Lecture 28

[[lecture-data]]

2024-11-06

5. Chapter 5

last time we did

Theorem

Suppose $| | \cdot | |$ is a matrix norm on $M_{n}$ . Then for all $A \in M_{n}$ , $ρ (A) \leq | | A | |$

(see matrix norms are bounded below by the spectral radius)

Lemma

Suppose $| | \cdot | |$ is a matrix norm on $M_{n}$ . Let $S \in M_{n}$ be invertible. Then $| | \cdot | |_{S}$ defined by (for all $A \in M_{n}$ ) is

| | A | |_{S} = | | S^{- 1} A S | |

And this is also a matrix norm on $M_{n}$ .

(demonstration)

For all $A, B \in M_{n}$ and $α \in C^{n}$ we have

$A \neq 0 ⟹ S^{- 1} A S \neq 0 ⟹ | | A | |_{S} = | | S^{- 1} A S | | \geq 0$
$| | α A | |_{S} = | | α S^{- 1} A S | | = | α | \cdot | | S^{- 1} A S | | = | α | \cdot | | A | |_{S}$
$| | A + B | |_{S} = | | S^{- 1} (A + B) S | | = | | S^{- 1} A S + S^{- 1} B S | | \leq | | S^{- 1} A S | | + | | S^{- 1} B S | | = | | A | |_{S} + | | B | |_{S}$
$| | A B | |_{S} = | | S^{- 1} (A B) S | | = | | S^{- 1} A S S^{- 1} B S | | \leq | | S^{- 1} A S | | \cdot | | S^{- 1} B S | | = | | A | |_{S} \cdot | | B | |_{S}$

(see invertible matrix norms)

Theorem

Let $A \in M_{n}$ be fixed. For all $ϵ > 0$ , there exists a matrix norm $| | \cdot | |$ on $M_{n}$ such that $| | A | | \leq ρ (A) + ϵ$ .

( in particular, this implies that $inf_{matrix norms | | \cdot | |} | | A | | = ρ (A)$ )

Proof

Let $A = S J S^{- 1}$ be a Jordan decomposition. For any $ϵ > 0$ , define $D$ diagonal such that $D_{i i} = ϵ^{i}$ .

Note that $D^{- 1} J D = \hat{J}$ where $\hat{J}$ has the same diagonals as $J$ , but each of the 1s turn into $ϵ$ . And this has $| | \cdot | |_{1, 1} \leq ρ (A) + ϵ$ .

So define invertible matrix norm ${| | A | |_{1, 1}}_{S D} = | | D^{- 1} S^{- 1} A S D | |_{1, 1} = | | D^{- 1} J D | |_{1, 1} \leq ρ (A) + ϵ$

(see we can find a matrix norm that evaluates arbitrarily close to the spectral radius)

Theorem

For any $A \in M_{n}$ , $lim_{k \to \infty} A^{k} = 0$ if and only if $ρ (A) < 1$

Proof

( $⟹$ ) if $ρ (A) < 1$ , then by since we can find a matrix norm that evaluates arbitrarily close to the spectral radius we have some norm such that $| | A | | < 1$ . So we have
$| | A^{k} - 0 | | = | | A | |^{k} \to 0$ as $k \to \infty$ by norm equivalence!

$(⟸)$ Conversely, suppose $A^{k} \to 0$ as $k \to \infty$ . Then any eigenvalue and eigenvector pair, say $λ, x \neq 0$ . Then $A^{k} x = λ^{k} x \to 0$ as $k \to \infty$ . Since $x \neq 0$ , we must have $| λ | < 1$ . Thus $ρ (A) < 1$ .

(see matrices are nilpotent in the limit when spectral radius is less than 1)

Theorem

For any $A \in M_{n}$ and any matrix norm on $M_{n}$ , we have $lim_{k \to \infty} | | A^{k} | |^{1 / k} = ρ (A)$ .

Proof

$ρ (A)^{k} = ρ (A^{k}) \leq | | A^{k} | | ⟹ ρ (A) \leq | | A^{k} | |^{1 / k}$
since matrix norms are bounded below by the spectral radius, then we simply take the $k$ th root.

Let $ϵ > 0$ be given. Then $ρ (\frac{1}{ρ (A) + ϵ} A) < 1$ . Thus since matrices are nilpotent in the limit when spectral radius is less than 1, we see that ${(\frac{1}{ρ (A) + ϵ} A)}^{k} \to 0$ . ie, there exists some $M$ such that for all $k \geq M$ we have that $| | {(\frac{1}{ρ (A) + ϵ} A)}^{k} | | < 1$ .

ie. $| | A^{k} | | < (ρ (A) + ϵ)^{k}$ . ie $| | A^{k} | |^{k} \leq ρ (A) + ϵ$

Since $ϵ$ is arbitrary, the result follows.

(see the limit of the powered norm is the spectral radius)

Theorem

Suppose $V, | | \cdot | |$ is a NLS. It is complete if and only if all absolutely convergent sequences converge.

ie ${x^{(i)}}_{i = 1}^{\infty} \subseteq V, \sum_{i = 1}^{\infty} | | x^{(i)} | | < \infty ⟹ \sum_{i = 1}^{\infty} x^{(i)}$ converges to a vector in $V$ .

(without proof)

(see normed linear spaces are complete if and only if all absolutely convergent series converge)

Corollary

For all $A \in M_{n}$ define $e^{A} = \sum_{i = 0}^{\infty} \frac{1}{i!} A^{i}$ . This is well-defined. (see matrix exponential)

Proof

Let $| | \cdot | |$ be any matrix norm. Then

\sum_{i=0}^\infty \left\lvert \left\lvert \frac{1}{i!} A^i \right\rvert \right\rvert \leq \sum_{i=0}^\infty \frac{1}{i!}\lvert \lvert A \rvert \rvert {#i} = e^{\lvert \lvert A \rvert \rvert }

by homogeneity and submultiplicity

When $i = 0$ : can I say that $| | A^{0} | | \leq | | A | |^{0} = 1$ ? Well, no. But it still converges. (as long as $n$ is finite). I can also choose an induced matrix norm and using norm equivalence it is ok :)

(see the matrix exponential is well-defined)