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Let $T: V \rightarrow V$ be a self adjoint linear map where $V$ is an inner product space.

Known fact: With respect to any orthonormal basis- the the matrix for $T$ is conjugate symmetric. When dealing with a real vectorspace just symmetric.

However is it true that if there exists some basis (not necessarily orthonormal) such that the matrix for $T$ is symmetric then $T$ is self adjoint?

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No. Choose any diagonalizable operator $T \colon V \rightarrow V$ which is not self-adjoint. This means that there is a basis $\mathcal{B} = (v_1, \ldots, v_n)$ for $V$ consisting of eigenvectors of $T$ but the basis $\mathcal{B}$ is not orthonormal. Then $[T]_{\mathcal{B}}$ is diagonal and in particular, symmetric, but $T$ won't be self-adjoint. For example, you can take $V = \mathbb{R}^2$ with the standard inner-product and $T(x,y) = (2x + y, 3y)$.

In fact, $T \colon V \rightarrow V$ is diagonalizable with real eigenvalues if and only if there exists a basis $\mathcal{B}$ of $V$ such that $[T]_{\mathcal{B}}$ is Hermitian.

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