0 & 0 & \dots & 1 & 0 & \dots & 0 \\ 0 & 0 & \dots & 0 & 0 & \dots & 0 \\ ⋮ \\ 0 & 0 & \dots & 0 & 0 & \dots & 0 \end{matrix})$

This is a block partial-identity form.

$(\begin{array}{cc} I & Z \\ Z & Z \end{array})$

Proof Gauss-Jordan reduce the given matrix and combine all the row reduction matrices to make $P$ . Then use the leading entries to do column reduction and finish by swapping the columns to put the leading ones on the diagonal. Combine the column reduction matrices into $Q$ .

QED

Example 2.8 We illustrate the proof by finding $P$ and $Q$ for this matrix.

$(\begin{matrix} 1 & 2 & 1 & - 1 \\ 0 & 0 & 1 & - 1 \\ 2 & 4 & 2 & - 2 \end{matrix})$

First Gauss-Jordan row-reduce.

$(\begin{matrix} 1 & - 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}) (\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ - 2 & 0 & 1 \end{matrix}) (\begin{matrix} 1 & 2 & 1 & - 1 \\ 0 & 0 & 1 & - 1 \\ 2 & 4 & 2 & - 2 \end{matrix}) = (\begin{matrix} 1 & 2 & 0 & 0 \\ 0 & 0 & 1 & - 1 \\ 0 & 0 & 0 & 0 \end{matrix})$

Then column-reduce, which involves right-multiplication.

$(\begin{matrix} 1 & 2 & 0 & 0 \\ 0 & 0 & 1 & - 1 \\ 0 & 0 & 0 & 0 \end{matrix}) (\begin{matrix} 1 & - 2 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}) (\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{matrix}) = (\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{matrix})$

Finish by swapping columns.

$(\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}) (\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}) = (\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix})$

Finally, combine the left-multipliers together as $P$ and the right-multipliers together as $Q$ to get $P H Q$ .

$(\begin{matrix} 1 & - 1 & 0 \\ 0 & 1 & 0 \\ - 2 & 0 & 1 \end{matrix}) (\begin{matrix} 1 & 2 & 1 & - 1 \\ 0 & 0 & 1 & - 1 \\ 2 & 4 & 2 & - 2 \end{matrix}) (\begin{matrix} 1 & 0 & - 2 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 \end{matrix}) = (\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix})$

Corollary 2.9 Matrix equivalence classes are characterized by rank: two same-sized matrices are matrix equivalent if and only if they have the same rank.

Proof Two same-sized matrices with the same rank are equivalent to the same block partial-identity matrix.

QED

Example 2.10 The $2 \times 2$ matrices have only three possible ranks: zero, one, or two. Thus there are three matrix equivalence classes.

All $2 \times 2$ matrices:

The two-by-two matrices are partitioned into three matrix-equivalence classes. Stars mark canonical representatives: the zero matrix, the diagonal matrix with entries 1 and 0, and the identity matrix.

Three equivalence

classes

Each class consists of all of the $2 \times 2$ matrices with the same rank. There is only one rank zero matrix. The other two classes have infinitely many members; we’ve shown only the canonical representative.

One nice thing about the representative in Theorem 2.7 is that we can completely understand the linear map when it is expressed in this way: where the bases are $B = ⟨ {\vec{β}}_{1}, \dots, {\vec{β}}_{n} ⟩$ and $D = ⟨ {\vec{δ}}_{1}, \dots, {\vec{δ}}_{m} ⟩$ then the map’s action is

$c_{1} {\vec{β}}_{1} + \dots + c_{k} {\vec{β}}_{k} + c_{k + 1} {\vec{β}}_{k + 1} + \dots + c_{n} {\vec{β}}_{n} \mapsto c_{1} {\vec{δ}}_{1} + \dots + c_{k} {\vec{δ}}_{k} + \vec{0} + \dots + \vec{0}$

where $k$ is the rank. Thus we can view any linear map as a projection.

${(\begin{matrix} c_{1} \\ ⋮ \\ c_{k} \\ c_{k + 1} \\ ⋮ \\ c_{n} \end{matrix})}_{B} ⟼ {(\begin{matrix} c_{1} \\ ⋮ \\ c_{k} \\ 0 \\ ⋮ \\ 0 \end{matrix})}_{D}$

Exercises

Exercise 2.11 Worked answer

Recommended. Decide if these are matrix equivalent.
1. $(\begin{matrix} 1 & 3 & 0 \\ 2 & 3 & 0 \end{matrix})$ , $(\begin{matrix} 2 & 2 & 1 \\ 0 & 5 & - 1 \end{matrix})$
2. $(\begin{matrix} 0 & 3 \\ 1 & 1 \end{matrix})$ , $(\begin{matrix} 4 & 0 \\ 0 & 5 \end{matrix})$
3. $(\begin{matrix} 1 & 3 \\ 2 & 6 \end{matrix})$ , $(\begin{matrix} 1 & 3 \\ 2 & - 6 \end{matrix})$
Back to Exercise 2.11

