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!

• I suppose $A$ is self-adjoint? – Calvin Khor Mar 26 '16 at 1:35
• in the case of a finite dimensional space over $\Bbb R$, self-adjoint is more commonly known as symmetric. This is then easy to google e.g. maths.manchester.ac.uk/~peter/MATH10212/notes10 – Calvin Khor Mar 26 '16 at 1:57
• You say that as if there isn't a proof in that PDF. – Calvin Khor Mar 26 '16 at 2:23
• It is hard to believe you did not find a proof of the Spectral Theorem, which is one of the most important theorems in linear algebra and functional analysis. What books did you check? – bartgol Mar 26 '16 at 5:03

Suppose $V$ is a finite-dimensional inner product space over the real numbers, and suppose $A$ is a symmetric linear operator on $V$. One way to find an eigenvector of $A$ is to minimize $F(v)=(Av,v)$ subject to the constraint that $\|v\|=1$. Such a minimum exists because you can use an orthonormal basis of $V$ and turn this into a minimization problem for a continuous function $F$ over the unit spherical shell in $\mathbb{R}^{n}$, which is compact. A continuous function on a compact set in $\mathbb{R}^n$ has a minimum.
Suppose the minimum of $F$ on $S=\{ v \in V : \|v\|=1 \}$ is $\lambda$, and suppose $F(v_0)=\lambda$ for some unit vector $v_0$. Then you can show that $\lambda$ is an eigenvalue and $(A-\lambda I)v_0=0$. This is because you must have $$\lambda \le \frac{(Aw,w)}{(w,w)},\;\;\; w\ne 0 \\ 0 \le ((A-\lambda I)w,w),\;\;\; w \in V.$$ That means $[x,y]=((A-\lambda I)x,y)$ has all of the properties of an inner product except that $[x,x]=0$ may occur even if $x \ne 0$. You can check that $[w,w] \ge 0$ because of the above, and it is easy to check that $[x,y]$ is linear in both $x$ and $y$. That's enough in order for the Cauchy-Schwarz inequality to hold: $$|[x,y]| \le [x,x]^{1/2}[y,y]^{1/2}\\ |((A-\lambda I)x,y)| \le ((A-\lambda I)x,x)^{1/2}((A-\lambda I)y,y)^{1/2}$$ But you know that $[v_0,v_0]=((A-\lambda I)v_0,v_0)=0$ because of how $v_0$ was constructed. Therefore, $$((A-\lambda I)v_0,y)=0, \;\;\; y \in V.$$ This is true for all $y\in V$. Taking $y=(A-\lambda I)v_0$ gives $(A-\lambda I)v_0=0$.
Finally, you can use finite induction to get everything. That's because you can now reduce a symmetric operator $A$ on a subspace of one lower dimension. This is because $V' = \{ v : (v,v_0)=0 \}$ is invariant under $A$. To see why, suppose $v \perp v_0$. Then $(Av,v_0) = (v,Av_0)=(v,\lambda v_0)=\lambda(v,v_0)=0$. So $A : V'\rightarrow V'$ and the restriction $A'$ of $A$ to $V'$ is symmetric. Keep repeating the above procedure and eventually you end up with an orthonormal basis of eigenvectors of $V$.
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 $A$.