Geometry of Determinants

The prior section develops the determinant algebraically, by considering formulas satisfying certain conditions. This section complements that with a geometric approach. Beyond its intuitive appeal, an advantage of this approach is that while we have so far only considered whether or not a determinant is zero, here we shall give a meaning to the value of the determinant. (The prior section treats the determinant as a function of the rows but this section focuses on columns.)

Determinants as Size Functions

This parallelogram picture is familiar from the construction of the sum of the two vectors.

Two vectors from a common origin span a parallelogram. Column-vector labels beside their tips give coordinates (x1,y1) and (x2,y2). No coordinate axes or dashed projections are drawn.

Definition 1.1 In $ℝ^{n}$ the box (or parallelepiped) formed by $⟨ {\vec{v}}_{1}, \dots, {\vec{v}}_{n} ⟩$ is the set ${t_{1} {\vec{v}}_{1} + \dots + t_{n} {\vec{v}}_{n} ∣ t_{1}, \dots, t_{n} \in [0 \dots 1]}$ .

Thus the parallelogram above is the box formed by $⟨ (\binom{x_{1}}{y_{1}}), (\binom{x_{2}}{y_{2}}) ⟩$ . A three-space box is shown in Example 1.4.

We can find the area of the above box by drawing an enclosing rectangle and subtracting away areas not in the box.

The parallelogram lies inside an axis-aligned rectangle. Solid guide segments partition the outside into six labelled regions A through F. Labels x1,x2,y1,y2 support the adjacent area subtraction.

$\begin{array}{l} area of parallelogram \\ = area of rectangle - area of A - area of B \\ - \dots - area of F \\ = (x_{1} + x_{2}) (y_{1} + y_{2}) - x_{2} y_{1} - x_{1} y_{1} / 2 \\ - x_{2} y_{2} / 2 - x_{2} y_{2} / 2 - x_{1} y_{1} / 2 - x_{2} y_{1} \\ = x_{1} y_{2} - x_{2} y_{1} \end{array}$

That the area equals the value of the determinant

$| \begin{matrix} x_{1} & x_{2} \\ y_{1} & y_{2} \end{matrix} | = x_{1} y_{2} - x_{2} y_{1}$

is no coincidence. The definition of determinants contains four properties that we know lead to a unique function for each dimension $n$ . We shall argue that these properties make good postulates for a function that measures the size of boxes in $n$ -space.

For instance, such a function should have the property that multiplying one of the box-defining vectors by a scalar will multiply the size by that scalar.

A shaded parallelogram is spanned by a horizontal vector v and an upward-sloping vector w. This is the unscaled region. The first vector is lengthened to k v with k=1.4, while w remains unchanged. Solid sides enclose the enlarged parallelogram; the original region is shaded for comparison.

Shown here is $k = 1.4$ . On the right the rescaled region is in solid lines with the original region shaded for comparison.

That is, we can reasonably expect that $size (\dots, k \vec{v}, \dots) = k \cdot size (\dots, \vec{v}, \dots)$ . Of course, this condition is one of those in the definition of determinants.

Another property of determinants that should apply to any function measuring the size of a box is that it is unaffected by row combinations. Here are before-combining and after-combining boxes (the scalar shown is $k = - 0.35$ ).

Before the shear, horizontal vector v and upward-sloping vector w span the shaded parallelogram. After the shear, v remains horizontal and the second vector is labelled k v+w and slopes left. The original parallelogram is shaded; the new one has the same base and height. The frozen diagram uses k=-0.65; adjacent original prose prints -0.35. Neither has been silently changed.

The box formed by $v$ and $k \vec{v} + \vec{w}$ slants differently than the original one but the two have the same base and the same height, and hence the same area. So we expect that size is not affected by a shear operation $size (\dots, \vec{v}, \dots, \vec{w}, \dots) = size (\dots, \vec{v}, \dots, k \vec{v} + \vec{w}, \dots)$ . Again, this is a determinant condition.

We expect that the box formed by unit vectors has unit size

The two unit vectors e1 and e2 form the horizontal and vertical sides of the unit square.

and we naturally extend that to any $n$ -space $size ({\vec{e}}_{1}, \dots, {\vec{e}}_{n}) = 1$ .

Condition (2) of the definition of determinant is redundant, as remarked following the definition. We know from the prior section that for each $n$ the determinant exists and is unique so we know that these postulates for size functions are consistent and that we do not need any more postulates. Therefore, we are justified in interpreting $det ({\vec{v}}_{1}, \dots, {\vec{v}}_{n})$ as giving the size of the box formed by the vectors.

Remark 1.2 Although condition (2) is redundant it raises an important point. Consider these two.