Answer.
1. Yes, each has rank two.
2. Yes, they have the same rank.
3. No, they have different ranks.
Exercise 2.12 Worked answer

Which of these are matrix equivalent to each other?
1. $(\begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{matrix})$
2. $(\begin{matrix} 1 & 3 \\ - 1 & - 3 \end{matrix})$
3. $(\begin{matrix} - 5 & 1 & 0 \\ - 1 & 0 & 1 \end{matrix})$
4. $(\begin{matrix} 0 & - 1 \\ 0 & 5 \end{matrix})$
5. $(\begin{matrix} 1 & 0 & 1 \\ 2 & 0 & 2 \\ 1 & 3 & 1 \end{matrix})$
6. $(\begin{matrix} 3 & 1 & 0 \\ 9 & 3 & 0 \\ - 3 & - 1 & 0 \end{matrix})$
Back to Exercise 2.12

Answer. Group the matrices into classes characterized by the condition that all matrices in the same class are the same size and the same rank.
1. $3 \times 3$ , rank $2$
2. $2 \times 2$ , rank $1$
3. $2 \times 3$ , rank $2$
4. $3 \times 3$ , rank $2$
5. $3 \times 3$ , rank $1$
Exercise 2.13 Worked answer

Recommended. Find the canonical representative of the matrix equivalence class of each matrix.
1. $(\begin{matrix} 2 & 1 & 0 \\ 4 & 2 & 0 \end{matrix})$
2. $(\begin{matrix} 0 & 1 & 0 & 2 \\ 1 & 1 & 0 & 4 \\ 3 & 3 & 3 & - 1 \end{matrix})$
Back to Exercise 2.13

Answer. We need only compute the rank of each.
1. $(\begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 0 \end{matrix})$
2. $(\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{matrix})$
Exercise 2.14 Worked answer

Suppose that, with respect to

$B = ℰ_{2} D = ⟨ (\begin{matrix} 1 \\ 1 \end{matrix}), (\begin{matrix} 1 \\ - 1 \end{matrix}) ⟩$

the transformation $t : ℝ^{2} \to ℝ^{2}$ is represented by this matrix.

$(\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix})$

Use change of basis matrices to represent $t$ with respect to each pair.
1. $\hat{B} = ⟨ (\begin{matrix} 0 \\ 1 \end{matrix}), (\begin{matrix} 1 \\ 1 \end{matrix}) ⟩$ , $\hat{D} = ⟨ (\begin{matrix} - 1 \\ 0 \end{matrix}), (\begin{matrix} 2 \\ 1 \end{matrix}) ⟩$
2. $\hat{B} = ⟨ (\begin{matrix} 1 \\ 2 \end{matrix}), (\begin{matrix} 1 \\ 0 \end{matrix}) ⟩$ , $\hat{D} = ⟨ (\begin{matrix} 1 \\ 2 \end{matrix}), (\begin{matrix} 2 \\ 1 \end{matrix}) ⟩$
Back to Exercise 2.14

Answer. Recall the diagram and the formula.

$\begin{matrix} ℝ_{wrt B}^{2} & \to_{T}^{t} & ℝ_{wrt D}^{2} \\ \begin{matrix} id & ↓ \end{matrix} & \begin{matrix} id & ↓ \end{matrix} \\ ℝ_{wrt \hat{B}}^{2} & \to_{\hat{T}}^{t} & ℝ_{wrt \hat{D}}^{2} \end{matrix} \hat{T} = {Rep}_{D, \hat{D}} (id) \cdot T \cdot {Rep}_{\hat{B}, B} (id)$
1. These two
  
  $(\begin{matrix} 1 \\ 1 \end{matrix}) = 1 \cdot (\begin{matrix} - 1 \\ 0 \end{matrix}) + 1 \cdot (\begin{matrix} 2 \\ 1 \end{matrix}) (\begin{matrix} 1 \\ - 1 \end{matrix}) = (- 3) \cdot (\begin{matrix} - 1 \\ 0 \end{matrix}) + (- 1) \cdot (\begin{matrix} 2 \\ 1 \end{matrix})$
  
  show that
  
  ${Rep}_{D, \hat{D}} (id) = (\begin{matrix} 1 & - 3 \\ 1 & - 1 \end{matrix})$
  
  and similarly these two
  
  $(\begin{matrix} 0 \\ 1 \end{matrix}) = 0 \cdot (\begin{matrix} 1 \\ 0 \end{matrix}) + 1 \cdot (\begin{matrix} 0 \\ 1 \end{matrix}) (\begin{matrix} 1 \\ 1 \end{matrix}) = 1 \cdot (\begin{matrix} 1 \\ 0 \end{matrix}) + 1 \cdot (\begin{matrix} 0 \\ 1 \end{matrix})$
  
  give the other nonsingular matrix.
  
  ${Rep}_{\hat{B}, B} (id) = (\begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix})$
  
  Then the answer is this.
  
