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.


  • 1
    $\begingroup$ Do you work with infinite-dimensioned $V$? Or finite-dimensioned? $\endgroup$ Apr 24, 2017 at 11:21
  • $\begingroup$ finite dimensional $\endgroup$
    – mcmapple
    Apr 24, 2017 at 11:47

1 Answer 1


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: $$ \langle T x , x \rangle = \sum_i \sum_j \langle \lambda_i \alpha_i v_i , \alpha_j v_j \rangle = \sum_i \sum_j \lambda_i \alpha_i \alpha_j \langle v_i , v_j \rangle = \sum_i \lambda_i \alpha_i^2 \ge \sum_i \alpha \alpha_i^2 = \alpha \langle x , x \rangle $$


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