Suppose $F=\mathbb{R}$. Let $A: V\to V$ (where $V$ is a finite dimensional inner product space over $F$) so that $A=A^*$ ("self-adjoint"), then there exists an orthonormal basis of eigenvectors and eigenvalues are real.
(I come across the above problem when trying to understand how one could represent adjoint operators via their orthonormal base. I cannot find a complete proof in any books (from linear algebra to functional analysis)
I hope someone could help in providing a full proof.)
Thanks!
Best Answer
This is actually pretty easy. For simplicity, let us assume that $V = \mathbb R^n$ and that $A$ is a symmetric matrix. First, $A$ has without doubt complex eigenvalues because the characteristic polynomial has complex zeros. Then you see that each eigenvalue of $A$ must be real. Indeed, if $\lambda$ is an eigenvalue of $A$ with corresponding eigenvector $x\in\mathbb C^n$, then $$ (\lambda - \overline{\lambda})\langle x,x\rangle = \langle\lambda x,x\rangle - \langle x,\lambda x\rangle = \langle Ax,x\rangle - \langle x,Ax\rangle = 0. $$ Thus, $\operatorname{Im}\lambda = 0$ and $\lambda$ is real. Note that we have used the complex inner product here, i.e., $\langle x,y\rangle = \sum_i x_i\overline{y_i}$. It also follows that $x$ can be chosen to have only real entries (if $x$ is not real, choose $x + \overline{x}$ instead).
Now, let $\lambda$ be an eigenvalue of $A$ with corresponding normalized eigenvector $x_1\in\mathbb R^n$. Then $\operatorname{span}\{x_1\}$ is invariant under $A$ and it also easily seen that also $(\operatorname{span}\{x_1\})^\perp$ is invariant under $A$. Now, restrict $A$ on the latter space and do the same thing again (which you can because the restriction is again symmetric/selfadjoint). Thus, you find an orthonormal basis of eigenvectors of $A$.