Lecture 15

[[lecture-data]]

2024-09-30

Announcement

Exam 1 covers chapter 1 through chapter 3.

3. Chapter 3

We are pretty much done with chapter 3

Suppose $A \in M_{n}$ such that $A^{2} = A$ . What can we say about this matrix?

its eigenvalues are only either 0 or 1
It is diagonalizable since the minimal poly implies that the maximum jordan block size is 1

The polynomial $t^{2} - t$ is an annihilating polynomial. This follows immediately from the given information. Thus we know that

q_{A} (t) = {\begin{cases} t \\ t - 1 \\ t (t - 1) \end{cases}

No matter what, the minimal polynomial is a product of distinct linear factors. Hence, $A$ is diagonalizable. (see a matrix is diagonalizable if and only if all the jordan blocks are of size 1 from last lecture)

p_{A} (t) = {\begin{cases} t^{n} if q_{A} (t) = t \\ (t - 1)^{n} if q_{A} (t) = t - 1 \\ t^{k} (t - 1)^{n - k} if q_{A} (t) = t (t - 1) \end{cases}

In the first case, there is only one possible $A$ (when all eigenvalues are 0) and this is the zero matrix.
In the second case, there is also only one possible diagonal matrix for $A$ and this is the identity matrix

Since $A$ is diagonalizable, we can write $A = S D S^{- 1}$ and let $D$ have all 1s together on the diagonal. Then

A [\begin{matrix} s^{(1)} | & s^{(2)} | & \dots | & s^{(n)} \end{matrix}] = [\begin{matrix} s^{(1)} | & s^{(2)} | & \dots | & s^{(n)} \end{matrix}] [\begin{matrix} 1 & 0 & \dots \\ 0 & 1 & \dots \\ ⋮ & ⋱ \end{matrix}]

The columns of $S$ form a basis for $C^{n}$ . Suppose the algebraic multiplicity of $1$ is $r$ . The eigenspace of $1$ call it ${eigspace}_{1}$ are the first $r$ columns of $S = {s^{(1)}, \dots, s^{(r)}}$
and the eigenspace of 0 called ${eigspace}_{0}$ are the last $n - r$ columns of $S = {s^{(r + 1)}, \dots, s^{(n)}}$

for all $x \in C^{n}$ , there exists a unique $v \in {eigspace}_{1}$ and $w \in {eigspace}_{0}$ such that $x = v + w$

But note that $A x = A (v + w) = v$ . (This is called projection 🙂)
ie

Theorem

if $A \in M_{n}$ is idempotent, ie $A^{2} = A$ then $A$ is a projection matrix.
(see idempotent matrices are projection matrices)

Note

The space ${eigspace}_{0}$ is also called the nullspace.

And in this case, ${eigspace}_{1}$ is the range of $A$ .

What if $A$ was also normal ? What else would that tell us? Then $A$ is hermitian. The eigenvalues of $A$ are only 1 and 0, so they are all real and the spectral theorem for hermitian matrices states this as as an equivalent condition.

4. Chapter 4

All about hermitian matrices and courant-fischer theorem

$A \in M_{n}$ is hermitian. Since all eigenvalues are real, we can order them such that $λ_{1} (A) \leq λ_{2} (A) \leq \dots \leq λ_{n} (A)$ .

Lemma

Suppose $U \in M_{n}$ is unitary and $S = {x \in C^{n} : | | x | |_{2} = 1}$ . Then $U : S \to S$ is bijective.

Proof

Think about it. Recall that unitary matrices define isometries.

Field of Values

Let $A \in M_{n}$ . The field of values of $A$ is defined as $$F(A) := \left{ \frac{x^*Ax}{x^*x} : x \in \mathbb{C}^n \neq 0 \right} = { x^*Ax : x \in \mathbb{C}^n, \lvert \lvert x \rvert \rvert_{2} = 1 }$$

This is a set of scalars and it is defined for any matrix.

(see field of values)

Theorem

Let $A \in M_{n}$ be normal. The field of values of
$A = H$ $(σ (A)) = {\sum_{i = 1}^{n} α_{i} λ_{i} : λ_{i} \in σ (A), \sum_{i = 1}^{n} α_{i} = 1}$

(where $H$ denotes the convex hull)

Proof

$A$ normal $⟹ A$ is unitarily diagonalizable, say $A = U D U^{*}$ . Then $F (A) = {x^{*} A x : x \in C^{n}, | | x | |_{2} = 1} = {x^{*} U D U^{*} x : x \in C^{n}, | | x | |_{2} = 1}$ . Let $y = U^{*} x$ . Then $F (A) = {y^{*} D y : y \in C^{n}, | | y | |_{2} = 1}$ but these are all the vectors $y$ on the unit sphere! And this is exactly

{\sum_{i = 1}^{n} {\bar{y}}_{i} y_{i} λ_{i} : \sum_{i = 1}^{n} | y_{i} |^{2} = 1}

So call $| y_{i} |^{2} = α_{i}$ and we get precisely all convex combinations of the eigenvalues.

(see the field of values for a normal matrix is the convex hull of the spectrum)