Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

1 Answer 1

up vote 1 down vote accepted

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 $$\lambda_2=\max_{0\neq x\perp v_1}\frac{v^\star Av}{v^\star v}$$

share|improve this answer
    
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
    
Still can't see how to prove 2. Can you please, help me? –  Guy Dec 27 '12 at 16:15
    
Ok im gonna do the calculation, but maybe there is a more straightforard one, because this one is so annoying. –  Tomás Dec 27 '12 at 16:52
show 2 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.