Let $V$ be a nontrivial finitely-generated inner product space and let $\alpha\in \operatorname{End}{(V)}$ be orthogonally diagonalizable and satisfy the condition that each of its eigenvalues is real. Is $\alpha$ necessarily self-adjoint?

Let $\lambda_1,\dots,\lambda_n\in\mathbb{R}$ be distinct nonzero eigenvalues of $\alpha.$

I guess the answer is yes, since $\alpha$ has an orthogonal basis composed of its distinct eigenvectors, let's name this basis $B=\{v_1,\dots,v_n\}$, where $v_1,\dots,v_n$ are the distinct eigenvectors of $\alpha$ associated with the eigenvalues $\lambda_1,\dots,\lambda_n$. Given $0\neq v,\:w\in V$, there exist scalars $\alpha_1,\dots,\alpha_n$ and $\beta_1\dots,\beta_n $ not all of which equal $0$ satisfying $v=\sum_{i=1}^{n}\alpha_iv_i$ and $w=\sum_{i=0}^{n}\beta_iv_i$, Hence $$\langle\alpha(v),w\rangle=\langle \sum_{i=1}^{n}\alpha_i\alpha(v_i),w\rangle=\sum_{i=1}^n\alpha_i\bar\beta_i\langle\alpha(v_i),v_i\rangle=\sum_{i=1}^n\alpha_i\bar\beta_i\langle v_i,\alpha(v_i)\rangle=\langle v,\alpha(w\rangle).$$

So $\alpha$ is self-adjoint, however, it's not necessarily positive definite, since its eigenvalues are not necessarily positive. Am I correct?


Your solution makes sense - another, slightly more abstract approach is to note that if $\alpha$ is as you describe it, one can write


with $U$ orthogonal and $D$ diagonal and real. Thus:



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