A matrix $A \in K^{n \times n}$ is diagonalizable with orthonormal basis on $K$ if there exists $P \in K^{n \times n}$ orthogonal such that $P^*AP = D$ where $D \in K^{n \times n}$ is a diagonal matrix.
I was asked if the following statement is true or false (and it is not a homework): A matrix $A \in \mathbb R^{n \times n}$, diagonalizable with orthonormal basis on $\mathbb R$ is necessarily symmetric.
I think the statement is true because : $A=PDP^T \implies A^T=(P^T)^TD^TP^T=PDP^T=A \implies A$ is symmetric.
But it think this would mean that the spectral theorem is an equivalence and not an implication. The spectral theorem says if A is symmetric, there exists an orthonormal basis of V consisting of eigenvectors of A. This is not an "if and only if".