# 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.

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 $$\left=\left<\sum a_i\lambda_i v_i,\hat{v}\right>\leq \lambda_{\text{max}}\left<\sum a_iv_i,\hat{v}\right>=\lambda_{\text{max}}\left.$$ We derive the result for $$\lambda_{\text{min}}$$ similarly.

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.

• Isn't $A$ assumed a general linear operator and not a matrix? – Melody Jan 13 at 23:17
• @Melody Linear operators on finite-dimensional spaces can be represented by matrices and vice versa – angryavian Jan 13 at 23:19
• thanks, but why $\langle Av, v \rangle = v^* U D U^* v$? – Peyman Jan 13 at 23:20
• @angryavian I know that, but technically they aren't the same thing though. One is basis free. Maybe I'm just being too nitpickty – Melody Jan 13 at 23:21
• @Melody It's ok, you are right. I should have prefaced my answer by choosing an arbitrary basis before mentioning matrices. – angryavian Jan 13 at 23:22