$\| \begin{matrix} 4 & 1 \\ 2 & 3 \end{matrix} \| = 10$	$\| \begin{matrix} 1 & 4 \\ 3 & 2 \end{matrix} \| = - 10$

Swapping the columns changes the sign. On the left, starting with $\vec{u}$ and following the arc inside the angle to $\vec{v}$ (that is, going counterclockwise), we get a positive size. On the right, starting at $\vec{v}$ and going to $\vec{u}$ , and so following the clockwise arc, gives a negative size. The sign returned by the size function reflects the orientation or sense of the box. (We see the same thing if we picture the effect of scalar multiplication by a negative scalar.)

Definition 1.3 The volume of a box is the absolute value of the determinant of a matrix with those vectors as columns.

Example 1.4 By the formula that takes the area of the base times the height, the volume of this parallelepiped is $12$ . That agrees with the determinant.

$| \begin{matrix} 2 & 0 & - 1 \\ 0 & 3 & 0 \\ 2 & 1 & 1 \end{matrix} | = 12$

Taking the vectors in a different order changes the sign but not the magnitude.

$| \begin{matrix} 0 & 2 & - 1 \\ 3 & 0 & 0 \\ 1 & 2 & 1 \end{matrix} | = - 12$

Theorem 1.5 A transformation $t : ℝ^{n} \to ℝ^{n}$ changes the size of all boxes by the same factor, namely, the size of the image of a box $| t (S) |$ is $| T |$ times the size of the box $| S |$ , where $T$ is the matrix representing $t$ with respect to the standard basis.

That is, the determinant of a product is the product of the determinants $| T S | = | T | \cdot | S |$ .

The two sentences say the same thing, first in map terms and then in matrix terms. This is because $| t (S) | = | T S |$ , as both give the size of the box that is the image of the unit box $ℰ_{n}$ under the composition $t \circ s$ , where the maps are represented with respect to the standard basis. We will prove the second sentence.

Proof First consider the case that $T$ is singular and thus does not have an inverse. Observe that if $T S$ is invertible then there is an $M$ such that $(T S) M = I$ , so $T (S M) = I$ , and so $T$ is invertible. The contrapositive of that observation is that if $T$ is not invertible then neither is $T S$ — if $| T | = 0$ then $| T S | = 0$ .

Now consider the case that $T$ is nonsingular. Any nonsingular matrix factors into a product of elementary matrices $T = E_{1} E_{2} \dots E_{r}$ . To finish this argument we will verify that $| E S | = | E | \cdot | S |$ for all matrices $S$ and elementary matrices $E$ . The result will then follow because $| T S | = | E_{1} \dots E_{r} S | = | E_{1} | \dots | E_{r} | \cdot | S | = | E_{1} \dots E_{r} | \cdot | S | = | T | \cdot | S |$ .

There are three types of elementary matrix. We will cover the $M_{i} (k)$ case; the $P_{i, j}$ and $C_{i, j} (k)$ checks are similar. The matrix $M_{i} (k) S$ equals $S$ except that row $i$ is multiplied by $k$ . The third condition of determinant functions then gives that $| M_{i} (k) S | = k \cdot | S |$ . But $| M_{i} (k) | = k$ , again by the third condition because $M_{i} (k)$ is derived from the identity by multiplication of row $i$ by $k$ . Thus $| E S | = | E | \cdot | S |$ holds for $E = M_{i} (k)$ .

QED

Example 1.6 Application of the map $t$ represented with respect to the standard bases by

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

will double sizes of boxes, e.g., from this

$| \begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix} | = 3$

to this

$| \begin{matrix} 3 & 3 \\ - 4 & - 2 \end{matrix} | = 6$

Corollary 1.7 If a matrix is invertible then the determinant of its inverse is the inverse of its determinant $| T^{- 1} | = 1 / | T |$ .

Proof $1 = | I | = | T T^{- 1} | = | T | \cdot | T^{- 1} |$

QED

Exercises

Exercise 1.8 Supplied answer

Is

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

inside of the box formed by these three?

$(\begin{matrix} 3 \\ 3 \\ 1 \end{matrix}) (\begin{matrix} 2 \\ 6 \\ 1 \end{matrix}) (\begin{matrix} 1 \\ 0 \\ 5 \end{matrix})$

Back to Exercise 1.8

Answer. Solving

$c_{1} (\begin{matrix} 3 \\ 3 \\ 1 \end{matrix}) + c_{2} (\begin{matrix} 2 \\ 6 \\ 1 \end{matrix}) + c_{3} (\begin{matrix} 1 \\ 0 \\ 5 \end{matrix}) = (\begin{matrix} 4 \\ 1 \\ 2 \end{matrix})$

gives the unique solution $c_{3} = 11 / 57$ , $c_{2} = - 40 / 57$ and $c_{1} = 99 / 57$ . Because $c_{1} > 1$ , the vector is not in the box.
Exercise 1.9 Supplied answer