  $\hat{T} = (\begin{matrix} 1 & - 3 \\ 1 & - 1 \end{matrix}) (\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix}) (\begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix}) = (\begin{matrix} - 10 & - 18 \\ - 2 & - 4 \end{matrix})$
  
  Although not strictly necessary, a check is reassuring. Arbitrarily fixing
  
  $\vec{v} = (\begin{matrix} 3 \\ 2 \end{matrix})$
  
  we have that
  
  ${Rep}_{B} (\vec{v}) = {(\begin{matrix} 3 \\ 2 \end{matrix})}_{B} {(\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix})}_{B, D} {(\begin{matrix} 3 \\ 2 \end{matrix})}_{B} = {(\begin{matrix} 7 \\ 17 \end{matrix})}_{D}$
  
  and so $t (\vec{v})$ is this.
  
  $7 \cdot (\begin{matrix} 1 \\ 1 \end{matrix}) + 17 \cdot (\begin{matrix} 1 \\ - 1 \end{matrix}) = (\begin{matrix} 24 \\ - 10 \end{matrix})$
  
  Doing the calculation with respect to $\hat{B}, \hat{D}$ starts with
  
  ${Rep}_{\hat{B}} (\vec{v}) = {(\begin{matrix} - 1 \\ 3 \end{matrix})}_{\hat{B}} {(\begin{matrix} - 10 & - 18 \\ - 2 & - 4 \end{matrix})}_{\hat{B}, \hat{D}} {(\begin{matrix} - 1 \\ 3 \end{matrix})}_{\hat{B}} = {(\begin{matrix} - 44 \\ - 10 \end{matrix})}_{\hat{D}}$
  
  and then checks that this is the same result.
  
  $- 44 \cdot (\begin{matrix} - 1 \\ 0 \end{matrix}) - 10 \cdot (\begin{matrix} 2 \\ 1 \end{matrix}) = (\begin{matrix} 24 \\ - 10 \end{matrix})$
2. These two
  
  $(\begin{matrix} 1 \\ 1 \end{matrix}) = \frac{1}{3} \cdot (\begin{matrix} 1 \\ 2 \end{matrix}) + \frac{1}{3} \cdot (\begin{matrix} 2 \\ 1 \end{matrix}) (\begin{matrix} 1 \\ - 1 \end{matrix}) = - 1 \cdot (\begin{matrix} 1 \\ 2 \end{matrix}) + 1 \cdot (\begin{matrix} 2 \\ 1 \end{matrix})$
  
  show that
  
  ${Rep}_{D, \hat{D}} (id) = (\begin{matrix} 1 / 3 & - 1 \\ 1 / 3 & 1 \end{matrix})$
  
  and these two
  
  $(\begin{matrix} 1 \\ 2 \end{matrix}) = 1 \cdot (\begin{matrix} 1 \\ 0 \end{matrix}) + 2 \cdot (\begin{matrix} 0 \\ 1 \end{matrix}) (\begin{matrix} 1 \\ 0 \end{matrix}) = - 1 \cdot (\begin{matrix} 1 \\ 0 \end{matrix}) + 0 \cdot (\begin{matrix} 0 \\ 1 \end{matrix})$
  
  show this.
  
  ${Rep}_{\hat{B}, B} (id) = (\begin{matrix} 1 & 1 \\ 2 & 0 \end{matrix})$
  
  With those, the conversion goes in this way.
  
  $\hat{T} = (\begin{matrix} 1 / 3 & - 1 \\ 1 / 3 & 1 \end{matrix}) (\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix}) (\begin{matrix} 1 & 1 \\ 2 & 0 \end{matrix}) = (\begin{matrix} - 28 / 3 & - 8 / 3 \\ 38 / 3 & 10 / 3 \end{matrix})$
  
  As in the prior item, a check provides some confidence that we did this calculation without mistakes. We can for instance, fix the vector
  
  $\vec{v} = (\begin{matrix} - 1 \\ 2 \end{matrix})$
  
  (this is arbitrary, taken from thin air). Now we have
  
  ${Rep}_{B} (\vec{v}) = (\begin{matrix} - 1 \\ 2 \end{matrix}) {(\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix})}_{B, D} {(\begin{matrix} - 1 \\ 2 \end{matrix})}_{B} = {(\begin{matrix} 3 \\ 5 \end{matrix})}_{D}$
  
  and so $t (\vec{v})$ is this vector.
  
  $3 \cdot (\begin{matrix} 1 \\ 1 \end{matrix}) + 5 \cdot (\begin{matrix} 1 \\ - 1 \end{matrix}) = (\begin{matrix} 8 \\ - 2 \end{matrix})$
  
  With respect to $\hat{B}, \hat{D}$ we first calculate
  
  ${Rep}_{\hat{B}} (\vec{v}) = (\begin{matrix} 1 \\ - 2 \end{matrix}) {(\begin{matrix} - 28 / 3 & - 8 / 3 \\ 38 / 3 & 10 / 3 \end{matrix})}_{\hat{B}, \hat{D}} {(\begin{matrix} 1 \\ - 2 \end{matrix})}_{\hat{B}} = {(\begin{matrix} - 4 \\ 6 \end{matrix})}_{\hat{D}}$
  
  and, sure enough, that is the same result for $t (\vec{v})$ .
  
