Lecture 08

[[lecture-data]]

2024-09-13

Readings

2. Chapter 2

primarily about Schur's theorem

Schur's Theorem

Suppose $A \in M_{n}$ has eigenvalues $λ_{1}, λ_{2}, \dots λ_{n}$ in any order. Then there exists a unitary matrix $U$ and upper triangular matrix $T$ such that

A = U T U^{*}

ie, for any square matrix, you can find a unitary similarity to an upper triangular matrix. Since $A \sim T$ and $T$ upper triangular, we can set the diagonal of $T$ to be exactly $λ_{1}, λ_{2}, \dots λ_{n}$

Additionally, if $A$ is real-valued and the eigenvalues of $A$ are all real, then $U$ and $T$ can be found to be real-valued.
(see Schur's theorem)

Proof

Let $x$ be a normalized eigenvector with eigenvalue $λ_{1}$ and let $U \in M_{n}$ be a unitary such that $x$ is the first column.
(last time Lecture 07, we talked about being able to find such a $U$ )

Then $$\begin{aligned}
U^AU &= \begin{bmatrix}
x^ \
u_{2}^* \
\dots \
u_{n}^*
\end{bmatrix} A \begin{bmatrix}
x & u_{2} & \dots & u_{n}
\end{bmatrix} \
&= \begin{bmatrix}
\lambda_{1} & * & \dots & * \
0 & \
\vdots & \text{"B"}\
0
\end{bmatrix}
\end{aligned}$$

So $σ (B) = {λ_{2}, \dots λ_{n}}$ and we can repeat this process:

Take $y$ be a normalized eigenvector associated with eigenvalue $λ_{2}$ of $B$ and let $V \in M_{n - 1}$ be unitary with first column $y$ .

Then

\begin{aligned} {(U [\begin{array}{c} I_{1} & 0 \\ 0 & V \end{array}])}^{*} A (U [\begin{array}{c} I_{1} & 0 \\ 0 & V \end{array}]) & = [\begin{array}{c} I & 0 \\ 0 & V^{*} \end{array}] U^{*} A U [\begin{array}{c} I & 0 \\ 0 & V \end{array}] \\ = [\begin{array}{c} λ_{1} & * \\ 0 & V^{*} B V \end{array}] \\ = [\begin{array}{c} λ_{1} & * & * \\ 0 & λ_{2} & * \\ ⋮ & ⋮ & "C" \\ 0 & 0 \end{array}] \end{aligned}

and $σ (C) = {λ_{3}, \dots, λ_{n}}$

Continuing this process, we get a unitary matrix which is the product of each of the unitary matrices that we found to get $Ω^{*} A Ω = T$ upper triangular, where each of the entries of the diagonal of $T$ are exactly the eigenvalues of $A$ . And this is the Schur Decomposition $◼$

Additionally, if $A$ is real and all the eigenvalues are real, then all steps can be done over $R$ $◼$

Theorem

Suppose $F$ $\subset M_{n}$ are pairwise commuting. Then they are simultaneously unitarily upper triangularizable.

(see pairwise commuting matrices are simultaneously upper triangularizable)

Proposition

If $A, B \in M_{n}$ commute, then there exists a bijection $π : {1, \dots, n} \to {1, \dots, n}$ such that

σ (A + B) = {λ_{i} (A) + λ_{π (i)} (B) : i = 1, \dots, n}

σ (A B) = {λ_{i} (A) λ_{π (i)} (B) : i = 1, \dots, n}

Proof

Since $A$ and $B$ commute, we can say $A = U T_{A} U^{*}$ and $B = U T_{B} U^{*}$ where $U \in M_{n}$ unitary and $T_{A}, T_{B} \in M_{n}$ upper triangular. Then

A + B = U T_{A} U^{*} + U T_{B} U^{*} = U (T_{A} + T_{B}) U^{*}

A B = U T_{A} U^{*} U T_{B} U^{*} = U [T_{A} T_{B}] U^{*}

Then, let $π$ be the bijection that maps the (* * * * )

Matrix Polynomial

Suppose we have a complex-valued polynomial

p (t) = a_{m} t^{m} + a_{m - 1} t^{m - 1} + \dots + a_{0} = a_{m} (t - λ_{1}) (t - λ_{2}) \dots (t - λ_{m})

Then for any $A \in M_{n}$ any matrix, define the matrix polynomial

p (A) := a_{m} A^{m} + a_{m - 1} A^{m - 1} + \dots + a_{0} I = a_{m} (A - λ_{1} I) (A - λ_{2} I) \dots (A - λ_{m} I)

(see matrix polynomial)

Suppose $A = U T U^{*}$ . Then

A^{k} = U T U^{*} U T U^{*} \dots U T U^{*} = U T^{k} U^{*} $ $ w h e r e $ T^{k} $ i s u p p e r t r i a n g u l a r w i t h d i a g o n a l t e r m s $ t_{i i}^{k} $ . T h u s, f o r a m a t r i x p o l y n o m i a l, w e h a v e $ $ p (A) = U [p (T)] U^{*} = U [\begin{matrix} p (t_{11}) & * & \dots & * \\ 0 & p (t_{22}) & ⋱ & ⋮ \\ ⋮ & ⋱ & ⋱ & * \\ 0 & \dots & 0 & p (t_{n n}) \end{matrix}] U^{*}

and we can say that $σ (p (A)) = {p (λ) : λ \in σ (A)}$

Theorem

Every matrix is upper triangularizable

going back to householder transformation things.

Say $w \in F^{n}$ is unit length $w^{*} w = 1$ . What are the eigenvalues and eigenvectors of $w w^{*}$ ?

$[w w^{*}] w = w$
so the eigenvectors of $w w^{*}$ are $span (w)$ with eigenvalue $1$
Anything orthogonal to $w$ will have eigenvalue $0$

So $- 2 w w^{*}$ has eigenvalues of $w w^{*}$ multiplied by $- 2$ with same eigenvectors
And $I - 2 w w^{*}$ has the eigenvalues $1 - 2 (σ (A))$

Suppose $x, y \in R^{n}$ with $x \neq y$ and $| | x | |_{2} = | | y | |_{2}$ and define $w := \frac{1}{| | x - y | |} (x - y)$ . Now, note that $(x + y)^{T} (x - y) = x^{T} x - y^{T} y + x^{y} - x^{T} y = 0$
$⟹ x - y \in span (w)$ is an eigenvector of $w w^{*}$ with eigenvalue $- 1$
$⟹ x + y \in w^{⊥}$ is an eigenvector of $w w^{*}$ with eigenvalue $+ 1$

Writing

$x = \frac{1}{2} (x - y) + \frac{1}{2} (x + y)$
$y = - (\frac{1}{2} (x - y) + \frac{1}{2} (x + y))$ we can see that

H_{w} x = y and H_{w} y = x

(ie, we are "flipping" about an axis)