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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.

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  • $\begingroup$ Hint $\|X\|=\|X^T\|$. $\endgroup$ – user251257 Aug 20 '15 at 16:10
  • $\begingroup$ if you consider $Y^T Y$, try the singular value decomposition of $X$. $\endgroup$ – user251257 Aug 20 '15 at 16:23
  • $\begingroup$ Can you please elaborate? $\endgroup$ – Ester Aug 20 '15 at 16:29
  • $\begingroup$ Assume $X=VSU^T$ where $U$ and $V$ are orthogonal and $S$ has the same size as $X$ and has non negative diagonal elements. Build a eigenvalue decomposition of $Y^TY$ using $X=VSU^T$. $\endgroup$ – user251257 Aug 20 '15 at 16:34
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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 $$ \left\|\pmatrix{A\\&B} v\right\|^2 = \left\|\pmatrix{aAx \\bBy} \right\|^2 = a^2\|Ax\|^2 + b^2\|Bx\|^2 \leq (a^2 + b^2) [\max\{\|Ax\|,\|Bx\|\}]^2 $$

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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 $\|Y\| = \|X\|$.

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