Lecture 09

[[lecture-data]]

2024-09-16

Readings

• a

2. Chapter 2

Note

Suppose I have two $[k+(n-k)]\times [k+(n-k)]$ block matrices

\begin{bmatrix} 0 & * \ 0 & T \end{bmatrix} \begin{bmatrix} D_{1} & * \ 0 & D_{2} \end{bmatrix} = \begin{bmatrix} 0 & | \begin{bmatrix} 0 \ \vdots \ 0 \end{bmatrix} & * \ 0 & | \begin{bmatrix} 0 \ \vdots \ 0 \end{bmatrix} & * \ \end{bmatrix} = \begin{bmatrix} 0 & * \ 0 & T_{*} \end{bmatrix}$$

Where $T$ is upper triangular and each $D_i$ is diagonal. Then the resulting matrix is $[(k+1)+(n-(k+1))]\times [(k+1)+(n-(k+1))]$ with $T_{\ast }$ also upper triangular.

Cayley-Hamilton

Let $A\in M_n$. Then $p_A(A)=0$

Note

This is a homework problem for diagonalizable matrices, but you are going to do it using the tools that you already have.

(see Cayley-Hamilton)

$A=UT{U}^{\ast }$ where $U\in M_n$ is unitary and $T\in M_n$ upper triangular per Schur's theorem. Say that we have characteristic polynomial

Let

$p_A(t)=(t-{\lambda }_1)(t-{\lambda }_2)\dots (t-{\lambda }_n)$

Such that $t_{ii}={\lambda }_i$. Now, we have from our definition of the matrix polynomial

p_{A}(A) &= Up_{A}(T) U^* \ &= U[(T-\lambda_{1}I)(T-\lambda_{2}I)\dots(T-\lambda_{n}I)]U^* \ &= U \begin{bmatrix} 0 & * & * \ 0 & * & * \ 0 & 0 & \ddots \end{bmatrix}

\begin{bmatrix}* & * & * \ 0 & 0 & * \ 0 & 0 & \ddots \end{bmatrix} \begin{bmatrix}* & * & * \ 0 & \ddots & * \ 0 & 0 & 0 \end{bmatrix}U^* \ &= U [0] U^* \ &= 0 \end{aligned}$$

Corrolary

Suppose $A\in M_n$ is invertible and $p_A(t)=t^n+a_{n-1}{t}^{n-1}+\cdots +a_0$ Then $A^{-1}=\frac{(-1{)}^{n+1}}{\det A}[A^{n-1}+a_{n-1}{A}^{n-2}+a_{n-2}{A}^{n-3}+\cdots +a_{1}I]$

$A$ invertible $⟹\det A\ne 0$ and $\det A=(-1{)}^n{a}_0$. By Cayley-Hamilton, we have

A[A^{n-1}+a_{n-1}A^{n-2}+a_{n-2}A^{n-3}+\dots+a_{1}I] &= -a_{0}I \ A\left[ \frac{1}{-a_{0}} [A^{n-1}+a_{n-1}A^{n-2}+a_{n-2}A^{n-3}+\dots+a_{1}I] \right] &= I \ \implies \left[ \frac{1}{-a_{0}} [A^{n-1}+a_{n-1}A^{n-2}+a_{n-2}A^{n-3}+\dots+a_{1}I] \right] &= A^{-1} \end{aligned}$$

(see matrices are polynomials of their inverses)

Theorem

Suppose $A\in M_n$. Then $\forall ϵ>0$, there exists $S\in M_n$ invertible and $D\in M_n$ diagonal, and $E\in M_n$ with $||E|{|}_F<ϵ$ such that $A=S(D+E)S^{-1}$

”$A$ is almost similar to a diagonal matrix”

$A=UT{U}^{\ast }$ where $U\in M_n$ unitary, $T\in M_n$ upper triangular per Schur. For all $\delta >0$, we can define

Let

\delta^{-1} & 0 & \dots & 0 \ 0 & \delta^{-2} & \ddots & \vdots \ \vdots & \ddots & \ddots & 0 \ 0 & \dots & 0 & \delta^{-n} \end{bmatrix} T \begin{bmatrix} \delta^{1} & 0 & \dots & 0 \ 0 & \delta^{2} & \ddots & \vdots \ \vdots & \ddots & \ddots & 0 \ 0 & \dots & 0 & \delta^{n} \end{bmatrix}$$