Recommended. Find the volume of the region defined by the vectors.
1. $⟨ (\begin{matrix} 1 \\ 3 \end{matrix}), (\begin{matrix} - 1 \\ 4 \end{matrix}) ⟩$
2. $⟨ (\begin{matrix} 2 \\ 1 \\ 0 \end{matrix}), (\begin{matrix} 3 \\ - 2 \\ 4 \end{matrix}), (\begin{matrix} 8 \\ - 3 \\ 8 \end{matrix}) ⟩$
3. $⟨ (\begin{matrix} 1 \\ 2 \\ 0 \\ 1 \end{matrix}), (\begin{matrix} 2 \\ 2 \\ 2 \\ 2 \end{matrix}), (\begin{matrix} - 1 \\ 3 \\ 0 \\ 5 \end{matrix}), (\begin{matrix} 0 \\ 1 \\ 0 \\ 7 \end{matrix}) ⟩$
Back to Exercise 1.9

Answer. For each, find the determinant and take the absolute value.
1. $7$
2. $0$
3. $- 58$

Exercise 1.10 Supplied answer

Recommended. In this picture the rectangle on the left is defined by the points $(3, 1)$ and $(1, 2)$ . Apply the matrix to get the rectangle on the right, defined by $(7, 1)$ and $(4, 2)$ . Why doesn’t this contradict Theorem 1.5?

	$⟶_{\begin{array}{l} \end{array}}^{(\begin{matrix} 2 & 1 \\ 0 & 1 \end{matrix})}$
area is $2$	determinant is $2$	area is $3$

Back to Exercise 1.10

Answer. We have drawn that picture to mislead. The picture on the left is not the box formed by two vectors. If we slide it to the origin then it becomes the box formed by this sequence.

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

Then the image under the action of the matrix is the box formed by this sequence.

$⟨ (\begin{matrix} 1 \\ 1 \end{matrix}), (\begin{matrix} 4 \\ 0 \end{matrix}) ⟩$

which has an area of $4$ .

Exercise 1.11 Supplied answer

Recommended. Find the volume of this region.

Back to Exercise 1.11

Answer. Move the parallelepiped to start at the origin, so that it becomes the box formed by

$⟨ (\begin{matrix} 3 \\ 0 \end{matrix}), (\begin{matrix} 2 \\ 1 \end{matrix}) ⟩$

and now the absolute value of this determinant is easily computed as $3$ .

$| \begin{matrix} 3 & 2 \\ 0 & 1 \end{matrix} | = 3$
Exercise 1.12 Supplied answer

Recommended. Suppose that $| A | = 3$ . By what factor do these change volumes?
1. $A$
2. $A^{2}$
3. $A^{- 2}$
Back to Exercise 1.12

Answer.
1. $3$
2. $9$
3. $1 / 9$
Exercise 1.13 Supplied answer

Recommended. Consider the linear transformation $t$ of $ℝ^{3}$ represented with respect to the standard bases by this matrix.

$(\begin{matrix} 1 & 0 & - 1 \\ 3 & 1 & 1 \\ - 1 & 0 & 3 \end{matrix})$
1. Compute the determinant of the matrix. Does the transformation preserve orientation or reverse it?
2. Find the size of the box defined by these vectors. What is its orientation?
  
  $(\begin{matrix} 1 \\ - 1 \\ 2 \end{matrix}) (\begin{matrix} 2 \\ 0 \\ - 1 \end{matrix}) (\begin{matrix} 1 \\ 1 \\ 0 \end{matrix})$
3. Find the images under $t$ of the vectors in the prior item and find the size of the box that they define. What is the orientation?
Back to Exercise 1.13

Answer.
1. Gauss’s Method
  
  $⟶_{\begin{array}{l} ρ_{1} + ρ_{3} \end{array}}^{- 3 ρ_{1} + ρ_{2}} (\begin{matrix} 1 & 0 & - 1 \\ 0 & 1 & 4 \\ 0 & 0 & 2 \end{matrix})$
  
  gives the determinant as $+ 2$ . The sign is positive so the transformation preserves orientation.
2. The size of the box is the value of this determinant.
  
  $| \begin{matrix} 1 & 2 & 1 \\ - 1 & 0 & 1 \\ 2 & - 1 & 0 \end{matrix} | = + 6$
  
  The orientation is positive.
3. Since this transformation is represented by the given matrix with respect to the standard bases, and with respect to the standard basis the vectors represent themselves, to find the image of the vectors under the transformation just multiply them, from the left, by the matrix.
  
  $(\begin{matrix} 1 \\ - 1 \\ 2 \end{matrix}) \mapsto (\begin{matrix} - 1 \\ 4 \\ 5 \end{matrix}) (\begin{matrix} 2 \\ 0 \\ - 1 \end{matrix}) \mapsto (\begin{matrix} 3 \\ 5 \\ - 5 \end{matrix}) (\begin{matrix} 1 \\ 1 \\ 0 \end{matrix}) \mapsto (\begin{matrix} 1 \\ 4 \\ - 1 \end{matrix})$
  
  Then compute the size of the resulting box.
  
