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Math Help - Equivalence of vector norms

  1. #1
    Member Mollier's Avatar
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    Equivalence of vector norms

    Hi.

    problem:
    For each of the following, verify the inequality and give an example of a nonzero vector or matrix for which equality is achieved. In this problem x is an m-vector and A is an m \times n matrix.

    (a) ||x||_{\infty} \leq ||x||_2,

    (b) ||x||_2 \leq \sqrt{m} ||x||_{\infty},

    (c) ||A||_{\infty} \leq \sqrt{n} ||A||_2,

    (d) ||A||_2 \leq \sqrt{m} ||A||_{\infty}.

    attempt:

    (a) I rewrite the inequality so it is easier for me to read:
    (max_i |x_i|)^2 \leq (x^2_1+\cdots +x^2_m)
    Let x_k be the largest component of x. Then (max_i|x_i|)^2=x^2_k.
    If the remaining components are zero, the inequality becomes an equality. If one or more of the remaining components are nonzero, it becomes an inequality.
    Example of vector: x=[x_k,0,\cdots,0].

    I'm sure there are better ways to verify this.

    (b) Again, I rewrite:

    x^2_1+\cdots+x^2_m \leq m(max_i|x_i|)^2.
    Let x_k be the largest component of x.
    Since mx^2_k=x^2_k+\cdots +x^2_k (m-times) and since x_i for  i\neq k are smaller than x_k, the inequality holds.
    It's an equality for all vectors x with m=1

    I am not at all sure what to do with (c) and (d).
    Any hints are appreciated.
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  2. #2
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    Is ||A||_p the regular p norm when A is regarded as a vector in \mathbb{R}^{nm}?
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  3. #3
    Member Mollier's Avatar
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    Quote Originally Posted by maddas View Post
    Is ||A||_p the regular p norm when A is regarded as a vector in \mathbb{R}^{nm}?
    ||A||_p=\left( \sum^m_{i=1} \sum^n_{j=1} |a_{ij}|^p \right)^{1/p}

    Is this what you mean?
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  4. #4
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    Yes. A matrix is a block of nm numbers, so it can be identified with a vector in R^{nm}, and its norm is the same as the norm of this vector, so this makes (a) stronger than (c) I believe. For (d), consider an n by 1 matrix all of whose entries are C>0. Then (d) asserts nmC^2 \leq mC^2 which is false if n>1, so perhaps there was a typo? If m is replaced by nm then (d) follows from (b) in the same way that (c) follows from (a).
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  5. #5
    Member Mollier's Avatar
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    If I look at the matrix A as a mn-dimensional vector and then calculate the, say 2-norm, I am calculating the Forbenius norm. ||A||_2 is the matrix norm induced by vector norms, and so is not the same as the Forbenius norm.
    Am I way off here?
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