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This is a probem from "Matrix theory: basic results and techniques" page 147

Let $A$ be a matrix. If there is a Hermitian matrix $X$ such that $\left(\begin{array}{cc}I+X&A\\A^*&I-X\end{array}\right)$ is positive semidefinite. Then $$|(Ay,y)|\le (y,y), \hbox{ for all $y$}.$$

How to prove this?

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By "posi" you mean that all the eigenvalues are positive? – Aaron Jun 30 '11 at 0:38
@Aaron: I believe that is correct seeing that the question ask's to show the matrix is $\ge~0.$ – night owl Jun 30 '11 at 0:43
Sorry, I missed some characters. – Sunni Jun 30 '11 at 1:00
@Sunny: Please try to make post titles informative, reflecting some of the content of the post. "Ask a problem from 'Matrix theory'" is none of these, much like "I have a question" doesn't really say much... – Arturo Magidin Jun 30 '11 at 3:07
up vote 4 down vote accepted

By hypothesis we have in particular $$\begin{bmatrix} y^{\ast} & r^{\ast}y^{\ast} \end{bmatrix}\begin{bmatrix} I+X&A\\A^*&I-X\end{bmatrix}\begin{bmatrix} y \\ ry \end{bmatrix} \geq 0 \qquad \text{for all } y \text{ and an arbitrary scalar } r.$$ Expanding the above gives $$y^{\ast}y + y^{\ast}Xy + y^{\ast}A ry +r^{\ast}y^{\ast}A^{\ast}y + r^{\ast}r y^{\ast}y - rr^{\ast}y^{\ast}Xy \geq 0.$$ Now choose $r$ such that $|r| = 1$ and $r(y,Ay) = - |(Ay,y)|$. Plugging this into the last inequality gives the desired result.

That $X$ is hermitian is only used to ensure that it makes sense to speak of positive semi-definiteness.

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Bravo, nice proof. I roginally thought to write $A=(I+X)^{1/2}C(I-X)^{1/2}$ for some contraction $C$. Then.... – Sunni Jun 30 '11 at 1:31
@Sunni: Thanks! I first tried to work with $\begin{bmatrix}y \\ y\end{bmatrix}$, but this didn't quite give the right thing. So I decided to plug in a scalar and the simplest approach seemed to work. – t.b. Jun 30 '11 at 1:35

This is a proof, but not as good as Theo's.

The condition is equivalent to $A=(I+X)^{1/2}C(I-X)^{1/2}$ for some contraction $C$.

$|(Ay,y)|=|((I+X)^{1/2}C(I-X)^{1/2}y,y)|\le (|(I+X)^{1/2}(I-X)^{1/2}|y,y) \le (\frac{I+X+I-X}{2}y,y)=(y,y)$, for all $y$. $|Y|$ means the absolute value of $Y$.

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That's very nice! (meta: please call me Theo) – t.b. Jun 30 '11 at 2:03

I found another proof, slightly different from Theo's.

$$\begin{bmatrix} y^{\ast} & 0\\0& y^* \end{bmatrix}\begin{bmatrix} I+X&A\\A^*&I-X\end{bmatrix}\begin{bmatrix} y & 0\\0& y \end{bmatrix}=\begin{bmatrix} ((I+X)y,y) &(Ay,y) \\(A^*y,y)&((I-X)y,y) \end{bmatrix} $$ is a $2\times 2$ positive semidefinite matrix. Therefore \begin{eqnarray*}|(Ay,y)|^2\le (I+X)y,y)((I-X)y,y)\ \le \left(\frac{(I+X)y,y)+((I-X)y,y)}{2}\right)^2=(y,y)^2\end{eqnarray*} Done.

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You could have edited your previous answer and added this, instead of posting a new answer... – Arturo Magidin Jun 30 '11 at 4:31
Good suggestion, next time, I shall do this. – Sunni Jun 30 '11 at 4:34
That's also nice. So I think we can believe the result :) I edited your other answer to change my name. – t.b. Jun 30 '11 at 4:36
Why did you delete your other answer? – t.b. Jun 30 '11 at 4:38
@Theo: Hmmm.. probably my fault; I suggested he could have simply added the second proof to the first posta (meaning it for future reference). And then he deleted the old one... – Arturo Magidin Jun 30 '11 at 4:44

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