Lecture 08

[[lecture-data]]

2024-09-13

Readings

• a

2. Chapter 2

primarily about Schur’s theorem

Schur's Theorem

Suppose $A\in M_n$ has eigenvalues ${\lambda }_1,{\lambda }_2,\dots {\lambda }_n$ in any order. Then there exists a unitary matrix $U$ and upper triangular matrix $T$ such that $A=UT{U}^{\ast }$

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 ${\lambda }_1,{\lambda }_2,\dots {\lambda }_n$

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 ${\lambda }_1$ 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 $\begin{array}{rl}{U}^{\ast }AU& =\left[\begin{array}{c}{x}^{\ast }\\ u_2^{\ast }\\ \dots \\ u_n^{\ast }\end{array}\right]A\left[\begin{array}{cccc}x& u_2& \dots & u_n\end{array}\right]\\ & =\left[\begin{array}{cccc}{\lambda }_1& \ast & \dots & \ast \\ 0& & & \\ ⋮& \text{"B"}& & \\ 0& & & \end{array}\right]\end{array}$

So $\sigma (B)=\{{\lambda }_2,\dots {\lambda }_n\}$ and we can repeat this process:

Take $y$ be a normalized eigenvector associated with eigenvalue ${\lambda }_2$ of $B$ and let $V\in M_{n-1}$ be unitary with first column $y$.

Then

\left(U \begin{bmatrix} I_{1} & 0 \\ 0 & V \end{bmatrix}\right)^* A \left(U \begin{bmatrix} I_{1} & 0 \\ 0 & V \end{bmatrix}\right) &= \begin{bmatrix} I & 0 \\ 0 & V^* \end{bmatrix}U^*AU\begin{bmatrix} I & 0 \\ 0 & V \end{bmatrix} \\ &= \begin{bmatrix} \lambda_{1} & * \\ 0 & V^*BV \end{bmatrix} \\ &= \begin{bmatrix} \lambda_{1} & * & * \\ 0 & \lambda_{2} & * \\ \vdots & \vdots & \text{"C"} \\ 0 & 0 \end{bmatrix} \end{aligned}$$ and $\sigma(C) = \{ \lambda_{3},\dots,\lambda_{n} \}$ Continuing this process, we get a unitary matrix which is the product of each of the unitary matrices that we found to get $\Omega^* A\Omega = T$ upper triangular, where each of the entries of the diagonal of $T$ are exactly the eigenvalues of $A$. And this is the [[Concept Wiki/Schur's theorem\|Schur Decomposition]] $\blacksquare$ Additionally, if $A$ is real and all the eigenvalues are real, then all steps can be done over $\mathbb{R}$ $\blacksquare$

Theorem

Suppose $\text{scr}F$ $\subset M_n$ 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 $\pi :\{1,\dots ,n\}\to \{1,\dots ,n\}$ such that

$\sigma (A+B)=\{{\lambda }_i(A)+{\lambda }_{\pi (i)}(B):i=1,\dots ,n\}$ $\sigma (AB)=\{{\lambda }_i(A){\lambda }_{\pi (i)}(B):i=1,\dots ,n\}$

$A$ and $B$ commute, we can say $A=U{T}_A{U}^{\ast }$ and $B=U{T}_B{U}^{\ast }$ where $U\in M_n$ unitary and $T_A,T_B\in M_n$ upper triangular. Then $A+B=U{T}_A{U}^{\ast }+U{T}_B{U}^{\ast }=U(T_A+T_B)U^{\ast }$ $AB=U{T}_A{U}^{\ast }U{T}_B{U}^{\ast }=U[T_A{T}_B]U^{\ast }$

Since

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

Matrix Polynomial

Suppose we have a complex-valued polynomial $p(t)=a_m{t}^m+a_{m-1}{t}^{m-1}+\cdots +a_0=a_m(t-{\lambda }_1)(t-{\lambda }_2)\dots (t-{\lambda }_m)$

Then for any $A\in M_n$ any matrix, define the matrix polynomial

$p(A):=a_m{A}^m+a_{m-1}{A}^{m-1}+\cdots +a_{0}I=a_m(A-{\lambda }_{1}I)(A-{\lambda }_{2}I)\dots (A-{\lambda }_{m}I)$

(see matrix polynomial)

Suppose $A=UT{U}^{\ast }$. Then $A^k=UT{U}^{\ast }UT{U}^{\ast }\dots UT{U}^{\ast }=U{T}^k{U}^{\ast }$ where $T^k$ is upper triangular with diagonal terms $t_{ii}^k$.

Thus, for a matrix polynomial, we have

p(t_{11}) & * & \dots & * \\ 0 & p(t_{22}) & \ddots & \vdots \\ \vdots & \ddots & \ddots & *\\ 0 & \dots & 0 & p(t_{nn}) \end{bmatrix} U^*$$ and we can say that $\sigma(p(A)) = \{ p(\lambda) : \lambda \in \sigma(A) \}$ >[!theorem] > >Every matrix is upper triangularizable going back to [[Concept Wiki/householder transformation]] things. Say $w \in \mathbb{F}^n$ is unit length $w^*w = 1$. What are the eigenvalues and eigenvectors of $ww^*$ ? - $[ww^*]w = w$ - so the eigenvectors of $ww^*$ are $\text{span}(w)$ with eigenvalue $1$ - Anything orthogonal to $w$ will have eigenvalue $0$ So $-2ww^*$ has eigenvalues of $ww^*$ multiplied by $-2$ with same eigenvectors And $I-2ww^*$ has the eigenvalues $1-2(\sigma(A))$ Suppose $x,y \in \mathbb{R}^n$ with $x \neq y$ and $\lvert \lvert x \rvert \rvert_{2} = \lvert \lvert y \rvert \rvert_{2}$ and define $w:= \frac{1}{\lvert \lvert x-y \rvert \rvert}(x-y)$. Now, note that $(x+y)^T(x-y) = x^Tx - y^Ty + x^y - x^Ty = 0$ $\implies x-y \in \text{span}(w)$ is an eigenvector of $ww^*$ with eigenvalue $-1$ $\implies x+y \in w^\perp$ is an eigenvector of $ww^*$ with eigenvalue $+1$ Writing - $x = \frac{1}{2}(x-y) + \frac{1}{2}(x+y)$ - $y = -(\frac{1}{2}(x-y) +\frac{1}{2}(x+y))$ we can see that $$H_{w}x = y \;\;\text{ and }\;\;H_{w}y = x$$ (ie, we are "flipping" about an axis)