Norm of a Block Matrix Let $X\in M_{m,n}(R)$ and $l=m+n$. Now consider the block matrix
$$
Y=\left[
\begin{array}{cc}
0 &X \\ \
X^T &0  
\end{array}\right]
$$
where $Y\in M_l(R)$.
I want to show that $||X||=||Y||$ where for any $A\in M_{m,n}(R)$,
we define $||A||=max\{||Ax||: x\in R^n,||x||=1\}$ and $R^n$ has the standard Euclidean inner product.
I have got upto $$
||Y||^2=||Y^TY||=|| \left[
\begin{array}{cc}
XX^T &0 \\ \
0 &X^TX  
\end{array}\right]||
$$
Thanks for any help.
 A: Note that: 


*

*$\|X^TX\| = \|XX^T\| = \sigma_1(X)^2 = \|X\|^2$.

*$\left\|\pmatrix{A\\&B} \right\| = \max\{\|A\|,\|B\|\}$


From there, we can quickly reach the desired conclusion.
In order to prove the second result:  Suppose that $x,y$ are unit vectors, and $a,b \geq 0$ are such that $a^2 + b^2 = 1$.  Then the vector
$$
v = \pmatrix{ax\\by}
$$
is a unit vector.  Moreover, every $v$ can be broken up in such a fashion.  We then have
$$
\left\|\pmatrix{A\\&B} v\right\|^2 = \left\|\pmatrix{aAx \\bBy} \right\|^2 = a^2\|Ax\|^2 + b^2\|Bx\|^2 \leq (a^2 + b^2) [\max\{\|Ax\|,\|Bx\|\}]^2
$$
A: Let $u\in\mathbb R^m$, $v\in\mathbb R^n$, and $w = (u,v)$. Then, we have
\begin{align} 
\|Yw\| &= \left\| \begin{bmatrix} 0 & X \\ X^T & 0 \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} \right\| \\
&= \left\| \begin{bmatrix} Xv \\ X^T u \end{bmatrix} \right\| \\
&= \sqrt{ \|Xv\|^2 + \|X^T u\|^2 } \\
&\le \|X\| \sqrt{ \|v\|^2 + \|u\|^2 } \\
&= \| X \| \|w\|.
\end{align}
That is, $\|Y\|$ is bounded from above by $\| X \|$.
Now, for $u\ne 0$ with $\|X^T u\| = \|X^T\|\|u\|$ and $v=0$, we additionally have
$$ \| Yw \| = \| X^T u \| = \|X^T \|\|u\| = \|X \| \|w\|. $$
Thus, the upper bound $\|X\|$ is attained and we obtain $\|Y\| = \|X\|$.
