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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.

2 Answers 2