In a previous exam the following question was asked which I was unable to answer due to the lack of knowledge of self-adjoint operators.

Let $S$ be a self-adjoint operator on a real finite dimensional vectorspace $V$ with inner product $\langle \cdot , \cdot \rangle$

Prove that: For every eigenvalue $k$ of $S$, $0 ≤ k ≤ 1$ iff $0 ≤ \langle Sv, v \rangle ≤ \langle v, v \rangle$ for every $v \in V$

  • $\begingroup$ $Sv = kv$ no? Can you not then use the properties of an inner product to deduce $$\langle kv, v \rangle = k \langle v, v \rangle \le \langle v, v \rangle$$ $\endgroup$ – mattos May 27 '16 at 14:33
  • $\begingroup$ are you working in a finite dimensional vector space? $\endgroup$ – User8128 May 27 '16 at 14:39
  • $\begingroup$ Oops thanks, Im an idiot $\endgroup$ – Bahbi May 27 '16 at 14:39
  • $\begingroup$ The problem is that it needs to hold for every $v$ and not every $v$ is an eigenvector. That is, for arbitrary $v$ there may not be an eigenvalue $k$ so that $Sv = kv$. $\endgroup$ – User8128 May 27 '16 at 14:51

A self-adjoint operator $S : X \to X$ (where $X$ is an inner product space) is an operator such that for all $x,y \in X$, we have $$\langle Sx,y \rangle = \langle x,Sy\rangle.$$ This is a generalization of a real, symmetric matrix.

One important property of such operators is that the eigenvalues of a self-adjoint operator are necessarily real. Indeed, if $k$ is any eigenvalue with corresponding (normalized) eigenvector $v$, we see $$k = k\langle v,v \rangle = \langle kv, v \rangle = \langle Sv, v \rangle = \langle v,Sv \rangle = \langle v, kv \rangle = \overline k \langle v, v \rangle = \overline k$$ showing that $k$ is real.

Another important property (perhaps the most important property) of self-adjoint operators is that the eigenvectors of a self-adjoint operator can be taken to form an orthonormal basis for the ambient space (here I am assuming you are working in a finite dimensional space, but a similar statement still holds in infinite dimension, we just need to generalize the idea of a basis a bit and we need completeness). That is, we can take $k_1, \ldots, k_n$ to be the eigenvalues of $S$ (possible with repetitions) with corresponding orthonormal eigenvectors $v_1,\ldots, v_n$ forming a basis for $X$. Then for any $v \in X$, there are scalars $\alpha_1, \ldots, \alpha_n$ so that $v = \alpha_1 v_1 + \cdots + \alpha_nv_n.$ Using linearity of the inner product, we see $$\langle v, v\rangle = \sum^n_{i=1} \sum^n_{j=1} \alpha_i \overline \alpha_j \langle v_i, v_j \rangle.$$ But by orthonormality, $\langle v_i, v_j \rangle = 0$ when $i \neq j$ and $\langle v_i, v_i \rangle = 1$. Thus the above sum becomes $$\langle v, v\rangle = \sum^n_{i=1} \alpha_i \overline \alpha_i = \sum^n_{i=1} \lvert \alpha_i \rvert^2.$$ Similarly, since $$Sv = S(\alpha_1v_1 + \cdots \alpha_n v_n) = \alpha_1 k_1 v_1 + \cdots + \alpha_n k_n v_n $$we have $$\langle Sv, v\rangle = \sum^n_{i=1} \sum^n_{j=1} k_i \alpha_i \overline \alpha_j \langle v_i, v_j \rangle = \sum^n_{i=1} k_i \lvert \alpha_i \rvert^2.$$ Clearly if $k_i \ge 0$ for all $i=1,\ldots, n$ then $$\langle Sv, v\rangle = \sum^n_{i=1} k_i \lvert \alpha_i \rvert^2 \ge 0.$$ Also, if $k_i \le 1$ for all $i = 1,\ldots, n$, then $$\langle Sv, v\rangle = \sum^n_{i=1} k_i \lvert \alpha_i \rvert^2 \le \sum^n_{i=1} \lvert \alpha_i \rvert^2 = \langle v , v \rangle.$$ Conversely, if the given condition holds for all vectors $v$, then applying the condition to the eigenvectors gives $$0 \le \langle Sv_i, v_i \rangle \le \langle v_i, v_i \rangle \,\,\,\, \implies \,\,\,\, 0 \le \langle k_i v_i, v_i \rangle \le \langle v_i, v_i \rangle$$ whence pulling the $k_i$ out of the inner product gives $0 \le k_i \le 1.$


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