# A hermitian matrices property

Let $A$ be a hermitian matrix. Then all its eigenvalues are real. Let $\lambda_1 \geq \lambda_2 \geq ... \geq \lambda_n$ be the eigenvalues with associated eigenvectors $v_1, v_2. ..., v_n$, respectively.

I'd like to prove that $\displaystyle \max_{0\neq x\perp v_1} \Bigl({x^*Ax\over x^*x}\Bigr)$ exists and that $\displaystyle \max_{0\neq x\perp v_1} \Bigl({x^*Ax\over x^*x}\Bigr)=\lambda_2$.

Generalize the previous statement.

There's a suggestion to take $x=\alpha_1 v_1 + ...+ \alpha_n v_n$. Still can't make anything out of it.

Thanks.

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Ok try this one. Write $v=c_2v_2+...+c_nv_n$. Then you have that (remember that $\lambda_i\leq\lambda_2$ for $i=2,...,n$) $$\frac{v^\star Av}{v^\star v}= \frac{\lambda_2c_2^2+...+\lambda_nc_n^2}{c_2^2+...+c_n^2}\leq\lambda _2$$
note that the equality is assumed if $v=v_2$, hence you have proved that $$\lambda_2=\max_{0\neq x\perp v_1}\frac{v^\star Av}{v^\star v}$$
1 - isn't true. Take $A=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ and $x=\begin{bmatrix} \sqrt 2\over 2 & \sqrt 2\over 2 \end{bmatrix}$. Notice that $x^*x=1$. We get ${x^*Ax}={3\over 2}> 1$. As for 2 it's even harder than the single case I'm trying to prove. –  Guy Dec 27 '12 at 14:59
There is something wrong with your argument because $\frac{x^\star Ax}{x^\star x}=\frac{x^\star}{\|x\|}A\big(\frac{x}{\|x\|}\big)$ –  Tomás Dec 27 '12 at 15:37
Yeah, there is, $A$ isn't hermitian. Sorry. –  Guy Dec 27 '12 at 15:45