Inequality for the norm of a matrix and the norm of its columns and rows.

Question

Let $\boldsymbol{M}$ be an $n \times m$ matrix. Let $\boldsymbol{M}_{i,:}$ be the $i^{th}$ row of $\boldsymbol{M}$ and $\boldsymbol{M}_{:,j}$ by the $j^{th}$ column. Is it the case that:
$$||\boldsymbol{M}_{i,:}||_2 \leq ||\boldsymbol{M}||_2$$
or
$$||\boldsymbol{M}_{:,j}||_2 \leq ||\boldsymbol{M}||_2$$
or both, where $||.||_2$ denotes both the $\ell_2$ norm of the columns and rows of $\boldsymbol{M}$, as well as the induced matrix norm of the entire matrix $\boldsymbol{M}$. It seems to me that this should certainly be true yet I should like to see an explicit proof.

Answer 1

I think you can prove this for the columns using the following definition for the induced matrix norm: $||M||_2 = sup_{||x|| = 1} ||Mx||_2$.

Let $M_{:j}$ be any column of $M$. Now for $x = (0, ..., 0, 1, 0, ..., 0)^T$ with the $1$ at position $j$, we get that $||x||_2 = 1$ and $Mx = M_{:j}$. So $||M||_2 = sup_{||x|| = 1} ||Mx||_2 \geq ||M_{:j}||_2$.

Now to prove it for the rows, I'll use the following definition for the induced matrix norm: $||M||_2 = sup_{x \not = 0} \frac{||Mx||_2}{||x||_2}$.

Consider any row $M_{i:}$ of M. Let $x = M_{i:}^T$. Then $Mx = MM_{i:}^T$ with $||MM_{i:}^T||_2 \geq ||M_{i:}||_2^2$. Thus $||M||_2 = sup_{x \not = 0} \frac{||Mx||_2}{||x||_2} \geq \frac{||M_{i:}||_2^2}{||M_{i:}||_2} = ||M_{i:}||_2$.