Proving Submultiplicativity on a Matrix Norm

Let $||A||=(\sum_{i=1}^{n}\sum_{j=1}^{n}{a_{ij}^p})^{1/p}$, and let p=2. Then prove that $\|AB\|\le \|A\|\|B\|$

I have looked at numerous proofs for this, and I don't see one that satisfies me fully. I don't want to use traces. I see proofs bringing an x into the left side, divide by $||Bx||$ then pull things out and cancel out and what not. But these don't involve using p=2. Thanks in advance

• If the problem is getting satisfied it would be better to see those proofs and the reason they don't satisfy you. Otherwise one might write yet another proof that doesn't satisfy you. – Pp.. Jan 29 '15 at 0:00
• Are you aware of the Cauchy Schwarz inequality? That makes for a quick and easy proof, if that's "satisfying" to you. – Omnomnomnom Jan 29 '15 at 0:50
• Yes I am, and I assume thats why p=2, I just can't connect the two though. Any help is appreciated thanks! – Jeff Jan 29 '15 at 0:54
• And maybe satisfying is the wrong word, they are all good proofs, and some of which I do understand, but would not be allowed to use in my analysis class. I didn't find anything that to my untrained eyes was applicable to this case – Jeff Jan 29 '15 at 0:55

1 Answer

Here is one proof:

Note that $\|A\|_F^2 = \sum_k \|Ae_k\|_2^2$. Any set of orthonormal vectors will do, this also shows that $\|A\|_2 \le \|A\|_F$.

Then $\|AB\|_F^2 = \sum_k \|ABe_k\|_2^2 \le \|A\|_2^2 \sum_k \|Be_k\|_2^2 = \|A\|_2^2 \|B\|_F^2 \le \|A\|_F^2 \|B\|_F^2$.

• is $e_k$ a value that will equal 1 when the whole summation is done? Also why is there the first $\le$? Thanks – Jeff Jan 29 '15 at 0:11
• $e_k$ is the $k$th standard unit vector. The above comment establishes the first $\le$, but here is a direct proof: Let $x$ be a vector such that $\|x\|_2 = 1$ and $\|Ax\|_2=\|A\|_2$. Then $\|A\|_2^2 =\|Ax\|_2^2 = \sum_k |a_k^T x|^2 \le \sum_k \|a_k\|_2^2 = \|A\|_F^2$. (The $a_k^T$ are the rows of $A$). – copper.hat Jan 29 '15 at 0:19