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Proof for induced matrix norm

  1. Sep 17, 2015 #1

    FOIWATER

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    Hi, I found a statement without a proof. It seems simple enough, but I am having trouble proving it because I am not positive about induced matrix norms. The statement is that $$||A^k|| \leq||A||^{k}$$ for some matrix A and positive integer k. I have found that the norm of a matrix is the supremum of the norm of Ax over the norm of x, but I do not know to which norms these refer?

    I am assuming euclidean norms. Since Ax gives us back a vector and x is itself a vector.

    So I have that:
    $$||A^k|| = \sup_{||x||=1}(||A^{k}x|| : ||x||=1)$$
    and
    $$||A||^{k} = \sup_{||x||=1}(||Ax|| : ||x||=1)^{k}$$

    Not sure what to do with them, though, any hints appreciated. I was thinking triangle inequality.. but I didn't really get anything from it. And I do not know if the triangle inequality applies to this matrix induced norm (although I think it applies to any operation that qualifies as a norm, since it defines norms).
     
  2. jcsd
  3. Sep 17, 2015 #2

    RUber

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    What if the x that gives the sup for A^k is not the same as the x that is the sup for A? A acting on any vector will always be less than or equal to its sup.
     
  4. Sep 17, 2015 #3

    RUber

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    To be more clear...
    Let ##X## be your maximizer for ##\|Ax\|##. Then
    ##\| A^k \| = \| A A A ... A \| = \sup_{\|x\|=1} \left( \| A A A ... Ax \| \right)##
    Let ##x_0 ## be any x such that ##\|x_0 \| = 1##.
    Then you know that ##\|Ax_0\| \leq \|AX\|=\|A\|. ##
    Let ##x_1 = Ax_0##. By similar argument, ##\|A\frac{x_1}{\|x_1\|}\|=\frac{\|Ax_1\|}{\|x_1\|} \leq \|AX\|\implies \|Ax_1\|\leq \|AX\|\|x_1\| \leq \|AX\|\|A\|=\|A\|^2 .##
    And by induction, you can clearly see that no matter what, you won't be able to break the inequality.
     
  5. Sep 17, 2015 #4

    FOIWATER

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    ∥AX∥=∥A∥ I am not sure of this statement? A times X should give a vector, not a matrix A?
     
  6. Sep 17, 2015 #5

    FOIWATER

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    This actually makes sense now thanks a lot!
     
    Last edited: Sep 17, 2015
  7. Sep 17, 2015 #6

    RUber

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    Right, but I defined ||AX|| to be ||A|| earlier.
    Of course, that is assuming that there is a maximum. You might have to keep the definition of ||A|| as the supremum to be proper...but the logic is the same.
     
  8. Sep 17, 2015 #7

    FOIWATER

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    That's good insight I hope to be able to work my way through problems like that some day
     
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