TITLE: The $2$-norm of a submatrix does not exceed the norm of the matrix
QUESTION [5 upvotes]: Let $A \in \mathbb{R}^{n\times n}$ and let $1\le  i_1 \le i_2 \le n$. Let $B$ be the submatrix with indices $(i_1:i_2,i_1:i_2)$. Prove that $\|B\|2\le \|A\|_2$.

Can I say since $B$ is submatrix of $A$ then $\|B\|_2\le \|A\|_2$? Or should I prove more? And how?

REPLY [2 votes]: If the norm was the Frobenius norm, you wouldn't have to say anything else. But apparently, you are to consider the induced norm corresponding to the Euclidean norm of vectors. 
To prove the claim, use the formula 
$$\|A\|_2 = \sup_{\|x\|,\|y\|\le 1} y^T Ax$$
If this supremum is further restricted to vectors $x,y$ supported on the index range $i_1:i_2$, the result is precisely $\|B\|_2$.