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Linear algebra+ linear operators

  • #1

Homework Statement



In [tex]R^{3}[/tex] ||x||= [tex]a_{1}[/tex]*|[tex]x_{1}[/tex]|+ [tex]a_{2}[/tex]*|[tex]x_{2}[/tex]|+ [tex]a_{3}[/tex]*|[tex]x_{3}[/tex]|. where [tex]a_{i}[/tex]>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


Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
Dick
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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
Thanks, but do you know how to proove it?
 
  • #4
By the way, do you want to say that it does nod depend on a_i??
 
  • #5
(i just need the real case)
 
  • #6
Dick
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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...
 
  • #7
Dick
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Homework Statement



In [tex]R^{3}[/tex] ||x||= [tex]a_{1}[/tex]*|[tex]x_{1}[/tex]|+ [tex]a_{2}[/tex]*|[tex]x_{2}[/tex]|+ [tex]a_{3}[/tex]*|[tex]x_{3}[/tex]|. where [tex]a_{i}[/tex]>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


Homework Statement





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.
 
  • #8
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
I think that it is a matrix operator..
 
  • #10
Dick
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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?
 
  • #11
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
Cauchy Shwartz: (x,y)<=||x||*||y||
 
  • #13
Dick
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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.
 
  • #14
HallsofIvy
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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.
 

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