Lecture 27

[[lecture-data]]

2024-11-04

Readings

• a

5. Chapter 5

Corollary of Hahn-Banach

Let $V,||\cdot ||$ be a NLS over $K$. If $x\in V$, then there exists $f\in V^{\ast }$ such that $||f|{|}_{V^{\ast }}=1$ and $f(x)=||x|{|}_V$

(see Hahn-Banach theorem)

Theorem

Let $||\cdot ||$ be a norm on $K^n$. Then $(||\cdot |{|}^D{)}^D=||\cdot ||$.

$n\in K^n$. Then $(||x|{|}^D{)}^D={\max }_{y\in K^n,||y|{|}^D=1}|y^{\ast }x|$ by definition of the dual norm. But we have

Let

\max_{y \in K^n, \lvert \lvert y \rvert \rvert^D=1} \lvert y^*x \rvert & \leq \max_{y \in K^n, \lvert \lvert y \rvert \rvert^D= 1 } \lvert \lvert x \rvert \rvert \cdot \lvert \lvert y \rvert \rvert ^D \\ &= \lvert \lvert x \rvert \rvert \\ &\text{ by H-B }, \exists y \in K^n, \lvert \lvert y \rvert \rvert ^D = 1, y^*x = \lvert \lvert x \rvert \rvert \\ \implies \max_{y \in K^n, \lvert \lvert y \rvert \rvert^D=1} \lvert y^*x \rvert &= \max_{y \in K^n, \lvert \lvert y \rvert \rvert^D= 1 } \lvert \lvert x \rvert \rvert \cdot \lvert \lvert y \rvert \rvert ^D \end{aligned}$$ (by [[Concept Wiki/Hahn-Banach theorem]] above)

(see the dual of the dual norm is the original norm)

Theorem

Let $||\cdot ||$ be a norm on $K^n$. For all $A\in M_n(K)$, $||A|{|}_{||\cdot ||,||\cdot ||}=||A^{\ast }|{|}_{||\cdot |{|}^D,||\cdot |{|}^D}$

matrix 1 norm is max column sum and matrix infinity norm is max row sum. But we can also see it via the dual of the dual norm is the original norm.

We can see this in

Proof

By definition, we have

\lvert \lvert A^* \rvert \rvert_{\lvert \lvert \cdot \rvert \rvert^D, \lvert \lvert \cdot \rvert \rvert^D} &= \max_{x \in \lvert \lvert x \rvert \rvert^D = 1} \lvert \lvert A^*x \rvert \rvert^D \\ &= \max_{x : \lvert \lvert x \rvert \rvert ^D = 1} [\max_{y: \lvert \lvert y \rvert \rvert =1} \lvert (A^*x)^*y \rvert ] \\ &= \max_{y: \lvert \lvert y \rvert \rvert =1} \max_{x : \lvert \lvert x \rvert \rvert ^D = 1} \lvert (Ay)^*x \rvert \\ &= \max_{y: \lvert \lvert y \rvert \rvert =1} (\lvert \lvert Ay \rvert \rvert ^D)^D \;\;\;(*) \\ & \max_{y : \lvert \lvert y \rvert \rvert =1} \lvert \lvert Ay \rvert \rvert \\ &= \lvert \lvert A \rvert \rvert _{\lvert \lvert \cdot \rvert \rvert, \lvert \lvert \cdot \rvert \rvert } \end{aligned}$$ - $(*)$ we have $Ay$ is a linear functional. But it is in the dual norm (for the dual norm!) (see [[Concept Wiki/the dual of the dual norm is the original norm]])

Matrix Norm

A norm $||\cdot ||$ on $M_n(K)$ over $K$ is a matrix norm if for all $A,B\in M_n$ we have $||AB||\le ||A||\cdot ||B||$ (this condition is called submultiplicity)

If, in particular, $||\cdot |{|}^{\prime }$ is any norm on $K^n$ and matrices in $M_n$ are $K^n,||\cdot |{|}^{\prime }\to K^n,||\cdot |{|}^{\prime }$, then the operator norm induced by that vector norm $||\cdot |{|}_{||\cdot |{|}^{\prime },||\cdot |{|}^{\prime }}$ is an induced matrix norm. (We have already shown this)

$||A|{|}_{2,2}$, $||A|{|}_{1,1},||A|{|}_{\infty ,\infty }$

Where did we already show this? We showed that $T:U\to V,S:V\to W$ satisfies $||S\circ T||\le ||S||\cdot ||T||$

(see matrix norm)

Exercise

1. $||\cdot |{|}_1$ is a matrix norm ie NTS $||AB|{|}_1\le ||A|{|}_1\cdot ||B|{|}_1$

2. $||\cdot |{|}_F$ is a matrix norm (frobenius norm)

3. $||\cdot |{|}_{\infty }$ is NOT a matrix norm

counterexample for (3) $A=B=$ matrix of all 1s. Then $AB$ is the matrix of all 2s.

Consider

Relationship between norms on square matrices

induced matrix norms $\subset$ matrix norms $\subset$ Norms on $M_n(K)$

• If we can prove convergence for general norm on matrices, then we also have convergence for matrix norms and induced matrix norms

Theorem

Suppose $||\cdot ||$ is a matrix norm on $M_n$. Then for all $A\in M_n$, we have $\rho (A)\le ||A||$.

$||A||={\max }_x\frac{||A||}{||x||}={\lambda }_{\max }$ (we simply choose some norm 1 eigenvector of $A$ with the maximum eigenvalue).

If the norm is an induced norm, then the result is trivial:

For any matrix norm, we let $x$ be an eigenvector associated with $\lambda$ an eigenvalue of maximum modulus. Set $B$ to be the $n\times n$ matrix with $x$ as each column. Then $||AB||=||Ax|Ax|\dots |Ax||=||\lambda B||\le ||A||\cdot ||B||$ Since $x\ne 0$ this implies that $B\ne 0⟹||B||\ne 0$. Thus we have that $\rho (A)=|\lambda |\le ||A||$

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

Preview: we will see next time that ${\inf }_{||\cdot ||\text{ matrix norm}}||A||=\rho (A)$