and $Q$ will have $ij$th entry equal to $t_{ij}{\delta }^{j-i}$

A &= UTU^* \ &= UITIU^* \ &= U\begin{bmatrix} \delta^{-1} & 0 & \dots & 0 \ 0 & \delta^{-2} & \ddots & \vdots \ \vdots & \ddots & \ddots & 0 \ 0 & \dots & 0 & \delta^{-n} \end{bmatrix}\begin{bmatrix} \delta^{1} & 0 & \dots & 0 \ 0 & \delta^{2} & \ddots & \vdots \ \vdots & \ddots & \ddots & 0 \ 0 & \dots & 0 & \delta^{n} \end{bmatrix}T\begin{bmatrix} \delta^{-1} & 0 & \dots & 0 \ 0 & \delta^{-2} & \ddots & \vdots \ \vdots & \ddots & \ddots & 0 \ 0 & \dots & 0 & \delta^{-n} \end{bmatrix}\begin{bmatrix} \delta^{1} & 0 & \dots & 0 \ 0 & \delta^{2} & \ddots & \vdots \ \vdots & \ddots & \ddots & 0 \ 0 & \dots & 0 & \delta^{n} \end{bmatrix}U^* \ &= U \begin{bmatrix} \delta^{-1} & 0 & \dots & 0 \ 0 & \delta^{-2} & \ddots & \vdots \ \vdots & \ddots & \ddots & 0 \ 0 & \dots & 0 & \delta^{-n} \end{bmatrix} Q \begin{bmatrix} \delta^{1} & 0 & \dots & 0 \ 0 & \delta^{2} & \ddots & \vdots \ \vdots & \ddots & \ddots & 0 \ 0 & \dots & 0 & \delta^{n} \end{bmatrix} U^* \ &= S Q S^{-1} \end{aligned}$$

Where $S=U\left[\begin{array}{cccc}{\delta }^{-1}& 0& \dots & 0\\ 0& {\delta }^{-2}& \ddots & ⋮\\ ⋮& \ddots & \ddots & 0\\ 0& \dots & 0& {\delta }^{-n}\end{array}\right]$ Note that $Q=\left[\begin{array}{cccc}{t}_{11}& 0& \dots & 0\\ 0& t_{22}& \ddots & ⋮\\ ⋮& \ddots & \ddots & 0\\ 0& \dots & 0& t_{nn}\end{array}\right]+\mathcal{O}(\delta )$ Which is what was to be shown. $■$

(see matrices are almost similar to diagonal matrices)

Theorem

Let $A\in M_n$. Then for all $ϵ>0$, there exists $E\in M_n$ such that $||E|{|}_F<ϵ$ and $A+E$ is diagonalizable

“every matrix is almost diagonalizable”

Caution

This is similar to the last result, but NOT THE SAME.

$A\in M_n=UT{U}^{\ast }$ by Schur. Then there exists a diagonal matrix $D$ such that $||D|{|}_F<ϵ$ and $T+D$ has distinct diagonals. Then

Let

A + UDU^* &= UTU^* + UDU^* \ &= UTU^* + UDU^* \ &= U(T+D)U^* \ \end{aligned}$$

and this has all eigenvalues distinct by construction, and thus is diagonalizable! So let $E=UD{U}^{\ast }$. Then since $U$ is unitary $||E|{|}_F=||D|{|}_F<ϵ$

(see matrices are almost diagonalizable)

Last main topic of Chapter 2: Normal Matrices

Normal Matrices

$A\in M_n$ is normal precisely when $A{A}^{\ast }=A^{\ast }A$

Example

1. diagonal matrices
2. hermitian matrices (duh)
3. unitary matrices

(see normal matrix)

Lemma

Let $T\in M_n$ be upper triangular. Then $T$ is normal if and only if $T$ is diagonal.

Proof (informal) $T$ is normal upper triangular. Then $T{T}^{\ast }=T^{\ast }T$ since it is normal.

Suppose

$$\left[\begin{array}{c}-r_1-\\ -r_2-\\ ⋮\\ -r_n-\end{array}\right]\left[\begin{array}{cccc}|& |& \dots & |\\ r_1^{\ast }& r_2^{\ast }& \dots & r_n^{\ast }\\ |& |& \dots & |\end{array}\right]$$

&= \begin{bmatrix} | & | & \dots & | \ c_{1} & c_{2} & \dots & c_{n} \ | & | & \dots & |\end{bmatrix}\begin{bmatrix} - c_{1}^* -\ - c_{2}^-\ \vdots \ - c_{n}^ -\end{bmatrix} \end{aligned}$$

Where $t_i$ are increasing in “nonzero length”

Then the $i$th diagonal of $T^{\ast }T$, call it $[T^{\ast }T{]}_{ii}=c_i^{\ast }{c}_i=||c_i|{|}^2$ where $t_i$ is the $i$the COLUMN of $T$

And the $i$th diagonal of $T{T}^{\ast }$, call it $[T{T}^{\ast }{]}_{ii}=r_i{r}_i^{\ast }=||r_i|{|}^2$ where $r_i$ is the $i$th ROW of $T$ (this is why we can do the multiplication like this to get a scalar)

So looking at $T$, we know that the $i$th column and the $i$th row must have the same length.

• So for the first column, the length will be the first element. Which means the other elements of the first row must be zero.
• if we continue in the same manner through the rest of the matrix, we realize that the rest of the non-diagonal elements must be 0 also.

(see triangular matrices are normal iff diagonal)