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

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    Oh, only now did I notice your reply. Thanks.2012-12-28