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The norm $\lVert A \rVert$ is different from the norm $\lVert A(x)\rVert$, right?

Just making sure that I am interpreting questions regarding matrix norm correctly.

I am asked to compare the norm of two $3 \times 3$ matrices $A$ and $B$.

The answer is that $$\lVert B \rVert \leqslant \lVert A \rVert, $$ but I notice that every entry of $A$ is bigger than or equal to every entry of $B$. So, the intuition that the norm of A would be bigger leads to the correct answer, in this case. Is this a theorem, though? That would be useful.

Thanks,

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  • $\begingroup$ It probably is a theorem. Which matrix norm is that? $\endgroup$ – muaddib Jun 9 '15 at 0:18
  • $\begingroup$ Does "bigger than or equal" entries mean in absolute value or in ordering as real (?) numbers? $\endgroup$ – hardmath Jun 9 '15 at 0:19
  • $\begingroup$ @muaddib, no matrix norm is specified - so I inevitably ask the question of which norm to use, but I am only told that all norms are equivalent in a finite dimensional vector space. (So I guess we could just stick with the usual 2-norm = square root of the largest eigenvalue of A*A = ||A||.) $\endgroup$ – user246802 Jun 9 '15 at 0:21
  • $\begingroup$ @hardmath - in ordering, comparing entry a_ij vs entry b_ij. $\endgroup$ – user246802 Jun 9 '15 at 0:22
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    $\begingroup$ @hardmath - just checked again. both, in ordering and in absolute value. $\endgroup$ – user246802 Jun 9 '15 at 0:24
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Since $||A||_P = (\Sigma_j\Sigma_i|a_{ij}|^p)^{1/p}$, if for any pair $|a_{ij}| \ge |b_{ij}|$ holds, I think $||A||_p \ge ||B||_p$

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  • $\begingroup$ Ah, very nice @Dylan. You are essentially treating the matrix as one long row vector and are applying the vector p-norm to it. Thanks so much. $\endgroup$ – user246802 Jun 9 '15 at 0:36
  • $\begingroup$ Hi @Dylan - I was wondering how your proof above could translate to saying ||A(x)|| greater than or equal to ||B(x)||? Do you think this is automatic, once we have the inequality for the matrices themselves? $\endgroup$ – user246802 Jun 9 '15 at 3:48
  • $\begingroup$ @user246802 sor, not able to solve that. can you give the definition of ||A(x)| where x is a variable.| $\endgroup$ – Dylan Jun 9 '15 at 7:45

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