1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Induced weighted matrix norm

  1. Feb 20, 2016 #1
    1. The problem statement, all variables and given/known data
    The weighted vector norm is defined as
    ##||x||_W = ||Wx||##.
    W is an invertible matrix.

    The induced weighted matrix norm is induced by the above vector norm and is written as:
    ##||A||_W = sup_{x\neq 0} \frac{||Ax||_W}{||x||_W}##
    A is a matrix.

    Need to show ##||A||_W = ||WAW^{-1}||##


    2. Relevant equations


    3. The attempt at a solution
    ##||A||_W = sup_{x\neq 0} \frac{||Ax||_W}{||x||_W} = sup_{x\neq 0} \frac{||WAx||}{||Wx||} = sup_{x\neq 0} \frac{||WAW^{-1}Wx||}{||Wx||}##

    For an induced norm we know that:
    ##\frac{||WAW^{-1}Wx||}{||Wx||} \leq \frac{||WAW^{-1}||||Wx||}{||Wx||} = ||WAW^{-1}||##


    Here is where I am lost. I already have gotten the expression into the form desired, but I do not know how to make the connection between ##sup_{x\neq0}## and the ##\leq##. My thought is that we are taking the supremum of ##\frac{||Ax||_W}{||x||_W}##, so it's maximum possible value is ##||WAW^{-1}||##
    and thus
    ##||A||_W=||WAW^{-1}||##
    Is this the right logic?
     
  2. jcsd
  3. Feb 20, 2016 #2

    fresh_42

    Staff: Mentor

    The inequality holds for all ##x \neq 0## and therefore for the supremum, too.
     
  4. Feb 21, 2016 #3
    Yes I understand this, but I don't think this answers my question? Or maybe I do not actually understand?
     
  5. Feb 21, 2016 #4

    fresh_42

    Staff: Mentor

    Putting it together you already have:

    ##||A||_W = sup_{x\neq 0} \frac{||Ax||_W}{||x||_W} = sup_{x\neq 0} \frac{||WAx||}{||Wx||} = sup_{x\neq 0} \frac{||WAW^{-1}Wx||}{||Wx||} = sup_{x\neq 0} \frac{||WAW^{-1}Wx||}{||Wx||}##

    ## \leq sup_{x\neq 0} \frac{||WAW^{-1}||||Wx||}{||Wx||} = sup_{x\neq 0} \frac{||WAW^{-1}||||Wx||}{||Wx||} = sup_{x\neq 0} ||WAW^{-1}|| = ||WAW^{-1}||##

    At least I cannot see why this shouldn't hold.
     
  6. Feb 22, 2016 #5
    Yes. I don't see anything wrong with those steps either, but I am not sure how to eventually get to the answer
    ##||A||_W = ||WAW^{-1}||##. Right now we have it in the form ##||A||_W \leq ||WAW^{-1}||##
     
  7. Feb 22, 2016 #6

    fresh_42

    Staff: Mentor

    Right. I've forgotten the other direction. Give me some time to think about it.
     
  8. Feb 22, 2016 #7
    Thanks. I think it is some property of the "supremum" that I may be missing.
     
  9. Feb 22, 2016 #8

    fresh_42

    Staff: Mentor

    If we have ##||A|| = sup_{x\neq 0} \frac{||Ax||}{||x||}## which is true for the matrix norm induced by the vector norm, then it's easy (I think).

    ##||A||_W = sup_{x\neq 0} \frac{||Ax||_W}{||x||_W} = sup_{x\neq 0} \frac{||WAx||}{||Wx||} = sup_{y\neq 0} \frac{||WAW^{-1}y||}{||y||} = ||WAW^{-1}||##
     
  10. Feb 22, 2016 #9
    Hmm, I do not understand the step
    ##sup_{y\neq0}\frac{||WAW^{-1}y||}{||y||} = ||WAW^{-1}||##

    Could you explain the equality?
     
  11. Feb 22, 2016 #10

    fresh_42

    Staff: Mentor

    The same as in the assumption that ##||A|| = sup_{x\neq 0}{\frac{||Ax||}{||x||}}## only with ##||WAW^{-1}||## instead of ##||A||## and ##y## instead of ##x##.
    Only point is whether this is right for the matrix norm given to you but usually it holds.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Induced weighted matrix norm
  1. Matrix norm (Replies: 5)

  2. Matrix Norms (Replies: 1)

  3. Induced norms (Replies: 0)

  4. Norm of matrix (Replies: 1)

  5. Norm of a Matrix (Replies: 4)

Loading...