2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.