I have a following question:
Let $A \in C^{n\times n}$ be Hermitian and $\lambda_\min$ be the smallest eigenvalue of $A$, i.e., $\lambda_\min = \min\{\lambda_1, \ldots, \lambda_n\}$. Show that $\lambda_{\min} \leq \min_j a_{j}$. Hint: use the properties of the Rayleigh quotient.
I got started with the problem, as follows:
Rayleigh quotient, for any vector $x$, is given as,
$$r(x) = \frac{\langle Ax,x\rangle}{\langle x,x\rangle}.$$
If $x$ is an eigenvector, $r(x) = \lambda$.
For a Hermitian matrix, $r(x)\in\mathbb R$. Also, for a Hermitian matrix, the eigenvalues are real.
Now, a Hermitian matrix can be diagonalized as:
$A = UDU^*$, where U is a unitary matrix of eigenvectors, and $D$ is a diagonal matrix of eigenvalues of $A$.
I am missing the final link, that I need to solve my problem.
Thanks.
Best Answer
We show that $\lambda_{\min}=\min_{x\neq 0}\frac{x^*Ax}{x^*x}$. Indeed, let $P$ an unitary matrix and $D$ a diagonal matrix such that $P^*DP=A$. We have $$x^*Ax=x^*P^*DPx=(Px)^*DPx$$ and $(Px)^*Px=x^*x$ hence $$\min_{x\neq 0}r(x)=\min_{x\neq 0}\frac{\langle Dx,x\rangle}{\langle x,x\rangle}.$$ Now it's easier to see that $\min_{x\neq 0}r(x)\geq \lambda_{\min}$. Thanks to that, note that $a_{jj}=r(e_j)$, where $e_j$ is the $j$-th element of the canonical basis of $\mathbb C^n$. So $a_{jj}=r(e_j)\geq \min_{x\neq 0}r(x)\geq \lambda_{\min}$.