  $| \begin{matrix} - 1 & 3 & 1 \\ 4 & 5 & 4 \\ 5 & - 5 & - 1 \end{matrix} | = + 12$
  
  The starting box is positively oriented, the transformation preserves orientations (since the determinant of the matrix is positive), and the ending box is also positively oriented.
Exercise 1.14 Supplied answer

By what factor does each transformation change the size of boxes?
1. $(\begin{matrix} x \\ y \end{matrix}) \mapsto (\begin{matrix} 2 x \\ 3 y \end{matrix})$
2. $(\begin{matrix} x \\ y \end{matrix}) \mapsto (\begin{matrix} 3 x - y \\ - 2 x + y \end{matrix})$
3. $(\begin{matrix} x \\ y \\ z \end{matrix}) \mapsto (\begin{matrix} x - y \\ x + y + z \\ y - 2 z \end{matrix})$
Back to Exercise 1.14

Answer. Express each transformation with respect to the standard bases and find the determinant.
1. $6$
2. $- 1$
3. $- 5$
Exercise 1.15 Supplied answer

What is the area of the image of the rectangle $[2. .4] \times [2. .5]$ under the action of this matrix?

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

Back to Exercise 1.15

Answer. The starting area is $6$ and the matrix changes sizes by $- 14$ . Thus the area of the image is $84$ .
Exercise 1.16 Supplied answer

If $t : ℝ^{3} \to ℝ^{3}$ changes volumes by a factor of $7$ and $s : ℝ^{3} \to ℝ^{3}$ changes volumes by a factor of $3 / 2$ then by what factor will their composition changes volumes?

Back to Exercise 1.16

Answer. By a factor of $21 / 2$ .
Exercise 1.17 Supplied answer

In what way does the definition of a box differ from the definition of a span?

Back to Exercise 1.17

Answer. For a box we take a sequence of vectors (as described in the remark, the order of the vectors matters), while for a span we take a set of vectors. Also, for a box subset of $ℝ^{n}$ there must be $n$ vectors; of course for a span there can be any number of vectors. Finally, for a box the coefficients $t_{1}$ , …, $t_{n}$ are in the interval $[0. .1]$ , while for a span the coefficients are free to range over all of $ℝ$ .
Exercise 1.18 Supplied answer

Does $| T S | = | S T |$ ? $| T (S P) | = | (T S) P |$ ?

Back to Exercise 1.18

Answer. Yes to both. For instance, the first is $| T S | = | T | \cdot | S | = | S | \cdot | T | = | S T |$ .
Exercise 1.19 Supplied answer

Show that there are no $2 \times 2$ matrices $A$ and $B$ satisfying these.

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

Back to Exercise 1.19

Answer. Because $| A B | = | A | \cdot | B | = | B A |$ and these two matrices have different determinants.
Exercise 1.20 Supplied answer
1. Suppose that $| A | = 3$ and that $| B | = 2$ . Find $| A^{2} \cdot B^{𝖳} \cdot B^{- 2} \cdot A^{𝖳} |$ .
2. Assume that $| A | = 0$ . Prove that $| 6 A^{3} + 5 A^{2} + 2 A | = 0$ .
Back to Exercise 1.20

Answer.
1. If it is defined then it is $(3^{2}) \cdot (2) \cdot (2^{- 2}) \cdot (3)$ .
2. $| 6 A^{3} + 5 A^{2} + 2 A | = | A | \cdot | 6 A^{2} + 5 A + 2 I |$ .
Exercise 1.21 Supplied answer

Recommended. Let $T$ be the matrix representing (with respect to the standard bases) the map that rotates plane vectors counterclockwise through $θ$ radians. By what factor does $T$ change sizes?

Back to Exercise 1.21

Answer. $| \begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix} | = 1$
Exercise 1.22 Supplied answer

Recommended. Must a transformation $t : ℝ^{2} \to ℝ^{2}$ that preserves areas also preserve lengths?

Back to Exercise 1.22

Answer. No, for instance the determinant of

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

is $1$ so it preserves areas, but the vector $T {\vec{e}}_{1}$ has length $2$ .
Exercise 1.23 Supplied answer

What is the volume of a parallelepiped in $ℝ^{3}$ bounded by a linearly dependent set?

Back to Exercise 1.23

Answer. It is zero.
Exercise 1.24 Supplied answer

Recommended. Find the area of the triangle in $ℝ^{3}$ with endpoints $(1, 2, 1)$ , $(3, - 1, 4)$ , and $(2, 2, 2)$ . (This asks for area, not volume. The triangle defines a plane; what is the area of the triangle in that plane?)

Back to Exercise 1.24

Answer. Two of the three sides of the triangle are formed by these vectors.

