Lecture 09

[[lecture-data]]

2024-09-16

Readings

2. Chapter 2

Note

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

[\begin{matrix} 0 & * \\ 0 & T \end{matrix}] [\begin{matrix} D_{1} & * \\ 0 & D_{2} \end{matrix}] = [\begin{matrix} 0 & | [\begin{matrix} 0 \\ ⋮ \\ 0 \end{matrix}] & * \\ 0 & | [\begin{matrix} 0 \\ ⋮ \\ 0 \end{matrix}] & * \end{matrix}] = [\begin{matrix} 0 & * \\ 0 & T_{*} \end{matrix}]

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 $T_{*}$ 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)

Proof

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

p_{A} (t) = (t - λ_{1}) (t - λ_{2}) \dots (t - λ_{n})

Such that $t_{i i} = λ_{i}$ . Now, we have from our definition of the matrix polynomial

\begin{aligned} p_{A} (A) & = U p_{A} (T) U^{*} \\ = U [(T - λ_{1} I) (T - λ_{2} I) \dots (T - λ_{n} I)] U^{*} \\ = U [\begin{array}{c} 0 & * & * \\ 0 & * & * \\ 0 & 0 & ⋱ \end{array}] [\begin{array}{c} * & * & * \\ 0 & 0 & * \\ 0 & 0 & ⋱ \end{array}] [\begin{array}{c} * & * & * \\ 0 & ⋱ & * \\ 0 & 0 & 0 \end{array}] 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} + \dots + 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} + \dots + a_{1} I]

Proof

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

\begin{aligned} A [A^{n - 1} + a_{n - 1} A^{n - 2} + a_{n - 2} A^{n - 3} + \dots + a_{1} I] + a_{0} I & = 0 \\ A [A^{n - 1} + a_{n - 1} A^{n - 2} + a_{n - 2} A^{n - 3} + \dots + a_{1} I] & = - a_{0} I \\ A [\frac{1}{- a_{0}} [A^{n - 1} + a_{n - 1} A^{n - 2} + a_{n - 2} A^{n - 3} + \dots + a_{1} I]] & = I \\ ⟹ [\frac{1}{- a_{0}} [A^{n - 1} + a_{n - 1} A^{n - 2} + a_{n - 2} A^{n - 3} + \dots + a_{1} I]] & = 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"

Proof

Let $A = U T U^{*}$ where $U \in M_{n}$ unitary, $T \in M_{n}$ upper triangular per Schur. For all $δ > 0$ , we can define

Q = [\begin{matrix} δ^{- 1} & 0 & \dots & 0 \\ 0 & δ^{- 2} & ⋱ & ⋮ \\ ⋮ & ⋱ & ⋱ & 0 \\ 0 & \dots & 0 & δ^{- n} \end{matrix}] T [\begin{matrix} δ^{1} & 0 & \dots & 0 \\ 0 & δ^{2} & ⋱ & ⋮ \\ ⋮ & ⋱ & ⋱ & 0 \\ 0 & \dots & 0 & δ^{n} \end{matrix}]

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

\begin{aligned} A & = U T U^{*} \\ = U I T I U^{*} \\ = U [\begin{array}{c} δ^{- 1} & 0 & \dots & 0 \\ 0 & δ^{- 2} & ⋱ & ⋮ \\ ⋮ & ⋱ & ⋱ & 0 \\ 0 & \dots & 0 & δ^{- n} \end{array}] [\begin{array}{c} δ^{1} & 0 & \dots & 0 \\ 0 & δ^{2} & ⋱ & ⋮ \\ ⋮ & ⋱ & ⋱ & 0 \\ 0 & \dots & 0 & δ^{n} \end{array}] T [\begin{array}{c} δ^{- 1} & 0 & \dots & 0 \\ 0 & δ^{- 2} & ⋱ & ⋮ \\ ⋮ & ⋱ & ⋱ & 0 \\ 0 & \dots & 0 & δ^{- n} \end{array}] [\begin{array}{c} δ^{1} & 0 & \dots & 0 \\ 0 & δ^{2} & ⋱ & ⋮ \\ ⋮ & ⋱ & ⋱ & 0 \\ 0 & \dots & 0 & δ^{n} \end{array}] U^{*} \\ = U [\begin{array}{c} δ^{- 1} & 0 & \dots & 0 \\ 0 & δ^{- 2} & ⋱ & ⋮ \\ ⋮ & ⋱ & ⋱ & 0 \\ 0 & \dots & 0 & δ^{- n} \end{array}] Q [\begin{array}{c} δ^{1} & 0 & \dots & 0 \\ 0 & δ^{2} & ⋱ & ⋮ \\ ⋮ & ⋱ & ⋱ & 0 \\ 0 & \dots & 0 & δ^{n} \end{array}] U^{*} \\ = S Q S^{- 1} \end{aligned}

Where $$ S = U\begin{bmatrix}
\delta^{-1} & 0 & \dots & 0 \
0 & \delta^{-2} & \ddots & \vdots \
\vdots & \ddots & \ddots & 0 \
0 & \dots & 0 & \delta^{-n}
\end{bmatrix}$$
Note that $$Q = \begin{bmatrix}
t_{11} & 0 & \dots & 0 \
0 & t_{22} & \ddots & \vdots \
\vdots & \ddots & \ddots & 0 \
0 & \dots & 0 & t_{nn}
\end{bmatrix} + \cal{O}(\delta)$$
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.

Proof

Let $A \in M_{n} = U T U^{*}$ by Schur. Then there exists a diagonal matrix $D$ such that $| | D | |_{F} < ϵ$ and $T + D$ has distinct diagonals. Then

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

and this has all eigenvalues distinct by construction, and thus is diagonalizable! So let $E = U D U^{*}$ . 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 $A A^{*} = A^{*} A$

Example

diagonal matrices
hermitian matrices (duh)
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)

Suppose $T$ is normal upper triangular. Then $T T^{*} = T^{*} T$ since it is normal.

\begin{aligned} T T^{*} & = T^{*} T \\ [\begin{array}{c} - r_{1} - \\ - r_{2} - \\ ⋮ \\ - r_{n} - \end{array}] [\begin{array}{c} | & | & \dots & | \\ r_{1}^{*} & r_{2}^{*} & \dots & r_{n}^{*} \\ | & | & \dots & | \end{array}] & = [\begin{array}{c} | & | & \dots & | \\ c_{1} & c_{2} & \dots & c_{n} \\ | & | & \dots & | \end{array}] [\begin{array}{c} - c_{1}^{*} - \\ - c_{2}^{*} - \\ ⋮ \\ - c_{n}^{*} - \end{array}] \end{aligned}

Where $t_{i}$ are increasing in "nonzero length"

Then the $i$ th diagonal of $T^{*} T$ , call it $[T^{*} T]_{i i} = c_{i}^{*} c_{i} = | | c_{i} | |^{2}$ where $t_{i}$ is the $i$ the COLUMN of $T$

And the $i$ th diagonal of $T T^{*}$ , call it $[T T^{*}]_{i i} = r_{i} r_{i}^{*} = | | r_{i} | |^{2}$ 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)