A counterintuitive bound on the frobenius norm

Mathematics — Alex @ 8:01 pm

In the process of investigating the \|\cdot\|_{\infty \rightarrow 2} approximation error, I came across an interesting question. From numerical examples, it seems that \|X\|_F \leq \sum_j \|X_j\|_p , where  1 \leq p \leq \infty but this baffles my intuition (which I’ll admit isn’t very sharp), especially for p=\infty. Here \|\cdot\|_F is the Frobenius norm, and X_j is the j-th column of X.

Hopefully I’ll have time to look at this later. Besides being counterinuitive, it seems potentially useful: consider a vector of composite length \ell = m n , you could partition this vector into m blocks of length n then apply this inequality to get a family of bounds on the euclidean length of the vector. Exploit the fact that \|X\|_F = \|X^T\|_F and you could derive some very weird bounds indeed.

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4 Comments »

  1. For the operator norm, the $p=\infty$ case is the Schur Lemma.
    But I doubt that it is true for Frobenius norm just the way you state it.

    Comment by Torus — 7/22/2008 @ 11:25 am
  2. It isn’t; my officemate just came up with a counterexample: \left( \begin{matrix} 1 & 2\\ 1 & 2 \end{matrix} \right)

    Comment by Alex — 7/22/2008 @ 11:34 am
  3. Your norm is very asymmetric, so you could try to symmetrize it to throw counterexamples like that. But I still suspect that the inequality for the symmetrized version will not hold.

    Comment by Torus — 7/22/2008 @ 12:15 pm
  4. As in, add in the sum of the p-norms of the rows? I doubt that’d work also.

    Comment by Alex — 7/22/2008 @ 1:51 pm

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