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Let $V$ be a finite-dimensional real inner product space and suppose $T$ is an endomorphism on $V$. Show that $(T+T^*)/2$ is self-adjoint and show that there is an orthonormal basis $\{\vec{v_1},...,\vec{v_n}\}$ of V consisting of eigenvectors of $(T+T^*)/2$ such that the eigenvalue corresponding to $\vec{v_i}$ is $(T\vec{v_i},\vec{v_i})$

I know how to prove the first part but I have no idea how to show that such basis exists.

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    $\begingroup$ Do you know the spectral theorem? $\endgroup$ – MathIsFun May 4 at 21:44
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    $\begingroup$ An operator on a real inner product space is orthogonally diagonalizable if and only if it is self-adjoint. If you don’t know the necessary properties of self-adjoint operators yet, this seems like not a terribly good exercise! As to why the eigenvalues satisfy that, note that if $v$ corresponds to $\lambda$, then $\lambda =\langle (T+T^*)/2(v),v\rangle = \frac{1}{2}(\langle Tv,v\rangle + \langle T^*v,v\rangle) = \frac{1}{2}(\langle Tv,v\rangle + \langle v,Tv\rangle)$, and go from there. $\endgroup$ – Arturo Magidin May 4 at 22:13
  • $\begingroup$ Now that I think of it, I do. It tells me that there is an orthonormal basis consisting of eigenvectors of $(T+T^*)/2$. What about the eigenvalue corresponding to $\vec{v_i}$ beqing equal to $(T\vec{v_i},\vec{v_i})$? $\endgroup$ – Montes May 4 at 22:14
  • $\begingroup$ You know $((T+T^*)/2)\overrightarrow{v_i}=\lambda_i \overrightarrow{v_i}$ for some $\lambda_i \in\mathbb{R}$. Try taking the inner product of both sides with $\overrightarrow{v_i}$. $\endgroup$ – MathIsFun May 4 at 22:32

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