# Counterexample to $\| X - Y \| \le \left\| \begin{pmatrix} X & 0 \\ 0 & Y \end{pmatrix} \right\|$?

I was reading the book Matrix Analysis and have encountered with the following exercise in Chapter IX. A Selection of Matrix Inequalities:

Let $X, Y$ be positive operators. Show that for every unitarily invariant norm $$\| X - Y \| \le \left\| \begin{pmatrix} X & 0 \\ 0 & Y \end{pmatrix} \right\|.$$

After some time that I spend proving it I started to plug-in different values for $X$ and $Y$ trying to understand the behaviour of the norms of these matrices. So, I picked the operator norm, which is indeed unitarily invariant norm.

Now, take positive $2\times 2$ matrices $$X = \begin{pmatrix} 1 & 1 \\ 0 & 0.5 \end{pmatrix}, \quad Y = \begin{pmatrix} 0.5 & 0 \\ 1 & 1\end{pmatrix}.$$ I calculated and triple double checked that the inequality above is violated with r.h.s. $\approx 1.46$ and l.h.s. $1.5$.

My question is: Can anyone verify whether this "counterexample" is true or not? If not, where is the mistake?

UPD: As it was correctly pointed out by @Ted Shifrin the operators should be Hermitian (symmetric) and the counterexample fails. @amsmath gives the proof for operator norm in comments. Any ideas how it can be proved for any unitarily invariant norm?

• I'm getting the same numbers as you. – Nate Eldredge Aug 12 '18 at 15:18
• What is the definition of a positive operator? – Ted Shifrin Aug 12 '18 at 15:22
• Just as I suspected: The definition (to which you linked me) requires that the operator be symmetric (hermitian). – Ted Shifrin Aug 12 '18 at 15:59
• $\left\|\begin{pmatrix}X & 0\\0 & Y\end{pmatrix}\right\| = \max\{\|X\|,\|Y\|\}$. So, $Y = -X\in\Bbb R$ should be sufficient as a counterexample. – amsmath Aug 12 '18 at 16:20
• For the operator norm you have $((X-Y)u,u) = (Xu,u)-(Yu.u)\le(Xu,u)\le\|X\|$ for all vectors $u$ with $\|u\|=1$. Similarly, $((Y-X)u.,u)\le\|Y\|$. Hence, $\|X-Y\| = \sup_{\|u\|=1}|((X-Y)u,u)|\le\max\{\|X\|,\|Y\|\}$. – amsmath Aug 12 '18 at 16:48