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

One way to find the area of this triangle is to find half the volume of the parallelogram formed by those two vectors and a length-one vector orthogonal to those two. To find the family of vectors orthogonal to those two we can start with the two relations

$(\begin{matrix} 1 \\ 0 \\ 1 \end{matrix}) \cdot (\begin{matrix} x \\ y \\ z \end{matrix}) = (\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}) (\begin{matrix} 2 \\ - 3 \\ 3 \end{matrix}) \cdot (\begin{matrix} x \\ y \\ z \end{matrix}) = (\begin{matrix} 0 \\ 0 \\ 0 \end{matrix})$

and solve the system

$\begin{array}{rcrcrcr} x & + & z & = & 0 \\ 2 x & - & 3 y & + & 3 z & = & 0 \end{array} ⟶_{\begin{array}{l} \end{array}}^{- 2 ρ_{1} + ρ_{2}} \begin{array}{rcrcrcr} x & + & z & = & 0 \\ - 3 y & + & z & = & 0 \end{array}$

to get this solution set.

${(\begin{matrix} - 1 \\ 1 / 3 \\ 1 \end{matrix}) z ∣ z \in ℝ}$

Here is a length one solution.

$\frac{1}{\sqrt{19 / 9}} \cdot (\begin{matrix} - 1 \\ 1 / 3 \\ 1 \end{matrix}) = (\begin{matrix} - 3 / \sqrt{19} \\ 1 / \sqrt{19} \\ 3 / \sqrt{19} \end{matrix})$

Thus the area of the triangle is half of the absolute value of this determinant.

$| \begin{matrix} 1 & 2 & - 3 / \sqrt{19} \\ 0 & - 3 & 1 / \sqrt{19} \\ 1 & 3 & 3 / \sqrt{19} \end{matrix} | = - 19 / \sqrt{19}$

Half of the absolute value is $\sqrt{19} / 2$ .
Exercise 1.25 Supplied answer

An alternate proof of Theorem 1.5 uses the definition of determinant functions.
1. Note that the vectors forming $S$ make a linearly dependent set if and only if $| S | = 0$ , and check that the result holds in this case.
2. For the $| S | \neq 0$ case, to show that $| T S | / | S | = | T |$ for all transformations, consider the function $d : ℳ_{n \times n} \to ℝ$ given by $T \mapsto | T S | / | S |$ . Show that $d$ has the first property of a determinant.
3. Show that $d$ has the remaining three properties of a determinant function.
4. Conclude that $| T S | = | T | \cdot | S |$ .
Back to Exercise 1.25

Answer.
1. Because the image of a linearly dependent set is linearly dependent, if the vectors forming $S$ make a linearly dependent set, so that $| S | = 0$ , then the vectors forming $t (S)$ make a linearly dependent set, so that $| T S | = 0$ , and in this case the equation holds.
2. We must check that if $T ⟶_{\begin{array}{l} \end{array}}^{k ρ_{i} + ρ_{j}} \hat{T}$ then $d (T) = | T S | / | S | = | \hat{T} S | / | S | = d (\hat{T})$ . We can do this by checking that combining rows first and then multiplying to get $\hat{T} S$ gives the same result as multiplying first to get $T S$ and then combining (because the determinant $| T S |$ is unaffected by the combining rows so we’ll then have that $| \hat{T} S | = | T S |$ and hence that $d (\hat{T}) = d (T)$ ). This check runs: after adding $k$ times row $i$ of $T S$ to row $j$ of $T S$ , the $j, p$ entry is $(k t_{i, 1} + t_{j, 1}) s_{1, p} + \dots + (k t_{i, r} + t_{j, r}) s_{r, p}$ , which is the $j, p$ entry of $\hat{T} S$ .
3. For the second property, we need only check that swapping $T ⟶_{\begin{array}{l} \end{array}}^{ρ_{i} \leftrightarrow ρ_{j}} \hat{T}$ and then multiplying to get $\hat{T} S$ gives the same result as multiplying $T$ by $S$ first and then swapping (because, as the determinant $| T S |$ changes sign on the row swap, we’ll then have $| \hat{T} S | = - | T S |$ , and so $d (\hat{T}) = - d (T)$ ). This check runs just like the one for the first property.
  
  For the third property, we need only show that performing $T ⟶_{\begin{array}{l} \end{array}}^{k ρ_{i}} \hat{T}$ and then computing $\hat{T} S$ gives the same result as first computing $T S$ and then performing the scalar multiplication (as the determinant $| T S |$ is rescaled by $k$ , we’ll have $| \hat{T} S | = k | T S |$ and so $d (\hat{T}) = k d (T)$ ). Here too, the argument runs just as above.
  
