Lecture 25 - Starting 8.C

We talked about stuff from the end of 8.B. See that section specifically to get info on what you missed.

Starting 8.C: Characteristic Polynomial and Minimal Polynomial

Let's start with the definition:

characteristic polynomial

Suppose $T \in L (V)$ , where $V$ is a complex vector space. Let $λ_{1}, . . ., λ_{m}$ be the distinct eigenvalues of $T$ with multiplicities $d_{1}, . ., d_{m}$ . Then the characteristic polynomial of $T$ is:

p_{T} (z) = (z - λ_{1})^{d_{1}} (z - λ_{2})^{d_{2}} \dots (z - λ_{m})^{d_{m}}

Where $p_{T}$ denotes this polynomial in relation to $T$ .

Fact

$\deg (p_{T}) = \dim (V)$ . This comes from Chapter 8 - Operators on Complex Vector Spaces#^aa3174.

Fact

The zeroes of the characteristic polynomial $p_{T}$ are precisely the eigenvalues of $T$ .

Axler would disapprove, but usually the characteristic polynomial is defined using the determinant:

p_{T} (z) = det (z I - T)

But it's harder to find these facts using the determinant definition. Hence why Axler does this.

Cayley-Hamilton Theorem

This is a big theorem, but it's a one liner as we've done all the leg work to get to this point:

Cayley-Hamilton Theorem

Every operator on a complex vector space is a root (namely a zero) of its characteristic polynomial.

What this says is that:

p_{T} (T) = {\vec{0}}_{T}

Proof
Use the definition with $z := T$ :

p_{T} (T) = (T - λ_{1} I)^{d_{1}} (T - λ_{2} I)^{d_{2}} \dots (T - λ_{m} I)^{d_{m}}

And if we plug in any vector $v \in V$ , then we know that:

V = ⨁_{i = 1}^{m} G (λ_{i}, T)

So then:

v = α_{1} v_{1} + \dots + α_{m} v_{m}

where each $v_{i} \in G (λ_{i}, T)$ . Notice that:

\begin{aligned} p_{T} (T) (v) & = (\prod_{i = 1}^{m} (T - λ_{i} I)^{d_{i}}) (\sum_{j = 1}^{m} v_{j}) \\ = \sum_{j = 1}^{m} \prod_{i = 1, i \neq j}^{m} (T - λ_{i} I)^{d_{i}} \underset{= {\vec{0}}_{T} since v_{j} \in G (λ_{j}, T)}{\underset{⏟}{(T - λ_{j} I)^{d_{j}}}} v_{j} \\ = \sum_{j = 1}^{m} \prod_{i = 1, i \neq j}^{m} (T - λ_{i} I)^{d_{i}} \vec{0} \\ = \sum_{j = 1}^{m} \vec{0} \\ = \vec{0} \end{aligned}

Thus $p_{T} (T) = {\vec{0}}_{T}$ as $v$ was arbitrary.
☐
This begs the question:

Question

Every operator $T \in L (V)$ is a zero of it's characteristic polynomial, namely $p_{T} (T) = {\vec{0}}_{T}$ . But is it possible that it could be the zero of a polynomial of lesser degree? Namely, could we reduce the degree of $p$ such that we get the same property?

As an example, consider:

M (T) = λ I = [\begin{matrix} λ & 0 & 0 \\ 0 & λ & 0 \\ 0 & 0 & λ \end{matrix}]

Notice then that:

M (T - λ I) = \vec{0} = [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}]

So then $λ$ is the only eigenvalue and has multiplicity 3. Thus we can write down:

p_{T} (z) = (z - λ)^{3}

We know that $p_{T} (T) = \vec{0}$ as we expect. But we don't need the 3rd power in this case for the Cayley Hamilton Theorem to be true. We could instead have:

μ (z) = (z - λ) \Rightarrow μ (T) = T - λ I = {\vec{0}}_{T}

Heck we could've even done a squared term and gotten something similar. But finding the smallest degree is of interest, and is called the minimal polynomial.

minimal polynomial

Suppose $T \in L (V)$ where $V$ is a complex vector space. The minimal polynomial of $T$ is defined to be the unique monic polynomial (the leading coefficient is equal to 1) $μ_{T}$ , of smallest degree, such that:

μ_{T} (T) = {\vec{0}}_{T}

Fact

$\deg (μ_{T}) \leq \deg (p_{T}) = \dim (V)$
Any polynomial $s$ with $s (T) = {\vec{0}}_{T}$ can only happen iff $s$ is a multiple of $μ_{T}$ .
$p_{T}$ is a multiple of $μ_{T}$
The zeroes of $μ_{T}$ are precisely the eigenvalues of $T$ .