# If every entry of a matrix A is bigger than every entry of a matrix B, is norm A bigger than norm B?

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,

• It probably is a theorem. Which matrix norm is that? – muaddib Jun 9 '15 at 0:18
• Does "bigger than or equal" entries mean in absolute value or in ordering as real (?) numbers? – hardmath Jun 9 '15 at 0:19
• @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||.) – user246802 Jun 9 '15 at 0:21
• @hardmath - in ordering, comparing entry a_ij vs entry b_ij. – user246802 Jun 9 '15 at 0:22
• @hardmath - just checked again. both, in ordering and in absolute value. – user246802 Jun 9 '15 at 0:24

Since $||A||_P = (\Sigma_j\Sigma_i|a_{ij}|^p)^{1/p}$, if for any pair $|a_{ij}| \ge |b_{ij}|$ holds, I think $||A||_p \ge ||B||_p$