Similarity

We have shown that for any homomorphism there are bases $B$ and $D$ such that the matrix representing the map has a block partial-identity form.

${Rep}_{B, D} (h) = (\begin{array}{cc} Identity & Zero \\ Zero & Zero \end{array})$

This representation describes the map as sending $c_{1} {\vec{β}}_{1} + \dots + c_{n} {\vec{β}}_{n}$ to $c_{1} {\vec{δ}}_{1} + \dots + c_{k} {\vec{δ}}_{k} + \vec{0} + \dots + \vec{0}$ , where $n$ is the dimension of the domain and $k$ is the dimension of the range. Under this representation the action of the map is easy to understand because most of the matrix entries are zero.

This chapter considers the special case where the domain and codomain are the same. Here we naturally ask for the domain basis and codomain basis to be the same. That is, we want a basis $B$ so that ${Rep}_{B, B} (t)$ is as simple as possible, where we take ‘simple’ to mean that it has many zeroes. We will find that we cannot always get a matrix having the above block partial-identity form but we will develop a form that comes close, a representation that is nearly diagonal.

Complex Vector Spaces

This chapter requires that we factor polynomials. But many polynomials do not factor over the real numbers; for instance, $x^{2} + 1$ does not factor into a product of two linear polynomials with real coefficients; instead it requires complex numbers $x^{2} + 1 = (x - i) (x + i)$ .

Consequently in this chapter we shall use complex numbers for our scalars, including entries in vectors and matrices. That is, we shift from studying vector spaces over the real numbers to vector spaces over the complex numbers. Any real number is a complex number and in this chapter most of the examples use only real numbers but nonetheless, the critical theorems require that the scalars be complex. So this first section is a review of complex numbers.

In this book our approach is to shift to this more general context of taking scalars to be complex for the pragmatic reason that we must do so in order to move forward. However, the idea of doing vector spaces by taking scalars from a structure other than the real numbers is an interesting and useful one. Delightful presentations that take this approach from the start are in [Halmos] and [Hoffman & Kunze].

Polynomial Factoring and Complex Numbers

This subsection is a review only. For a full development, including proofs, see [Ebbinghaus].

Consider a polynomial $p (x) = c_{n} x^{n} + \dots + c_{1} x + c_{0}$ with leading coefficient $c_{n} \neq 0$ . We say that it is a degree $n$ polynomial. If $n = 0$ then $p$ is a constant polynomial $p (x) = c_{0}$ . Constant polynomials that are not the zero polynomial, $c_{0} \neq 0$ , have degree zero. We define the zero polynomial to have degree $- \infty$ .

Remark 1.1 Defining the degree of the zero polynomial to be $- \infty$ allows the equation $degree (f g) = degree (f) + degree (g)$ to hold for all polynomials.

Just as integers have a division operation—e.g., ‘ $4$ goes $5$ times into $21$ with remainder $1$ ’—so do polynomials.

Theorem 1.2 (Division Theorem for Polynomials) Let $p (x)$ be a polynomial. If $d (x)$ is a non-zero polynomial then there are quotient and remainder polynomials $q (x)$ and $r (x)$ such that

$p (x) = d (x) \cdot q (x) + r (x)$

where the degree of $r (x)$ is strictly less than the degree of $d (x)$ .

The point of the integer statement ‘ $4$ goes $5$ times into $21$ with remainder $1$ ’ is that the remainder is less than $4$ —while $4$ goes $5$ times, it does not go $6$ times. Similarly, the final clause of the polynomial division statement is crucial.

Example 1.3 If $p (x) = 2 x^{3} - 3 x^{2} + 4 x$ and $d (x) = x^{2} + 1$ then $q (x) = 2 x - 3$ and $r (x) = 2 x + 3$ . Note that $r (x)$ has a lower degree than does $d (x)$ .

Corollary 1.4 The remainder when $p (x)$ is divided by $x - λ$ is the constant polynomial $r (x) = p (λ)$ .

Proof The remainder must be a constant polynomial because it is of degree less than the divisor $x - λ$ . To determine the constant, take the theorem’s divisor $d (x)$ to be $x - λ$ and substitute $λ$ for $x$ .

QED

If a divisor $d (x)$ goes into a dividend $p (x)$ evenly, meaning that $r (x)$ is the zero polynomial, then $d (x)$ is a called a factor of $p (x)$ . Any root of the factor, any $λ \in ℝ$ such that $d (λ) = 0$ , is a root of $p (x)$ since $p (λ) = d (λ) \cdot q (λ) = 0$ .

Corollary 1.5 If $λ$ is a root of the polynomial $p (x)$ then $x - λ$ divides $p (x)$ evenly, that is, $x - λ$ is a factor of $p (x)$ .

Proof By the above corollary $p (x) = (x - λ) \cdot q (x) + p (λ)$ . Since $λ$ is a root, $p (λ) = 0$ so $x - λ$ is a factor.

QED

A repeated root of a polynomial is a number $λ$ such that the polynomial is evenly divisible by $(x - λ)^{n}$ for some power larger than one. The largest such power is called the multiplicity of $λ$ .

Finding the roots and factors of a high-degree polynomial can be hard. But for second-degree polynomials we have the quadratic formula: the roots of $a x^{2} + b x + c$ are these

$λ_{1} = \frac{- b + \sqrt{b^{2} - 4 a c}}{2 a} λ_{2} = \frac{- b - \sqrt{b^{2} - 4 a c}}{2 a}$

(if the discriminant $b^{2} - 4 a c$ is negative then the polynomial has no real number roots). A polynomial that cannot be factored into two lower-degree polynomials with real number coefficients is said to be irreducible over the reals.