  $- 4 \cdot (\begin{matrix} 1 \\ 2 \end{matrix}) + 6 \cdot (\begin{matrix} 2 \\ 1 \end{matrix}) = (\begin{matrix} 8 \\ - 2 \end{matrix})$
Exercise 2.15 Worked answer

What sizes are $P$ and $Q$ in the equation $\hat{H} = P H Q$ ?

Back to Exercise 2.15

Answer. Where $H$ and $\hat{H}$ are $m \times n$ , the matrix $P$ is $m \times m$ while $Q$ is $n \times n$ .
Exercise 2.16 Worked answer

Recommended. Consider the spaces $V = 𝒫_{2}$ and $W = ℳ_{2 \times 2}$ , with these bases.

$\begin{matrix} B = ⟨ 1, 1 + x, 1 + x^{2} ⟩ D = ⟨ (\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix}), (\begin{matrix} 0 & 0 \\ 1 & 1 \end{matrix}), (\begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix}), (\begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix}) ⟩ \\ \hat{B} = ⟨ 1, x, x^{2} ⟩ \hat{D} = ⟨ (\begin{matrix} - 1 & 0 \\ 0 & 0 \end{matrix}), (\begin{matrix} 0 & - 1 \\ 0 & 0 \end{matrix}), (\begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix}), (\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix}) ⟩ \end{matrix}$

We will find $P$ and $Q$ to convert the representation of a map with respect to $B, D$ to one with respect to $\hat{B}, \hat{D}$
1. Draw the appropriate arrow diagram.
2. Compute $P$ and $Q$ .
Back to Exercise 2.16

Answer.
1. This is the arrow diagram.
  
  $\begin{matrix} V_{wrt B} & \to_{H}^{h} & W_{wrt D} \\ \begin{matrix} id & ↓ \end{matrix} & \begin{matrix} id & ↓ \end{matrix} \\ V_{wrt \hat{B}} & \to_{\hat{H}}^{h} & W_{wrt \hat{D}} \end{matrix}$
  
  The diagram gives $\hat{H} = P H Q$ where $Q = {Rep}_{\hat{B}, B} (id)$ and $P = {Rep}_{D, \hat{D}} (id)$ (remembering that the operation done first is written on the right).
2. We want $Q = {Rep}_{\hat{B}, B} (id)$ and $P = {Rep}_{D, \hat{D}} (id)$ . For $Q$ we do these calculations (done here by eye).
  
  ${Rep}_{B} (id (1)) = (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) {Rep}_{B} (id (x)) = (\begin{matrix} - 1 \\ 1 \\ 0 \end{matrix}) {Rep}_{B} (id (x^{2})) = (\begin{matrix} - 1 \\ 0 \\ 1 \end{matrix})$
  
  These calculations give $P$ .
  
  $\begin{matrix} {Rep}_{\hat{D}} (id ((\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix}))) = (\begin{matrix} 0 \\ 0 \\ 0 \\ 1 \end{matrix}) {Rep}_{\hat{D}} (id ((\begin{matrix} 0 & 0 \\ 1 & 1 \end{matrix}))) = (\begin{matrix} 0 \\ 0 \\ 1 \\ 1 \end{matrix}) \\ {Rep}_{\hat{D}} (id ((\begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix}))) = (\begin{matrix} 0 \\ - 1 \\ 1 \\ 1 \end{matrix}) {Rep}_{\hat{D}} (id ((\begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix}))) = (\begin{matrix} - 1 \\ - 1 \\ 1 \\ 1 \end{matrix}) \end{matrix}$
  
  This is the answer.
  
  $P = (\begin{matrix} 0 & 0 & 0 & - 1 \\ 0 & 0 & - 1 & - 1 \\ 0 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{matrix}) Q = (\begin{matrix} 1 & - 1 & - 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix})$
Exercise 2.17 Worked answer

Recommended. Find the change of basis matrices $Q$ and $P$ that will convert the representation of a $t : ℝ^{2} \to ℝ^{2}$ with respect to $B, D$ to one with respect to $\hat{B}, \hat{D}$ .

$B = ⟨ (\begin{matrix} 1 \\ 0 \end{matrix}), (\begin{matrix} 1 \\ 1 \end{matrix}) ⟩ D = ⟨ (\begin{matrix} 0 \\ 1 \end{matrix}), (\begin{matrix} - 1 \\ 0 \end{matrix}) ⟩ \hat{B} = ℰ_{2} \hat{D} = ⟨ (\begin{matrix} 1 \\ - 1 \end{matrix}), (\begin{matrix} 0 \\ 1 \end{matrix}) ⟩$

Back to Exercise 2.17

Answer. This is the arrow diagram.

