If $A = A^*$ then $ \lambda_{min} \leq\frac{\langle Av,v \rangle}{\langle v,v \rangle}\leq\lambda_{max} $ I'm leaning linear algebra and new to it. I have trouble with this problem and actually, I don't know what to do! any help or hint would be appreciated.

for the linear operator $A\in \ell(V)$ wich $V$ is an Inner product space with finite dimentions. if $A = A^*$ then show that for every $v \in V$ we have:
  $$ \lambda_{min} \leq\frac{\langle Av,v\rangle}{\langle v,v\rangle}\leq\lambda_{max}, v \neq 0$$ where the lambdas are eigenvalues.

 A: I will use matrix notation, since we are in a finite-dimensional vector space.
$A=A^*$ implies $A$ can be diagonalized by a unitary basis, i.e. $A=UDU^*$ where $U$ is unitary and $D$ is diagonal with diagonal entries $\lambda_1, \ldots, \lambda_n$. Then
$$\langle Av, v \rangle = v^* U D U^* v = \sum_{i=1}^n \lambda_i (U^* v)_i^2.$$
Use $\lambda_{\min} \le \lambda_i \le \lambda_{\max}$ for all $i$ as well as the fact that $\sum_{i=1}^n (U^* v)_i^2 = \langle U^*v, U^* v \rangle = \langle v, v\rangle$ to conclude.
A: Since $A=A^*$ and $A$ is a linear transformation in a finite dimensional inner product space we can represent the transformation by a Hermitian matrix $M$ with basis $\mathcal{B}$. Since $M$ is Hermitian it has a basis of eigenvectors $v_1,...,v_n$ with eigenvalues $\lambda_1,...,\lambda_n$. 
Let $[v]_\mathcal{B}=\hat{v}$. Then $\hat{v}=\sum a_i v_i$ for constants $a_i$, so $$\left<Av,v\right>=\left<\sum a_i\lambda_i v_i,\hat{v}\right>\leq \lambda_{\text{max}}\left<\sum a_iv_i,\hat{v}\right>=\lambda_{\text{max}}\left<v,v\right>.$$
We derive the result for $\lambda_{\text{min}}$ similarly.
