Minimum and maximum eigenvalues of $B$ 
Let $\lambda_{\min}$ and $\lambda_{\max}$ be the minimum and maximum eigenvalues of Hermitian matrix $B$. Show that if $y$ is a unit vector with respect to the standard inner product on $\mathbb{C}^{n\times1}$, then $\lambda_{\min} \leq y^* B y \leq \lambda_{\max}$.

Can someone help me prove this? Thanks!
 A: Assuming you meant that you are trying to prove $\lambda_{\min}\le y^{*}By\le\lambda_{\min}$, because $B$ is Hermitian, its eigenvectors are orthogonal (even orthonormal) and complete, and the eigenvalues are real. Let $\{\lambda_j,|\lambda_j\rangle\}$ be the eigendecomposition of $B$, and suppose that
\begin{align*}
B&=\sum_{j}\lambda_j |\lambda_j\rangle\langle\lambda_j|, \quad\text{and} \\
y&=\sum_{j}a_j|\lambda_j\rangle.
\end{align*}
Then
\begin{align*}
y^* By&=\left(\sum_i \overline{a}_i\langle\lambda_i|\right)\left(\sum_{j}\lambda_j |\lambda_j\rangle\langle\lambda_j|\right)\left(\sum_{k}a_k|\lambda_k\rangle\right) \\
&=\left(\sum_i \overline{a}_i\langle\lambda_i|\right)\left(\sum_{j,k}\lambda_ja_k|\lambda_j\rangle\langle\lambda_j|\lambda_k\rangle\right) \\
&=\left(\sum_i \overline{a}_i\langle\lambda_i|\right)\left(\sum_{k}\lambda_ka_k|\lambda_k\rangle\right) \\
&=\sum_{i,k} \overline{a}_i\lambda_k a_k\langle\lambda_i|\lambda_k\rangle \\
&=\sum_i \overline{a}_i\lambda_i a_i \\
&=\sum_i \lambda_i|a_i|^2.
\end{align*}
But because $y$ is a unit vector, 
$$\sum_i|a_i|^2=1,$$
from which we get that
$$\lambda_{\min}\le \sum_i \lambda_i|a_i|^2\le\lambda_{\max},$$
and the result follows.
