# Linear algebra+ linear operators

## Homework Statement

In $$R^{3}$$ ||x||= $$a_{1}$$*|$$x_{1}$$|+ $$a_{2}$$*|$$x_{2}$$|+ $$a_{3}$$*|$$x_{3}$$|. where $$a_{i}$$>0

What is ||A||(indused norm = sup||Ax|| as ||x||=1). (Suppose we know the coeffisients of the matrix/operator A)??

## The Attempt at a Solution

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Dick
Homework Helper
If A is given by a matrix, it's the square root of the largest eigenvalue of AA* (*=hermitian conjugate or transpose in the real case).

Thanks, but do you know how to proove it?

By the way, do you want to say that it does nod depend on a_i??

(i just need the real case)

Dick
Homework Helper
Thanks, but do you know how to proove it?
|Av|^2=v^(T)*A^T*A*v. A^T*A is a symmetric matrix, so it has a complete orthogonal set of eigenvectors. Can you show the eigenvalues are nonnegative? Now expand a general v in terms of those eigenvectors and go from there...

Dick
Homework Helper

## Homework Statement

In $$R^{3}$$ ||x||= $$a_{1}$$*|$$x_{1}$$|+ $$a_{2}$$*|$$x_{2}$$|+ $$a_{3}$$*|$$x_{3}$$|. where $$a_{i}$$>0

What is ||A||(indused norm = sup||Ax|| as ||x||=1). (Suppose we know the coeffisients of the matrix/operator A)??

## The Attempt at a Solution

Hey, that's not a matrix operator at all, is it? You just want to maximize the inner product of (a1,a2,a3) and x. That's much easier. Think Cauchy-Schwarz.

Unfortunately I do not understand why

|Av|^2=v^(T)*A^T*A*v is true.. Why does it not depend on a_i??
2. I don't think I know how to proove that the eigenvalues are nonnegative or where to go from there..

I think that it is a matrix operator..

Dick
Homework Helper
I think that it is a matrix operator..
What you wrote is the dot product of (a1,a2,a3) with (x1,x2,x3). That's a matrix operator only in the sense it's a 1x3 matrix. You want to maximize it. Will you look up the Cauchy-Schwarz inequality. Please?

The thing is that the dot product does not have the modules..
Even if I do maximize it.. I do not see where to go from there..

Cauchy Shwartz: (x,y)<=||x||*||y||

Dick
Homework Helper
Cauchy Shwartz: (x,y)<=||x||*||y||
Yes, and add "with equality holding only when y is a multiple of x". That's the part you want.

HallsofIvy
Homework Helper
Hey, that's not a matrix operator at all, is it? You just want to maximize the inner product of (a1,a2,a3) and x. That's much easier. Think Cauchy-Schwarz.
What you wrote is the dot product of (a1,a2,a3) with (x1,x2,x3). That's a matrix operator only in the sense it's a 1x3 matrix. You want to maximize it. Will you look up the Cauchy-Schwarz inequality. Please?
Dick, I think you have misinterpreted the question. What he wrote was the definition of norm (not an inner product between (a1, a2, a3) and (x1, x2, x3)- a1, a2, and a3 are fixed for all x) THEN he asked
"What is ||A||(indused norm = sup||Ax|| as ||x||=1). (Suppose we know the coeffisients of the matrix/operator A)??"

In other words, what norm does that defined norm on the vector space induce on the linear operators.

The definition of "induced norm" of A is ||A||= lub ||Ax|| for all x with norm 1.