  The fourth property, that if $T$ is $I$ then the result is $1$ , is obvious.
4. Determinant functions are unique, so $| T S | / | S | = d (T) = | T |$ , and so $| T S | = | T | | S |$ .
Exercise 1.26 Supplied answer

Give a non-identity matrix with the property that $A^{𝖳} = A^{- 1}$ . Show that if $A^{𝖳} = A^{- 1}$ then $| A | = \pm 1$ . Does the converse hold?

Back to Exercise 1.26

Answer. Any permutation matrix has the property that the transpose of the matrix is its inverse.

For the implication, we know that $| A^{𝖳} | = | A |$ . Then $1 = | A \cdot A^{- 1} | = | A \cdot A^{𝖳} | = | A | \cdot | A^{𝖳} | = {| A |}^{2}$ .

The converse does not hold; here is an example.

$(\begin{matrix} 3 & 1 \\ 2 & 1 \end{matrix})$
Exercise 1.27 Supplied answer

The algebraic property of determinants that factoring a scalar out of a single row will multiply the determinant by that scalar shows that where $H$ is $3 \times 3$ , the determinant of $c H$ is $c^{3}$ times the determinant of $H$ . Explain this geometrically, that is, using Theorem 1.5. (The observation that increasing the linear size of a three-dimensional object by a factor of $c$ will increase its volume by a factor of $c^{3}$ while only increasing its surface area by an amount proportional to a factor of $c^{2}$ is the Square-cube law [Wikipedia, Square-cube Law].)

Back to Exercise 1.27

Answer. Where the sides of the box are $c$ times longer, the box has $c^{3}$ times as many cubic units of volume.
Exercise 1.28 Supplied answer

We say that matrices $H$ and $G$ are similar if there is a nonsingular matrix $P$ such that $H = P^{- 1} G P$ (we will study this relation in Chapter Five). Show that similar matrices have the same determinant.

Back to Exercise 1.28

Answer. If $H = P^{- 1} G P$ then $| H | = | P^{- 1} | | G | | P | = | P^{- 1} | | P | | G | = | P^{- 1} P | | G | = | G |$ .
Exercise 1.29 Supplied answer

We usually represent vectors in $ℝ^{2}$ with respect to the standard basis so vectors in the first quadrant have both coordinates positive.

${Rep}_{ℰ_{2}} (\vec{v}) = (\begin{matrix} + 3 \\ + 2 \end{matrix})$

Moving counterclockwise around the origin, we cycle through four regions:

$\dots ⟶ (\begin{matrix} + \\ + \end{matrix}) ⟶ (\begin{matrix} - \\ + \end{matrix}) ⟶ (\begin{matrix} - \\ - \end{matrix}) ⟶ (\begin{matrix} + \\ - \end{matrix}) ⟶ \dots .$

Using this basis

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

gives the same counterclockwise cycle. We say these two bases have the same orientation.
1. Why do they give the same cycle?
2. What other configurations of unit vectors on the axes give the same cycle?
3. Find the determinants of the matrices formed from those (ordered) bases.
4. What other counterclockwise cycles are possible, and what are the associated determinants?
5. What happens in $ℝ^{1}$ ?
6. What happens in $ℝ^{3}$ ?
A fascinating general-audience discussion of orientations is in [Gardner].
Back to Exercise 1.29

Answer.
1. The new basis is the old basis rotated by $π / 4$ .
2. $⟨ (\begin{matrix} - 1 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ - 1 \end{matrix}) ⟩$ , $⟨ (\begin{matrix} 0 \\ - 1 \end{matrix}), (\begin{matrix} 1 \\ 0 \end{matrix}) ⟩$
3. In each case the determinant is $+ 1$ (we say that these bases have positive orientation).
4. Because only one sign can change at a time, the only other cycle possible is
  
  $\dots ⟶ (\begin{matrix} + \\ + \end{matrix}) ⟶ (\begin{matrix} + \\ - \end{matrix}) ⟶ (\begin{matrix} - \\ - \end{matrix}) ⟶ (\begin{matrix} - \\ + \end{matrix}) ⟶ \dots .$
  
  Here each associated determinant is $- 1$ (we say that such bases have a negative orientation).
5. There is one positively oriented basis $⟨ (1) ⟩$ and one negatively oriented basis $⟨ (- 1) ⟩$ .
6. There are $48$ bases ( $6$ half-axis choices are possible for the first unit vector, $4$ for the second, and $2$ for the last). Half are positively oriented like the standard basis on the left below, and half are negatively oriented like the one on the right
  
  In $ℝ^{3}$ positive orientation is sometimes called ‘right hand orientation’ because if a person places their right hand with their fingers curling from ${\vec{e}}_{1}$ to ${\vec{e}}_{2}$ then the thumb will point with ${\vec{e}}_{3}$ .
Exercise 1.30 Supplied answer

This question uses material from the optional Determinant Functions Exist subsection. Prove Theorem 1.5 by using the permutation expansion formula for the determinant.