$\begin{matrix} V_{wrt B} & \to_{T}^{t} & W_{wrt D} \\ \begin{matrix} id & ↓ \end{matrix} & \begin{matrix} id & ↓ \end{matrix} \\ V_{wrt \hat{B}} & \to_{\hat{T}}^{t} & W_{wrt \hat{D}} \end{matrix}$

Where $Q = {Rep}_{\hat{B}, B} (id)$ and $P = {Rep}_{D, \hat{D}} (id)$ the equation is $\hat{T} = P T Q$ .

These are the calculations for $Q = {Rep}_{\hat{B}, B} (id)$ (done by eye).

${Rep}_{B} (id ((\begin{matrix} 1 \\ 0 \end{matrix}))) = (\begin{matrix} 1 \\ 0 \end{matrix}) {Rep}_{B} (id ((\begin{matrix} 0 \\ 1 \end{matrix}))) = (\begin{matrix} - 1 \\ 1 \end{matrix})$

So we get this.

$Q = (\begin{matrix} 1 & - 1 \\ 0 & 1 \end{matrix})$

These calculations give $P = {Rep}_{D, \hat{D}} (id)$ .

${Rep}_{\hat{D}} (id ((\begin{matrix} 0 \\ 1 \end{matrix}))) = (\begin{matrix} 0 \\ 1 \end{matrix}) {Rep}_{\hat{D}} (id ((\begin{matrix} - 1 \\ 0 \end{matrix}))) = (\begin{matrix} - 1 \\ - 1 \end{matrix})$

Concatenate them to make the other matrix.

$P = (\begin{matrix} 0 & - 1 \\ 1 & - 1 \end{matrix})$
Exercise 2.18 Worked answer

Recommended. Find the $P$ and $Q$ to express $H$ via $P H Q$ as a block partial identity matrix.

$H = (\begin{matrix} 2 & 1 & 1 \\ 3 & - 1 & 0 \\ 1 & 3 & 2 \end{matrix})$

Back to Exercise 2.18

Answer. Gauss’s Method gives this.

$(\begin{matrix} 2 & 1 & 1 \\ 3 & - 1 & 0 \\ 1 & 3 & 2 \end{matrix}) ⟶_{\begin{array}{l} - (1 / 2) ρ_{1} + ρ_{3} \end{array}}^{- (3 / 2) ρ_{1} + ρ_{2}} ⟶_{\begin{array}{l} \end{array}}^{ρ_{2} + ρ_{3}} ⟶_{\begin{array}{l} - (2 / 5) ρ_{2} \end{array}}^{(1 / 2) ρ_{1}} (\begin{matrix} 1 & 1 / 2 & 1 / 2 \\ 0 & 1 & 3 / 5 \\ 0 & 0 & 0 \end{matrix})$

Column operations complete the job of reaching the canonical form for matrix equivalence.

$⟶_{\begin{array}{l} \end{array}}^{- (3 / 5) {col}_{2} + {col}_{3}} ⟶_{\begin{array}{l} - (1 / 5) {col}_{1} + {col}_{3} \end{array}}^{- (1 / 2) {col}_{1} + {col}_{2}} (\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{matrix})$

Then these are the two matrices.

$\begin{aligned} P & = (\begin{array}{c} 1 & 0 & 0 \\ 0 & - 2 / 5 & 0 \\ 0 & 0 & 1 \end{array}) (\begin{array}{c} 1 / 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}) (\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 1 \end{array}) (\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ - 1 / 2 & 0 & 1 \end{array}) (\begin{array}{c} 1 & 0 & 0 \\ - 3 / 2 & 1 & 0 \\ 0 & 0 & 1 \end{array}) \\ = (\begin{array}{c} 1 / 2 & 0 & 0 \\ 3 / 5 & - 2 / 5 & 0 \\ - 2 & 1 & 1 \end{array}) \\ Q & = (\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & - 3 / 5 \\ 0 & 0 & 1 \end{array}) (\begin{array}{c} 1 & - 1 / 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}) (\begin{array}{c} 1 & 0 & - 1 / 5 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}) = (\begin{array}{c} 1 & - 1 / 2 & - 1 / 5 \\ 0 & 1 & - 3 / 5 \\ 0 & 0 & 1 \end{array}) \end{aligned}$
Exercise 2.19 Worked answer

Recommended. Use Theorem 2.7 to show that a square matrix is nonsingular if and only if it is equivalent to an identity matrix.

Back to Exercise 2.19

Answer. Any $n \times n$ matrix is nonsingular if and only if it has rank $n$ , that is, by Theorem 2.7, if and only if it is matrix equivalent to the $n \times n$ matrix whose diagonal is all ones.
Exercise 2.20 Worked answer

Show that, where $A$ is a nonsingular square matrix, if $P$ and $Q$ are nonsingular square matrices such that $P A Q = I$ then $Q P = A^{- 1}$ .

Back to Exercise 2.20

Answer. If $P A Q = I$ then $Q P A Q = Q$ , so $Q P A = I$ , and so $Q P = A^{- 1}$ .
Exercise 2.21 Worked answer

Why does Theorem 2.7 not show that every matrix is diagonalizable (see Example 2.3)?

