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Problem:

A is an $n\times n$ complex matrix. Let $\lambda _\max(Y)$ and $\sigma _\max(Y)$ denote respectively the largest eigenvalue and the largest singular value of a square matrix $Y$. The question is to prove that: $$\lambda _\max\left(\frac{A+A^*}{2}\right)\leq \sigma _\max (A)$$

I started as follows: Since $A$ is Hermitian, then $(A+A^*)$ is Hermitian as well. Then: $$\lambda_\max\left(\frac{A+A^*}{2}\right)=\max_{\left \| x \right \|=1}x^*\left(\frac{A+A^*}{2}\right)x$$ where $x\in \mathbb{C}^n$

I need to prove the following to finish my proof: $$\max_{\left \| x \right \|=1}x^*\left(\frac{A+A^*}{2}\right)x\geq \max_{\left \| x \right \|=1}(x^*A^*Ax)^{\frac{1}{2}}\tag{1}$$

From the first sight, it looks like I have to use the fact that: $\sqrt{ab}\geq \frac{a+b}{2}$ for $a$ and $b$ positive, but this works only for numbers and not matrices.

I appreciate if someone can prove the inequality $(1)$.

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Are you sure you need to require that $A$ is hermitian? Because in that case, $(A+A^*)/2=A$. –  Martin Argerami Apr 23 '12 at 2:04
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And you also need an absolute value in the left-hand side (inside the max); otherwise the inequality is trivially false with $A=-1$. –  Martin Argerami Apr 23 '12 at 2:12
    
@Martin Argerami: You're right. $A$ is not Hermitian. I will edit the original statement. –  Boyan Klo Apr 23 '12 at 2:18
    
If $A$ is allowed to be non-hermitian, the inequality is again false in general. For example, if $A^*=-A$, then the left-hand side is zero, while the right-hand side is not. –  Martin Argerami Apr 23 '12 at 3:03
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1 Answer 1

up vote 3 down vote accepted

Your inequality is actually a well-known equality (in the case where $A$ is hermitian).

The right-hand side of your inequality is the operator norm of $A$, i.e. $$ \max_{\left \| x \right \|=1}(x^*A^*Ax)^{\frac{1}{2}}=\|A\|. $$ For the left-hand side we get, using the Cauchy-Schwarz inequality, $$ \max_{\left \| x \right \|=1}\left|x^*(\frac{A+A^*}{2})x\right|\leq\max_{\left \| x \right \|=1}\|x\|\,\left\|\frac{A+A^*}2\right\|\,\|x\|=\left\|\frac{A+A^*}2\right\|\\ \leq\frac{\|A\|+\|A^*\|}2=\frac{\|A\|+\|A\|}2=\|A\|. $$ Note that so far we don't use that $A$ is hermitian. When $A$ is hermitian, however, we can prove the reverse inequality, which requires a little more work.

Let $$ s=\max\{|x^*Ax|:\ \|x\|=1\} $$ Now take vectors $x,y$ with $\|x\|=\|y\|=1$. Then $$ (x\pm y)^*A(x\pm y)=x^*Ax+y^*Ay\pm2\text{Re}\,y^*Ax. $$ Subtracting one equation from the other we get $$ 4\text{Re}\,y^*Ax=(x+y)^*A(x+y)-(x-y)^*A(x-y) $$ (this is the real part of what is usually called the polarization identity). Now we get $$ 4\text{Re}\,y^*Ax\leq s\,\|x+y\|^2+s\,\|x-y\|^2=2s\,(\|x\|^2+\|y\|^2)=2s. $$ Fix $t$ such that $y^*Ax=e^{it}\,|y^*Ax|$. Note that we can do the above process for the vector $e^{-it}x$, and so this shows that $$ |y^*Ax|\leq s\ \ \ \ \text{ for any }x,y\text{ with }\|x\|=\|y\|=1. $$ Taking max over $y$ with $\|y\|=1$, we obtain $$ \|Ax\|\leq s \ \ \ \text{ for all } x \text{ with } \|x\|=1, $$ and thus $$ \|A\|\leq s. $$

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