Back to Exercise 1.30

Answer. We will compare $det ({\vec{s}}_{1}, \dots, {\vec{s}}_{n})$ with $det (t ({\vec{s}}_{1}), \dots, t ({\vec{s}}_{n}))$ to show that the second differs from the first by a factor of $| T |$ . We represent the $\vec{s}$ ’s with respect to the standard bases

${Rep}_{ℰ_{n}} ({\vec{s}}_{i}) = (\begin{matrix} s_{1, i} \\ s_{2, i} \\ ⋮ \\ s_{n, i} \end{matrix})$

and then we represent the map application with matrix-vector multiplication

$\begin{aligned} {Rep}_{ℰ_{n}} (t ({\vec{s}}_{i})) & = (\begin{array}{cccc} t_{1, 1} & t_{1, 2} & \dots & t_{1, n} \\ t_{2, 1} & t_{2, 2} & \dots & t_{2, n} \\ ⋮ \\ t_{n, 1} & t_{n, 2} & \dots & t_{n, n} \end{array}) (\begin{array}{c} s_{1, j} \\ s_{2, j} \\ ⋮ \\ s_{n, j} \end{array}) \\ = s_{1, j} (\begin{array}{c} t_{1, 1} \\ t_{2, 1} \\ ⋮ \\ t_{n, 1} \end{array}) + s_{2, j} (\begin{array}{c} t_{1, 2} \\ t_{2, 2} \\ ⋮ \\ t_{n, 2} \end{array}) + \dots + s_{n, j} (\begin{array}{c} t_{1, n} \\ t_{2, n} \\ ⋮ \\ t_{n, n} \end{array}) \\ = s_{1, j} {\vec{t}}_{1} + s_{2, j} {\vec{t}}_{2} + \dots + s_{n, j} {\vec{t}}_{n} \end{aligned}$

where ${\vec{t}}_{i}$ is column $i$ of $T$ . Then $det (t ({\vec{s}}_{1}), \dots, t ({\vec{s}}_{n}))$ equals $det (s_{1, 1} {\vec{t}}_{1} + s_{2, 1} {\vec{t}}_{2} + \dots + s_{n, 1} {\vec{t}}_{n}, \dots, s_{1, n} {\vec{t}}_{1} + s_{2, n} {\vec{t}}_{2} + \dots + s_{n, n} {\vec{t}}_{n})$ .

As in the derivation of the permutation expansion formula, we apply multilinearity, first splitting along the sum in the first argument

$\begin{matrix} det (s_{1, 1} {\vec{t}}_{1}, \dots, s_{1, n} {\vec{t}}_{1} + s_{2, n} {\vec{t}}_{2} + \dots + s_{n, n} {\vec{t}}_{n}) \\ + \dots + det (s_{n, 1} {\vec{t}}_{n}, \dots, s_{1, n} {\vec{t}}_{1} + s_{2, n} {\vec{t}}_{2} + \dots + s_{n, n} {\vec{t}}_{n}) \end{matrix}$

and then splitting each of those $n$ summands along the sums in the second arguments, etc. We end with, as in the derivation of the permutation expansion, $n^{n}$ summand determinants, each of the form $det (s_{i_{1}, 1} {\vec{t}}_{i_{1}}, s_{i_{2}, 2} {\vec{t}}_{i_{2}}, \dots, s_{i_{n}, n} {\vec{t}}_{i_{n}})$ . Factor out each of the $s_{i, j}$ ’s $= s_{i_{1}, 1} s_{i_{2}, 2} \dots s_{i_{n}, n} \cdot det ({\vec{t}}_{i_{1}}, {\vec{t}}_{i_{2}}, \dots, {\vec{t}}_{i_{n}})$ .

As in the permutation expansion derivation, whenever two of the indices in $i_{1}$ , …, $i_{n}$ are equal then the determinant has two equal arguments, and evaluates to $0$ . So we need only consider the cases where $i_{1}$ , …, $i_{n}$ form a permutation of the numbers $1$ , …, $n$ . We thus have

$det (t ({\vec{s}}_{1}), \dots, t ({\vec{s}}_{n})) = \sum_{permutations ϕ} s_{ϕ (1), 1} \dots s_{ϕ (n), n} det ({\vec{t}}_{ϕ (1)}, \dots, {\vec{t}}_{ϕ (n)}) .$

Swap the columns in $det ({\vec{t}}_{ϕ (1)}, \dots, {\vec{t}}_{ϕ (n)})$ to get the matrix $T$ back, which changes the sign by a factor of $sgn ϕ$ , and then factor out the determinant of $T$ .