Back to Exercise 2.21

Answer. By the definition following Example 2.3, a matrix $M$ is diagonalizable if it represents $M = {Rep}_{B, D} (t)$ a transformation with the property that there is some basis $\hat{B}$ such that ${Rep}_{\hat{B}, \hat{B}} (t)$ is a diagonal matrix—the starting and ending bases must be equal. But Theorem 2.7 says only that there are $\hat{B}$ and $\hat{D}$ such that we can change to a representation ${Rep}_{\hat{B}, \hat{D}} (t)$ and get a diagonal matrix. We have no reason to suspect that we could pick the two $\hat{B}$ and $\hat{D}$ so that they are equal.
Exercise 2.22 Worked answer

Must matrix equivalent matrices have matrix equivalent transposes?

Back to Exercise 2.22

Answer. Yes. Row rank equals column rank, so the rank of the transpose equals the rank of the matrix. Same-sized matrices with equal ranks are matrix equivalent.
Exercise 2.23 Worked answer

What happens in Theorem 2.7 if $k = 0$ ?

Back to Exercise 2.23

Answer. Only a zero matrix has rank zero.
Exercise 2.24 Worked answer

Show that matrix equivalence is an equivalence relation.

Back to Exercise 2.24

Answer. For reflexivity, to show that any matrix is matrix equivalent to itself, take $P$ and $Q$ to be identity matrices. For symmetry, if $H_{1} = P H_{2} Q$ then $H_{2} = P^{- 1} H_{1} Q^{- 1}$ (inverses exist because $P$ and $Q$ are nonsingular). Finally, for transitivity, assume that $H_{1} = P_{2} H_{2} Q_{2}$ and that $H_{2} = P_{3} H_{3} Q_{3}$ . Then substitution gives $H_{1} = P_{2} (P_{3} H_{3} Q_{3}) Q_{2} = (P_{2} P_{3}) H_{3} (Q_{3} Q_{2})$ . A product of nonsingular matrices is nonsingular (we’ve shown that the product of invertible matrices is invertible; in fact, we’ve shown how to calculate the inverse) and so $H_{1}$ is therefore matrix equivalent to $H_{3}$ .
Exercise 2.25 Worked answer

Recommended. Show that a zero matrix is alone in its matrix equivalence class. Are there other matrices like that?

Back to Exercise 2.25

Answer. By Theorem 2.7, a zero matrix is alone in its class because it is the only $m \times n$ of rank zero. No other matrix is alone in its class; any nonzero scalar product of a matrix has the same rank as that matrix.
Exercise 2.26 Worked answer

What are the matrix equivalence classes of matrices of transformations on $ℝ^{1}$ ? $ℝ^{3}$ ?

Back to Exercise 2.26

Answer. There are two matrix equivalence classes of $1 \times 1$ matrices—those of rank zero and those of rank one. The $3 \times 3$ matrices fall into four matrix equivalence classes.
Exercise 2.27 Worked answer

How many matrix equivalence classes are there?

Back to Exercise 2.27

Answer. For $m \times n$ matrices there are classes for each possible rank: where $k$ is the minimum of $m$ and $n$ there are classes for the matrices of rank $0$ , $1$ , …, $k$ . That’s $k + 1$ classes. (Of course, totaling over all sizes of matrices we get infinitely many classes.)
Exercise 2.28 Worked answer

Are matrix equivalence classes closed under scalar multiplication? Addition?

Back to Exercise 2.28

Answer. They are closed under nonzero scalar multiplication since a nonzero scalar multiple of a matrix has the same rank as does the matrix. They are not closed under addition, for instance, $H + (- H)$ has rank zero.
Exercise 2.29 Worked answer

Let $t : ℝ^{n} \to ℝ^{n}$ represented by $T$ with respect to $ℰ_{n}, ℰ_{n}$ .
1. Find ${Rep}_{B, B} (t)$ in this specific case.
  
  $T = (\begin{matrix} 1 & 1 \\ 3 & - 1 \end{matrix}) B = ⟨ (\begin{matrix} 1 \\ 2 \end{matrix}), (\begin{matrix} - 1 \\ - 1 \end{matrix}) ⟩$
2. Describe ${Rep}_{B, B} (t)$ in the general case where $B = ⟨ {\vec{β}}_{1}, \dots, {\vec{β}}_{n} ⟩$ .
Back to Exercise 2.29

Answer. Here is the picture.

