1
$\begingroup$

I am trying to prove that if $T:V \to V$ is a self-adjoint endomorphism on a finite dimensional real inner product space $V$ with the standard inner product, and each eigenvalue of $T$, $\lambda$ is greater than or equal to $\alpha \in \mathbb{R}$ then

$(T\overrightarrow{x},\overrightarrow{x}) \ge \alpha(\overrightarrow{x},\overrightarrow{x})$

I know from the spectral theorem for self adjoint endomorphisms that the eigenvectors of T form an orthanormal basis for V. But I am not sure how to proceed from here.

Any advice would be much appreciated.

Thanks!

$\endgroup$
  • 1
    $\begingroup$ Do you work with infinite-dimensioned $V$? Or finite-dimensioned? $\endgroup$ – TZakrevskiy Apr 24 '17 at 11:21
  • $\begingroup$ finite dimensional $\endgroup$ – mcmapple Apr 24 '17 at 11:47
1
$\begingroup$

Let $v_1, \dots, v_n$ be an orthonormal basis of $V$ formed by eigenvectors of $T$, where $T v_i = \lambda_i v_i$.

Write $x=\alpha_1 v_1 + \cdots + \alpha_n v_n$ and just compute: $$ \langle T x , x \rangle = \sum_i \sum_j \langle \lambda_i \alpha_i v_i , \alpha_j v_j \rangle = \sum_i \sum_j \lambda_i \alpha_i \alpha_j \langle v_i , v_j \rangle = \sum_i \lambda_i \alpha_i^2 \ge \sum_i \alpha \alpha_i^2 = \alpha \langle x , x \rangle $$

$\endgroup$

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.