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 $A\in M_n$, $\rho (A)\le ||A||$

(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 $||A|{|}_S=||S^{-1}AS||$ And this is also a matrix norm on $M_n$.

(demonstration) $A,B\in M_n$ and $\alpha \in {\mathbb{C}}^n$ we have

For all

1. $A\ne 0⟹S^{-1}AS\ne 0⟹||A|{|}_S=||S^{-1}AS||\ge 0$
2. $||\alpha A|{|}_S=||\alpha S^{-1}AS||=|\alpha |\cdot ||S^{-1}AS||=|\alpha |\cdot ||A|{|}_S$
3. $||A+B|{|}_S=||S^{-1}(A+B)S||=||S^{-1}AS+S^{-1}BS||\le ||S^{-1}AS||+||S^{-1}BS||=||A|{|}_S+||B|{|}_S$
4. $||AB|{|}_S=||S^{-1}(AB)S||=||S^{-1}AS{S}^{-1}BS||\le ||S^{-1}AS||\cdot ||S^{-1}BS||=||A|{|}_S\cdot ||B|{|}_S$

(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 $||A||\le \rho (A)+ϵ$.

( in particular, this implies that ${\inf }_{\text{matrix norms }||\cdot ||}||A||=\rho (A)$)

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

Let

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

So define invertible matrix norm ${||A|{|}_{1,1}}_{SD}=||D^{-1}{S}^{-1}ASD|{|}_{1,1}=||D^{-1}JD|{|}_{1,1}\le \rho (A)+ϵ$

(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 $\rho (A)<1$

$⟹$) if $\rho (A)<1$, 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 $\lambda ,x\ne 0$. Then $A^{k}x={\lambda }^{k}x\to 0$ as $k\to \infty$. Since $x\ne 0$, we must have $|\lambda |<1$. Thus $\rho (A)<1$.

(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 ${\lim }_{k\to \infty }||A^k|{|}^{1/k}=\rho (A)$ .

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

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

ie. $||A^k||<(\rho (A)+ϵ{)}^k$. ie $||A^k|{|}^k\le \rho (A)+ϵ$

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)

$||\cdot ||$ be any matrix norm. Then ${\sum }_{i=0}^{\infty }\left|\left|\frac{1}{i!}{A}^i\right|\right|\le {\sum }_{i=0}^{\infty }\frac{1}{i!}||A|{|}^i=e^{||A||}$

Let

by homogeneity and submultiplicity

When $i=0$ : can I say that $||A^0||\le ||A|{|}^0=1$ ? 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)