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Matrix Norm

  1. Oct 11, 2006 #1
    "Let [tex]Mat_n[/tex] denote the space of [tex]n\times n[/tex] matrices. For [tex]A\in Mat_n[/tex], define the norms [tex]||A||_1[/tex] as follows:

    [tex]||A||_1=\sup_{0\neq x\in\mathbb{R}^n}\frac{||Ax||}{||x||}[/tex],

    where ||x|| is the usual Euclidean norm.

    Prove that this norm is really a norm (triangle ineq, etc)"

    I don't know how to even prove that the supremum exists.
     
  2. jcsd
  3. Oct 11, 2006 #2

    Hurkyl

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    Supremums always exist. That's why the extended real numbers are so nice! :smile:

    I don't see why you would need to prove the supremum is finite, but you could always start by trying to solve simplified problems. (e.g. pick nice matrices, or restrict the supremum to a nice set of vectors) Or, you could try invoking interesting properties about the map x->Ax.
     
  4. Oct 11, 2006 #3

    StatusX

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    Would it be a norm if instead of looking at the sup over all x, you looked at some fixed x?
     
  5. Oct 11, 2006 #4

    AKG

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    If v is any nonzero vector, then v' = v/|v| is a parallel vector to v of unit norm. Observe that |Av|/|v| = |A(|v|v')|/|v| = ||v|Av'|/|v| (by linearity of A) = |v|(|Av'|/|v|) = |Av'| = |Av'|/1 = |Av'|/|v'|. So the above definition for the norm of A is equivalent to the supremum of |Ax| over all x of unit norm. The set of x of unit norm is the unit (n-1)-sphere, which is compact in the usual topology of Rn, and you can prove that the function which maps x in the unit sphere to |Ax| is continuous by showing that it is the composition of two functions, the Euclidean norm function and the function A, and that each of these functions are continuous. Since this composition is a continuous function from a compact set to R, the extreme value theorem tells you that it obtains a maximum, which in turn tells you that the supremum exists.

    Proving that this function |.|1 really does satisfy the norm axioms is easy, especially after realizing that:

    [tex]||A||_1 = \sup _{x\in \mathbb{R}^n,\, ||x|| = 1}||Ax||[/tex]
     
    Last edited: Oct 11, 2006
  6. Oct 11, 2006 #5

    AKG

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    No, it wouldn't be a norm if you looked at some fixed x. The map on the plane A(x,y) = x will map (0,1) to 0, so if we define the "norm" of A as |A(0,1)|, then we'd get a non-zero matrix with zero "norm", contradicting the norm axioms.

    Since a norm, by definition, has the reals as its codomain (and not the extended reals), you do have to verify that the supremum always exists.
     
  7. Oct 11, 2006 #6

    StatusX

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    Sorry, I meant would it satisfy the triangle inequality. That's the part I was trying to show how to prove.
     
  8. Oct 11, 2006 #7
    Ok, I got it. Thanks.
     
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