Linear algebra+ linear operators

[Mechmathian]

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)??

Homework Equations

The Attempt at a Solution

 

[Dick]

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).

 

[Mechmathian]

Thanks, but do you know how to proove it?

 

[Mechmathian]

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

 

[Mechmathian]

(i just need the real case)

 

[Dick]

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 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)??

Homework Equations

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.

 

[Mechmathian]

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..

 

[Mechmathian]

I think that it is a matrix operator..

 

[Dick]

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?

 

[Mechmathian]

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..

 

[Mechmathian]

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

 

[Dick]

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]

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.