$\begin{matrix} ℝ_{wrt ℰ_{2}}^{2} & \to_{T}^{t} & ℝ_{wrt ℰ_{2}}^{2} \\ \begin{matrix} id & ↓ \end{matrix} & \begin{matrix} id & ↓ \end{matrix} \\ ℝ_{wrt B}^{2} & \to_{\hat{T}}^{t} & ℝ_{wrt B}^{2} \end{matrix}$

There are two ways to move from the lower left to the lower right. The first is direct, using $\hat{T} = {Rep}_{B, B} (t)$ . The second moves up, then over, then down, using ${Rep}_{ℰ_{2}, B} (id) \cdot T \cdot {Rep}_{B, ℰ_{2}} (id)$ (remember that they get written right-to-left, so the “up” matrix is on the right, in order to have that when applied to a ${Rep}_{B} (\vec{v})$ the matrix applied first is ${Rep}_{B, ℰ_{2}} (id)$ ). We write $S$ for ${Rep}_{B, ℰ_{2}} (id)$ ; of the two we choose this one because it is easier to calculate. So we have $\hat{T} = S^{- 1} T S$ .
1. We have
  
  ${Rep}_{B, ℰ_{2}} (id) = S = (\begin{matrix} 1 & - 1 \\ 2 & - 1 \end{matrix})$
  
  and
  
  ${Rep}_{ℰ_{2}, B} (id) = {Rep}_{B, ℰ_{2}} (id)^{- 1} = S^{- 1} = {(\begin{matrix} 1 & - 1 \\ 2 & - 1 \end{matrix})}^{- 1} = (\begin{matrix} - 1 & 1 \\ - 2 & 1 \end{matrix})$
  
  and thus the answer is this.
  
  ${Rep}_{B, B} (t) = (\begin{matrix} - 1 & 1 \\ - 2 & 1 \end{matrix}) (\begin{matrix} 1 & 1 \\ 3 & - 1 \end{matrix}) (\begin{matrix} 1 & - 1 \\ 2 & - 1 \end{matrix}) = (\begin{matrix} - 2 & 0 \\ - 5 & 2 \end{matrix})$
  
  As a quick check, we can take a vector at random
  
  $\vec{v} = (\begin{matrix} 4 \\ 5 \end{matrix})$
  
  giving
  
  ${Rep}_{ℰ_{2}} (\vec{v}) = (\begin{matrix} 4 \\ 5 \end{matrix}) (\begin{matrix} 1 & 1 \\ 3 & - 1 \end{matrix}) (\begin{matrix} 4 \\ 5 \end{matrix}) = (\begin{matrix} 9 \\ 7 \end{matrix}) = t (\vec{v})$
  
  while the calculation with respect to $B, B$
  
  ${Rep}_{B} (\vec{v}) = (\begin{matrix} 1 \\ - 3 \end{matrix}) {(\begin{matrix} - 2 & 0 \\ - 5 & 2 \end{matrix})}_{B, B} {(\begin{matrix} 1 \\ - 3 \end{matrix})}_{B} = {(\begin{matrix} - 2 \\ - 11 \end{matrix})}_{B}$
  
  yields the same result.
  
  $- 2 \cdot (\begin{matrix} 1 \\ 2 \end{matrix}) - 11 \cdot (\begin{matrix} - 1 \\ - 1 \end{matrix}) = (\begin{matrix} 9 \\ 7 \end{matrix})$
2. As in the first item of this question
  
  $S = {Rep}_{B, ℰ_{2}} (id) = (\begin{array}{ccc} {\vec{β}}_{1} & \dots & {\vec{β}}_{n} \end{array}) {Rep}_{ℰ_{2}, B} (id) = S^{- 1} = {Rep}_{B, ℰ_{2}} (id)^{- 1}$
  
  so, writing $S$ for the matrix whose columns are the basis vectors, we have that ${Rep}_{B, B} (t) = \hat{T} = S^{- 1} T S$ .
Exercise 2.30 Worked answer
1. Let $V$ have bases $B_{1}$ and $B_{2}$ and suppose that $W$ has the basis $D$ . Where $h : V \to W$ , find the formula that computes ${Rep}_{B_{2}, D} (h)$ from ${Rep}_{B_{1}, D} (h)$ .
2. Repeat the prior question with one basis for $V$ and two bases for $W$ .
Back to Exercise 2.30

Answer.
1. The adapted form of the arrow diagram is this.
  
  $\begin{matrix} V_{wrt B_{1}} & \to_{H}^{h} & W_{wrt D} \\ \begin{matrix} id & ↓ & Q \end{matrix} & \begin{matrix} id & ↓ & P \end{matrix} \\ V_{wrt B_{2}} & \to_{\hat{H}}^{h} & W_{wrt D} \end{matrix}$
  
  Since there is no need to change bases in $W$ (or we can say that the change of basis matrix $P$ is the identity), we have ${Rep}_{B_{2}, D} (h) = {Rep}_{B_{1}, D} (h) \cdot Q$ where $Q = {Rep}_{B_{2}, B_{1}} (id)$ .
2. Here, this is the arrow diagram.
  
  $\begin{matrix} V_{wrt B} & \to_{H}^{h} & W_{wrt D_{1}} \\ \begin{matrix} id & ↓ & Q \end{matrix} & \begin{matrix} id & ↓ & P \end{matrix} \\ V_{wrt B} & \to_{\hat{H}}^{h} & W_{wrt D_{2}} \end{matrix}$
  
  We have that ${Rep}_{B, D_{2}} (h) = P \cdot {Rep}_{B, D_{1}} (h)$ where $P = {Rep}_{D_{1}, D_{2}} (id)$ .
Exercise 2.31 Worked answer
1. If two matrices are matrix equivalent and invertible, must their inverses be matrix equivalent?
2. If two matrices have matrix equivalent inverses, must the two be matrix equivalent?
3. If two matrices are square and matrix equivalent, must their squares be matrix equivalent?
4. If two matrices are square and have matrix equivalent squares, must they be matrix equivalent?
Back to Exercise 2.31

Answer.
1. Here is the arrow diagram, and a version of that diagram for inverse functions.
  
