Orthonormal basis of eigenvectors of a self-adjoint operator

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.

• Do you know the spectral theorem? – MathIsFun May 4 at 21:44
• 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. – Arturo Magidin May 4 at 22:13
• 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 $(T\vec{v_i},\vec{v_i})$? – Montes May 4 at 22:14
• 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 $\overrightarrow{v_i}$. – MathIsFun May 4 at 22:32