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Matrix Norms, and proving their characteristics

  1. Sep 14, 2005 #1
    I have a couple of questions on this matter...

    QUESTION 1.)
    I am to show a certain Lemma, that the definition of a matrix norm
    ||A|| = max(||x||<=1) ||Ax||
    obides with the definitions of a standard vector norm

    a.) ||A||>=0 for all A and ||A||=0 <=> A=0
    b.) ||tA|| = |t| ||A|| for any real number t
    c.) ||A+B|| <=||A||+||B|| for all A,B

    Ive shown b and c, although b is confusing me somewhat:
    b.) ||tA|| must satisfy the following
    ||tA|| = max(||x||<=1) ||tAx||
    = |t| max(||x||<=1) ||Ax||
    = |t| ||A||
    I thinkthis proves it, but still im not quite sure what would happen if t is negative from the beginning.
    Could someone tell me if ||tA|| is then equal to |t| ||A|| and why?

    The really important part of of question 1 is getting a hint of how to show a.), since i have no idea how to...

    Question 2
    Show that
    ||A|| = max(||x||!=0) ||Ax||/||x||

    From a definition in the book, I know that ||Ax||<=||A|| ||x||
    then i must have that
    ||A|| <= max(||x||!=0) ||A|| ||x||/||x||
    giving me that
    ||A|| <= max(||x||!=0) ||A||
    which proves it.
    Is this proof ok?

    Question 3
    The definition i use above is proven in the book, the first line of this proof says
    ||Ax|| = ||A (x/||x||) x||
    why is this so?
    Is it cause x/||x|| is a unit vector, and thus makes no difference when multiplied on A?

    Anyone who can help me, will receive an appropriate e-hug :-)

    Cheers
    Daniel
     
  2. jcsd
  3. Sep 14, 2005 #2

    AKG

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    Homework Helper

    Question 1:

    You know Ax is a vector right, and you know that the norm of a vector is greater than or equal to 0, with equality iff the vector is the zero-vector, right? So this is enough to show that ||A|| > 0. It remains to show that ||A|| = 0 iff A = 0.

    Assume ||A|| = 0
    Then max||x|| < 1||Ax|| = 0
    And then ||Ax|| = 0 for all x such that ||x|| < 1
    From here, you need to prove that A = 0, and the above should be enough to make that easy.

    On the other hand, assume A is 0 and prove that ||A|| = 0. That should be easy.

    Question 2.

    Your proof is not right. For one, you end up proving an inequality, and you're asked to prove an equality. But the overall approach is wrong. I assume that you are still working with the norm from the previous question, so:

    ||A|| = max||x|| < 1||Ax||

    You want to prove that ||A|| = max||x|| != 0||Ax||/||x||, so you want to prove that:

    max||x|| < 1||Ax|| = max||x|| != 0||Ax||/||x||

    Let m = ||x||, and define v as x/m, so x = mv

    max||x|| < 1||Ax||
    = maxm < 1m||Av||
    = m x maxm < 1||Av||

    max||x|| != 0||Ax||/||x||
    = maxm != 0m||Av||/m
    = maxm != 0||Av||
    = 1 x maxm != 0||Av||

    All you have to do is show:

    m x maxm < 1||Av|| = 1 x maxm != 0||Av||

    Question 3.

    Doesn't make any sense. On the right side of the equation, you seem to be "multiplying" two vectors because you have:

    (x/||x||) x

    Do you mean the dot product of (x/||x||) with x, or the cross product of those two vectors or, as I suspect, was this a typo?
     
  4. Sep 14, 2005 #3
    typo

    It was a typo :-)
    it should have been
    (x/||x||)||x||
    should make more sense now :-D
    Thanks for the answers!

    I have a very basic question also, I would like the following definition to be described in as plain words as possible, I want to be absolutely sure what it means...
    ||A|| = max(||x||<=1) ||Ax||
    I was trying to explain to a mate of mine, and found out I couldnt answer his following questions with certainty
    What i told him, was that u multiply the mxn matrix with a n-vector, and get a new m vector, then take the max value in this.
    He then asked:
    Why should we do the Ax multiplication? Is x to be chosen somehow, what is the idea of this?

    ||x||<=1 - this is based on ||x|| being the 2 norm right? the length of the vector?
     
  5. Sep 14, 2005 #4

    AKG

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    Certainly you can see that (x/||x||)||x|| = x.

    ||x|| is the length of x, yes. If x is a 2-vector, then the set of x satisfying ||x|| < 1 is just the unit circle, so you are basically seeing how A transforms the unit circle, then finding the point in the image of the circle under the transformation A that is farthest from the origin.
     
  6. Sep 14, 2005 #5

    Hurkyl

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    If it helps, one of the more important properties of an operator norm is that it satisfies:

    ||Tx|| <= ||T|| ||x||

    The definition of ||T|| is chosen so that ||T|| is large enough for this inequality to be valid, but no larger than necessary.

    In fact, I've seen the definition written in three different ways (I'm not sure if I've seen the one you wrote or not): (click the image to see how to make them)

    [tex]
    ||T|| := \sup_{x \neq 0} \frac{||Tx||}{||x||}
    [/tex]

    [tex]
    ||T|| := \sup_{||x|| \leq 1} ||Tx||
    [/tex]

    [tex]
    ||T|| := \sup_{||x|| = 1} ||Tx||
    [/tex]

    Exercise 1: prove that all three are equivalent.
    Exercise 2: explain why these definitions capture the notion I gave above for why we choose the operator norm the way we do.
     
  7. Sep 15, 2005 #6
    About the (x/||x||)||x|| being equal to x...
    Doh! Of course i can see that.... sorry :-S
     
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