$= \sum_{ϕ} s_{ϕ (1), 1} \dots s_{ϕ (n), n} det ({\vec{t}}_{1}, \dots, {\vec{t}}_{n}) \cdot sgn ϕ = det (T) \sum_{ϕ} s_{ϕ (1), 1} \dots s_{ϕ (n), n} \cdot sgn ϕ .$

As in the proof that the determinant of a matrix equals the determinant of its transpose, we commute the $s$ ’s to list them by ascending row number instead of by ascending column number (and we substitute $sgn (ϕ^{- 1})$ for $sgn (ϕ)$ ).

$= det (T) \sum_{ϕ} s_{1, ϕ^{- 1} (1)} \dots s_{n, ϕ^{- 1} (n)} \cdot sgn ϕ^{- 1} = det (T) det ({\vec{s}}_{1}, {\vec{s}}_{2}, \dots, {\vec{s}}_{n})$
Exercise 1.31 Supplied answer

Recommended.
1. Show that this gives the equation of a line in $ℝ^{2}$ through $(x_{2}, y_{2})$ and $(x_{3}, y_{3})$ .
  
  $| \begin{matrix} x & x_{2} & x_{3} \\ y & y_{2} & y_{3} \\ 1 & 1 & 1 \end{matrix} | = 0$
2. [Petersen] Prove that the area of a triangle with vertices $(x_{1}, y_{1})$ , $(x_{2}, y_{2})$ , and $(x_{3}, y_{3})$ is
  
  $\frac{1}{2} | \begin{matrix} x_{1} & x_{2} & x_{3} \\ y_{1} & y_{2} & y_{3} \\ 1 & 1 & 1 \end{matrix} | .$
3. [Math. Mag., Jan. 1973] Prove that the area of a triangle with vertices at $(x_{1}, y_{1})$ , $(x_{2}, y_{2})$ , and $(x_{3}, y_{3})$ whose coordinates are integers has an area of $N$ or $N / 2$ for some positive integer $N$ .
Back to Exercise 1.31

Answer.
1. An algebraic check is easy.
  
  $0 = x y_{2} + x_{2} y_{3} + x_{3} y - x_{3} y_{2} - x y_{3} - x_{2} y = x \cdot (y_{2} - y_{3}) + y \cdot (x_{3} - x_{2}) + x_{2} y_{3} - x_{3} y_{2}$
  
  simplifies to the familiar form
  
  $y = x \cdot (x_{3} - x_{2}) / (y_{3} - y_{2}) + (x_{2} y_{3} - x_{3} y_{2}) / (y_{3} - y_{2})$
  
  (the $y_{3} - y_{2} = 0$ case is easily handled).
  
  For geometric insight, this picture shows that the box formed by the three vectors. Note that all three vectors end in the $z = 1$ plane. Below the two vectors on the right is the line through $(x_{2}, y_{2})$ and $(x_{3}, y_{3})$ .
  
  The box will have a nonzero volume unless the triangle formed by the ends of the three is degenerate. That only happens (assuming that $(x_{2}, y_{3}) \neq (x_{3}, y_{3})$ ) if $(x, y)$ lies on the line through the other two.
2. This is how the answer was given in the cited source. We find the altitude through $(x_{1}, y_{1})$ of a triangle with vertices $(x_{1}, y_{1})$ $(x_{2}, y_{2})$ and $(x_{3}, y_{3})$ in the usual way from the normal form of the above:
  
  $\frac{1}{\sqrt{(x_{2} - x_{3})^{2} + (y_{2} - y_{3})^{2}}} | \begin{matrix} x_{1} & x_{2} & x_{3} \\ y_{1} & y_{2} & y_{3} \\ 1 & 1 & 1 \end{matrix} | .$
  
  Another step shows the area of the triangle to be
  
  $\frac{1}{2} | \begin{matrix} x_{1} & x_{2} & x_{3} \\ y_{1} & y_{2} & y_{3} \\ 1 & 1 & 1 \end{matrix} | .$
  
  This exposition reveals the modus operandi more clearly than the usual proof of showing a collection of terms to be identical with the determinant.
3. This is how the answer was given in the cited source. Let
  
  $D = | \begin{matrix} x_{1} & x_{2} & x_{3} \\ y_{1} & y_{2} & y_{3} \\ 1 & 1 & 1 \end{matrix} |$
  
  then the area of the triangle is $(1 / 2) | D |$ . Now if the coordinates are all integers, then $D$ is an integer.

References cited in this section

Wikipedia, Square-cube Law

The Square-cube law, http://en.wikipedia.org/wiki/Square-cube_law, 2011-Jan-17.

Gardner

Martin Gardner, The New Ambidextrous Universe, third revised edition, W. H. Freeman and Company, 1990.

Petersen

G. M. Petersen, Area of a Triangle, American Mathematical Monthly, volume 62 number 4 (Apr. 1955), p. 249.

Math. Mag., Jan. 1973

Marvin Bittinger (proposer), Quickie 578, Mathematics Magazine, volume 46 number 5 (Jan. 1973), p. 286, 296.