I'm trying to prove this:
Let V be a vector space over $\mathbb{R}$ with finite dimension and inner product and let $\rho: V \longrightarrow V$ be a linear operator for which there is an orthonormal basis consisting of eigenvectors of $\rho$, then $\rho$ is symmetric, i.e. it's self-adjoint over $\mathbb{R}.$
So I considered a matrix $A$ that is associated to $\rho$, then $A$ is orthonormal and has eigenvectors. So I have to prove that $A = A^*$, but I don't know how to relate those matrices. Any hint will be appreciated.
Best Answer
If I have understood your question properly, we have an orthogonal matrix $A$ that can be orthogonally diagonalized and we wish to show that $A$ is symmetric. Let B be an orthogonal matrix such that $$B^TAB=D$$ where Dis diagonal. Then $$A=(B^T)^{-1}DB^{-1}$$ $$=BDB^T.$$ Thus $$A^T=(B^T)^TD^TB^T$$ $$=BDB^T,$$ so $A=A^T.$