  $\begin{matrix} V_{wrt B} & \to_{H}^{h} & W_{wrt D} \\ \begin{matrix} id & ↓ & Q \end{matrix} & \begin{matrix} id & ↓ & P \end{matrix} \\ V_{wrt \hat{B}} & \to_{\hat{H}}^{h} & W_{wrt \hat{D}} \end{matrix} \begin{matrix} V_{wrt B} & \leftarrow_{H^{- 1}}^{h^{- 1}} & W_{wrt D} \\ \begin{matrix} id & ↓ & Q \end{matrix} & \begin{matrix} id & ↓ & P \end{matrix} \\ V_{wrt \hat{B}} & \leftarrow_{{\hat{H}}^{- 1}}^{h^{- 1}} & W_{wrt \hat{D}} \end{matrix}$
  
  Yes, the inverses of the matrices represent the inverses of the maps. That is, we can move from the lower right to the lower left by moving up, then left, then down. In other words, where $\hat{H} = P H Q$ (and $P, Q$ invertible) and $H, \hat{H}$ are invertible then ${\hat{H}}^{- 1} = Q^{- 1} H^{- 1} P^{- 1}$ .
2. Yes; this is the prior part repeated in different terms.
3. No, we need another assumption: if $H$ represents $h$ with respect to the same starting as ending bases $B, B$ , for some $B$ then $H^{2}$ represents $h \circ h$ . As a specific example, these two matrices are both rank one and so they are matrix equivalent
  
  $(\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}) (\begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix})$
  
  but the squares are not matrix equivalent—the square of the first has rank one while the square of the second has rank zero.
4. No. These two are not matrix equivalent but have matrix equivalent squares.
  
  $(\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}) (\begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix})$
Exercise 2.32 Worked answer

Square matrices are similar if they represent the same transformation, but each with respect to the same ending as starting basis. That is, ${Rep}_{B_{1}, B_{1}} (t)$ is similar to ${Rep}_{B_{2}, B_{2}} (t)$ .
1. Give a definition of matrix similarity like that of Definition 2.4.
2. Prove that similar matrices are matrix equivalent.
3. Show that similarity is an equivalence relation.
4. Show that if $T$ is similar to $\hat{T}$ then $T^{2}$ is similar to ${\hat{T}}^{2}$ , the cubes are similar, etc.
5. Prove that there are matrix equivalent matrices that are not similar.
Back to Exercise 2.32

Answer.
1. The arrow diagram suggests the definition.
  
  $\begin{matrix} V_{wrt B_{1}} & \to_{T}^{t} & V_{wrt B_{1}} \\ \begin{matrix} id & ↓ \end{matrix} & \begin{matrix} id & ↓ \end{matrix} \\ V_{wrt B_{2}} & \to_{\hat{T}}^{t} & V_{wrt B_{2}} \end{matrix}$
  
  Call matrices $T, \hat{T}$ similar if there is a nonsingular matrix $P$ such that $\hat{T} = P^{- 1} T P$ .
2. Take $P^{- 1}$ to be $P$ and take $P$ to be $Q$ .
3. This is as in Exercise 2.24. Reflexivity is obvious: $T = I^{- 1} T I$ . Symmetry is also easy: $\hat{T} = P^{- 1} T P$ implies that $T = P \hat{T} P^{- 1}$ (multiply the first equation from the right by $P^{- 1}$ and from the left by $P$ ). For transitivity, assume that $T_{1} = {P_{2}}^{- 1} T_{2} P_{2}$ and that $T_{2} = {P_{3}}^{- 1} T_{3} P_{3}$ . Then $T_{1} = {P_{2}}^{- 1} ({P_{3}}^{- 1} T_{3} P_{3}) P_{2} = ({P_{2}}^{- 1} {P_{3}}^{- 1}) T_{3} (P_{3} P_{2})$ and we are finished on noting that $P_{3} P_{2}$ is an invertible matrix with inverse ${P_{2}}^{- 1} {P_{3}}^{- 1}$ .
4. Assume $\hat{T} = P^{- 1} T P$ . For squares, ${\hat{T}}^{2} = (P^{- 1} T P) (P^{- 1} T P) = P^{- 1} T (P P^{- 1}) T P = P^{- 1} T^{2} P$ . Higher powers follow by induction.
5. These two are matrix equivalent but their squares are not matrix equivalent.
  
  $(\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}) (\begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix})$
  
  By the prior item, matrix similarity and matrix equivalence are thus different.

Change of Basis

Changing Representations of Vectors

Exercises

Changing Map Representations

Exercises

References cited in this section