Lecture 11 - Finishing 7.A

As some things for the class:

this is exam week.
exams will be less proof heavy, although there will be some more brief proofs.

Recall that:

Definition

$T$ self-adjoint means $T = T^{*}$ .
$T$ being normal means $T T^{*} = T^{*} T$ .

One thing we proved is:

Nice properties of normal operators.

$T \in L (V)$ is normal operator iff $‖ T v ‖ = ‖ T^{*} v ‖$ for all $v \in V$ .

We wanted to prove:

Eigenvectors for a normal operator

Suppose $T \in L (V)$ is normal, and a given eigenvector $v \in V$ for $T$ (ie: $T v = λ v$ where $v \neq \vec{0}$ ). Then $T^{*} v = \overset{―}{λ} v$ , so $v$ is an eigenvector of $T^{*}$ as well.

We do the proof:

Proof
We can show that $T - λ I$ is normal, via:

$\begin{aligned} (T - λ I) (T - λ I)^{*} & = (T - λ I)^{*} (T - λ I) & ? \\ T T^{*} - λ T^{*} - \overset{―}{λ} T + | λ |^{2} & = T^{*} T - λ T^{*} - \overset{―}{λ} T + | λ | \end{aligned}$

As such, then we can use the other lemma above to get:

(T - λ I) v = 0

for any eigenvector $v \neq 0$ with eigenvalue $λ$ . As such, then $‖ (T - λ I) v ‖ = 0$ . Thus, iff:

\begin{aligned} ‖ (T - λ I)^{*} v ‖ & = 0 \\ ‖ (T^{*} - \overset{―}{λ} I) v ‖ & = 0 \end{aligned}

So then $(T^{*} - \overset{―}{λ} I) v = 0$ and as such then $v$ is an eigenvector of $T^{*}$ with eigenvalue $\overset{―}{λ}$ .
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Lemma

Suppose $T$ is normal. Then eigenvectors of $T$ corresponding to different eigenvalues are orthogonal.

Proof
Consider two different $v_{1}, v_{2} \neq 0$ are eigenvectors, corresponding to $λ_{1}, λ_{2}$ respectively, where $λ_{1} \neq λ_{2}$ . Thus:

T v_{1} = λ_{1} v_{1}, T v_{2} = λ_{2} v_{2}

using Lecture 10 - Adjoint in More Detail#^bfabbc, then:

T^{*} v_{1} = \overset{―}{λ_{1}} v_{1}, T^{*} v_{2} = \overset{―}{λ_{2}} v_{2}

Notice that:

\begin{aligned} ⟨ T v_{1}, v_{2} ⟩ & = ⟨ v_{1}, T^{*} v_{2} ⟩ \\ ⟨ λ_{1} v_{1}, v_{2} ⟩ & = ⟨ v_{1}, \overset{―}{λ_{2}} v_{2} ⟩ \\ λ_{1} ⟨ v_{1}, v_{2} ⟩ & = λ_{2} ⟨ v_{1}, v_{2} ⟩ \\ (λ_{1} - λ_{2}) ⟨ v_{1}, v_{2} ⟩ \end{aligned}

But since $λ_{1} \neq λ_{2}$ then $λ_{1} - λ_{2} \neq 0$ so then clearly $⟨ v_{1}, v_{2} ⟩ = 0$ . Thus $v_{1}, v_{2}$ are orthogonal.
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So maybe we can make a basis with eigenvectors with are all orthogonal. This next section discusses when that happens.

7.B: The Spectral Theorem

Recall that an operator $T \in L (V)$ (assuming finite dimensional over $F$ ) is diagonalizable iff there is a basis of eigenvectors (an eigenbasis) of $V$ . Thus, it is a basis consisting of eigenvectors of $T$ .

If you have such a basis $β$ , then thinking about the matrix is just gonna have eigenvalues on the diagonal:

M (T, β) = [\begin{matrix} λ_{1}, & 0 & \dots & 0 \\ 0 & λ_{2} & \dots & 0 \\ ⋮ & ⋮ & ⋱ \\ 0 & 0 & \dots & λ_{n} \end{matrix}]

where there could be repetition on our $λ_{i}$ 's. The question is, when does $T$ have an orthonormal eigenbasis?