Theorem 1.6 Any constant or linear polynomial is irreducible over the reals. A quadratic polynomial is irreducible over the reals if and only if its discriminant is negative. No cubic or higher-degree polynomial is irreducible over the reals.

Corollary 1.7 Any polynomial with real coefficients factors into a product of linear and irreducible quadratic polynomials with real coefficients. That factorization is unique; any two factorizations have the same factors raised to the same powers.

Note the analogy with the prime factorization of integers. In both cases the uniqueness clause is very useful.

Example 1.8 Because of uniqueness we know, without multiplying them out, that $(x + 3)^{2} (x^{2} + 1)^{3}$ does not equal $(x + 3)^{4} (x^{2} + x + 1)^{2}$ .

Example 1.9 By uniqueness, if $c (x) = m (x) \cdot q (x)$ then where $c (x) = (x - 3)^{2} (x + 2)^{3}$ and $m (x) = (x - 3) (x + 2)^{2}$ , we know that $q (x) = (x - 3) (x + 2)$ .

While $x^{2} + 1$ has no real roots and so doesn’t factor over the real numbers, if we imagine a root—traditionally denoted $i$ , so that $i^{2} + 1 = 0$ —then $x^{2} + 1$ factors into a product of linears, $(x - i) (x + i)$ . When we adjoin this root $i$ to the reals and close the new system with respect to addition and multiplication then we have the complex numbers, $ℂ = {a + b i ∣ a, b \in ℝ and i^{2} = - 1}$ .

For a scalar $z \in ℂ$ , where $z = a + b i$ , we call $a$ the real part of $z$ , and $b$ the imaginary part. We often picture complex numbers on the complex plane, with $a$ plotted on the real axis, the horizontal axis, and $b$ plotted on the imaginary axis, the vertical axis. Note that the distance of the point from the origin is the length, $| a + b i | = \sqrt{a^{2} + b^{2}}$ .

Recall the definitions of the complex number addition and scalar multiplication

$(a + b i) + (c + d i) = (a + c) + (b + d) i r \cdot (a + b i) = (r a) + (r b) i$

(and consequently subtraction is $(a + b i) - (c + d i) = (a - c) + (b - d) i$ ). Recall also the definition of complex-complex multiplication.

$\begin{aligned} (a + b i) (c + d i) & = a c + a d i + b c i + b d (- 1) \\ = (a c - b d) + (a d + b c) i \end{aligned}$

Example 1.10 For instance, $(2 - 3 i) (4 - 0.5 i) = 6.5 - 13 i$ .

Over the complex numbers, any quadratic polynomial factors into linears.

$a x^{2} + b x + c = a \cdot (x - \frac{- b + \sqrt{b^{2} - 4 a c}}{2 a}) \cdot (x - \frac{- b - \sqrt{b^{2} - 4 a c}}{2 a})$

Example 1.11 The second degree polynomial $x^{2} + x + 1$ factors over the complex numbers into the product of two first degree polynomials.

$(x - \frac{- 1 + \sqrt{- 3}}{2}) (x - \frac{- 1 - \sqrt{- 3}}{2}) = (x - (- \frac{1}{2} + \frac{\sqrt{3}}{2} i)) (x - (- \frac{1}{2} - \frac{\sqrt{3}}{2} i))$

In $ℂ$ , in contrast with the reals, there are no irreducible quadratics. All polynomials factor completely into linears.

Theorem 1.12 (Fundamental Theorem of Algebra)

Polynomials with complex coefficients factor into linear polynomials with complex coefficients. The factorization is unique.

Complex Representations

With the above definitions for the complex numbers, all of the operations that we’ve used for real vector spaces carry over unchanged to vector spaces with complex scalars.

Example 2.1 Matrix multiplication is the same, although the computation can involve more arithmetic.

$\begin{matrix} (\begin{matrix} 1 + 1 i & 2 - 0 i \\ i & - 2 + 3 i \end{matrix}) (\begin{matrix} 1 + 0 i & 1 - 0 i \\ 3 i & - i \end{matrix}) \\ \begin{aligned} = (\begin{array}{c} (1 + 1 i) \cdot (1 + 0 i) + (2 - 0 i) \cdot (3 i) & (1 + 1 i) \cdot (1 - 0 i) + (2 - 0 i) \cdot (- i) \\ (i) \cdot (1 + 0 i) + (- 2 + 3 i) \cdot (3 i) & (i) \cdot (1 - 0 i) + (- 2 + 3 i) \cdot (- i) \end{array}) \\ = (\begin{array}{c} 1 + 7 i & 1 - 1 i \\ - 9 - 5 i & 3 + 3 i \end{array}) \end{aligned} \end{matrix}$

We shall carry over unchanged from the previous chapters everything that we can. For instance, we shall call this

$⟨ (\begin{matrix} 1 + 0 i \\ 0 + 0 i \\ ⋮ \\ 0 + 0 i \end{matrix}), \dots, (\begin{matrix} 0 + 0 i \\ 0 + 0 i \\ ⋮ \\ 1 + 0 i \end{matrix}) ⟩$

the standard basis for $ℂ^{n}$ as a vector space over $ℂ$ and again denote it $ℰ_{n}$ . Another example is that $𝒫_{n}$ will be the vector space of degree $n$ polynomials with coefficients that are complex.

References cited in this section

Halmos

Paul R. Halmos, Finite Dimensional Vector Spaces, second edition, Van Nostrand, 1958.

Hoffman & Kunze

Kenneth Hoffman, Ray Kunze, Linear Algebra, second edition, Prentice-Hall, 1971.

Ebbinghaus

H. D. Ebbinghaus, Numbers, Springer-Verlag, 1990.