How is the $p$-norm for square matrices submultiplicative for $p=1$ (Cauchy-Schwarz?)

In my linear algebra self-study I can't understand why the vector p-norm for square matrices is submultiplicative for p=1,2 in a sum norm.

I found this Stack Overflow post that discusses the case $1 \leqslant p \leqslant 2$ Vector $p$-norm for square matrices is submultiplicative for $1 \le p \le 2$ but it says "For $p=1$ and $p=2$ the result follows easily from the Cauchy-Schwarz inequality", and I don't understand how it follows. I'm particularly unable to see how Cauchy-Schwarz Inequality works for $p=1$.

For $p=1$ you don't need the Cauchy-Schwarz inequality. This is solely a consequence of the triangle inequality (and positivity) of the norm: Denote $A = (a_{ij})_{i,j=1,\dotsc,n}$ and $B=(b_{kl})_{k,l=1,\dotsc,n}$. Then \begin{align*} \lVert AB\rVert_p &= \sum_{i,l=1}^n \left\lvert \sum_{j=1}^n a_{ij}b_{jl} \right\rvert \le \sum_{i,l=1}^n\sum_{j=1}^n\lvert a_{ij}\rvert\cdot \lvert b_{jl}\rvert\\ &\le \sum_{i,l=1}^n\sum_{j=1}^n\left(\lvert a_{ij}\rvert\cdot \lvert b_{jl}\rvert + \sum_{\substack{k=1\\k\neq j}}^n \lvert a_{ij}\rvert \cdot \lvert b_{kl}\rvert\right)\\ &= \sum_{i,j=1}^n\lvert a_{ij}\rvert \cdot \sum_{k,l=1}^n\lvert b_{kl}\rvert = \lVert A\rVert_p\cdot \lVert B\rVert_p. \end{align*}
• Ah sorry, I misread what you meant. The term in the big parentheses equals $\sum_{k=1}^n \lvert a_{ij}\rvert\cdot \lvert b_{kl}\rvert$. Rearranging the sum yields the next expression. – Claudius Jan 21 '16 at 7:44
• No, the sum is not a typo. The third sum is exactly, what is need to get from $\lvert a_{ij}\rvert \cdot \lvert b_{jl}\rvert$ to $\sum_{k=1}^n\lvert a_{j}\rvert \cdot \lvert b_{kl}\rvert$. And the second Sigma is still the one from the first line. – Claudius Jan 21 '16 at 7:51