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?


$$
\|A\|\_2 \le \|A\|\_1
$$



Is the matrix induced L1-norm greater than the induced L2-norm?
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.)


