# Linear algebra+ linear operators

1. May 22, 2008

### Mechmathian

1. The problem statement, all variables and given/known data

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)??

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. May 22, 2008

### 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).

3. May 22, 2008

### Mechmathian

Thanks, but do you know how to proove it?

4. May 22, 2008

### Mechmathian

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

5. May 22, 2008

### Mechmathian

(i just need the real case)

6. May 22, 2008

### Dick

|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...

7. May 22, 2008

### Dick

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.

8. May 22, 2008

### 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..

9. May 22, 2008

### Mechmathian

I think that it is a matrix operator..

10. May 22, 2008

### Dick

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?

11. May 23, 2008

### 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..

12. May 23, 2008

### Mechmathian

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

13. May 23, 2008

### Dick

Yes, and add "with equality holding only when y is a multiple of x". That's the part you want.

14. May 24, 2008

### HallsofIvy

Staff Emeritus
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.