Is the matrix induced L1-norm greater than the induced L2-norm? For $A \in \mathbb R^{m \times n}$ and the induced norms:
$$
\| A \|_1 = \max_{x \ne 0} \frac{\|Ax\|_1}{\|x\|_1}
$$
$$
\| A \|_2 = \max_{x \ne 0} \frac{\|Ax\|_2}{\|x\|_2}
$$
... where:
$$
\|x\|_1 = \sum_{k=1}^n |x_k|
$$
$$
\|x\|_2 = \sqrt{\sum_{k=1}^n |x_k|^2}
$$
... does the following inequality hold in general?
$$
\|A\|_2 \le \|A\|_1
$$
 A: To avoid ambiguity I will write $\lVert\cdot\rVert_{p\to r}$ for the $\ell_p$-to-$\ell_r$-norm. Note that in general, $\lVert A\rVert_{1\to r} = \max_{1\leq j\leq n} \lVert (Ae_j)\rVert_r$.
Let $A$ be the $n\times n$ matrix whose top row has $1$ in every entry, and all other entries of the matrix are $0$. Then by the remark above, $\lVert A\rVert_{1\to 1} =1$. On the other hand,
$$ \lVert A\rVert_{2\to 2} \geq \lVert A^* \rVert_{2\to 2} \geq \lVert A^*e_1\rVert_2 = \sqrt{n} $$
giving a counterexample to your question. (In fact this lower bound is an equality, although this is not needed to answer the question.)
A: It certainly holds for all matrices where $|a_{ij}|=|a_{ji}|$ (meaning both symmetric and skew-symmetric matruces, among otbers). It is reported here that
$||A||_2^2\le||A||_1||A||_\infty,$
and where the condition above is satisfied $||A||_1=||A||_\infty$.

For all $n×n$ matrices we always have
$||A||_1\le n||A||_\infty$
$||A||_\infty\le n||A||_1$
and so the bound given above implies
$||A||_2\le\sqrt{n}||A||_1$
$||A||_2\le\sqrt{n}||A||_\infty.$