Let's investigate this for a second. Which ones do we know have this property?

Suppose $T$ has an orthonormal eigenbasis. Then $M (T, β)$ is as shown above, and thus is diagonal. But what about $M (T^{*})$ ? It also must be diagonal. This is because:

M (T^{*}) = M (T)^{*} = [\begin{matrix} \overset{―}{λ_{1}}, & 0 & \dots & 0 \\ 0 & \overset{―}{λ_{2}} & \dots & 0 \\ ⋮ & ⋮ & ⋱ \\ 0 & 0 & \dots & \overset{―}{λ_{n}} \end{matrix}]

Notice that:

M (T^{*} T) = M (T^{*}) M (T) = M (T) M (T^{*}) = M (T T^{*})

Thus $T$ is normal. Even better, all the $λ_{i}$ 's would have to be real since then $λ_{i} = \overset{―}{λ_{1}}$ which can only happen if the $λ_{i}$ 's are all real. And then $M (T) = M (T^{*})$ so $T = T^{*}$ . So if it's a real vector space, then we get both normality and self-adjointness together. Thus:

When $F = C$ , then $T$ is normal is equivalent to having an orthonormal eigenbasis.
When $F = R$ , then $T$ is self-adjoint is equivalent to having an orthonormal eigenbasis.

The Spectral Theorem(s)

Suppose $T \in L (V)$ , the following are equivalent:

$T$ is normal (when $F = C$ ), $T$ is self-adjoint (and thus also normal) (when $F = R$ )
$V$ has an orthonormal eigenbasis.
$T$ has a diagonal matrix representation w.r.t. some orthonormal basis.

Proof
We already know (2) is equivalent with (3) from Chapter 5. We already know that (c) implies (a) via our explanation prior. Thus, we just need to show that we can go from (a) to any of (b) or (c). It actually depends on our field $F$ which one to do, based on if we have a real-vector space or complex-vector space.

In a more specific case (and without loss of generality), we focus on (a) implies (c). Suppose $T$ is normal, and $V$ is a complex vector space. Because $V$ is over $C$ , then we know that there exists an orthonormal basis, namely $β = {e_{1}, . . ., e_{n}}$ such that:, $M (T)$ is UT:

M (T) = [\begin{matrix} a_{11} & a_{12} & \dots & a_{1 n} \\ 0 & a_{22} & \dots & a_{2 n} \\ ⋮ & ⋮ & ⋱ \\ 0 & 0 & \dots & a_{n n} \end{matrix}]

thus:

M (T^{*}) = M (T)^{*} = M (T) = [\begin{matrix} \overset{―}{a_{11}} & 0 & \dots & 0 \\ \overset{―}{a_{12}} & \overset{―}{a_{22}} & \dots & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ \\ \overset{―}{a_{1 n}} & \overset{―}{a_{2 n}} & \dots & a_{n n} \end{matrix}]

The first column of $M (T)$ is just $T e_{1}$ , and the same for $M (T^{*})$ is $T^{*} e_{1}$ .

Recall that since $‖ T v ‖ = ‖ T^{*} v ‖$ by an earlier theorem, so since:

v = c_{1} e_{1} + \dots + c_{n} e_{n} \Leftrightarrow ‖ v ‖^{2} = \sum_{i = 1}^{n} | c_{i} |^{2}

so then:

‖ T e_{1} ‖ = ‖ T e_{2} ‖ \Leftrightarrow | a_{11} |^{2} = | \overset{―}{a_{11}} |^{2} + \dots + | \overset{―}{a_{1 n}} |^{2}

But since $| a_{11} |^{2} = | \overset{―}{a_{11}} |^{2}$ then everything else must be 0's. As such, then $a_{12}, . . ., a_{1 n} = 0$ .

Repeat this process for all $n$ columns. As such, $M (T)$ is diagonal.
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