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Question: Prove or give a counterexample: If $T \in \mathcal{L}(V)$ is self-adjoint and there exists an orthonormal basis $e_1, \dots, e_n$ such that $\langle Te_j, e_j \rangle \geq 0$ for each $j$, then $T$ is a positive operator.

I'm struggling to come up with a counterexample for this. I have an idea of using $T \in \mathcal{L}(\mathbb{R}^2)$ given by $T(x,y) = (x, -y)$. So its basis is $ \mathcal{M}(T) = \begin{pmatrix}1 & 0\\0 & -1\end{pmatrix}$ and it's symmetric so it's self adjoint. How should I show the orthonormal part? And most importantly, do you think this counterexample work? Otherwise how should I approach this?

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Just take the diagonal entries as $1$ each and non-diagonal entries $2$ each ($2\times2$ matrix). If $\{e_1,e_2\}$ is the standard basis then you get a counterexample because the determinant is $-3$. (For a positive definite meatric determinant cannot be negative).

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  • $\begingroup$ I don't quite understand what you mean? Could you please clarify? $\endgroup$
    – user589759
    Nov 15, 2018 at 6:20
  • $\begingroup$ Eigen values of a positive operator are non-negative and determinant is the product of the eigen values. $\endgroup$ Nov 15, 2018 at 6:22
  • $\begingroup$ How would that relate to orthonormal basis? $\endgroup$
    – user589759
    Nov 15, 2018 at 6:24
  • $\begingroup$ $\langle Te_i,Te_j \rangle$ is nothing but the $(i,j)$ element of the matrix which represents $T$ relative to the basis $\{e_1,e_2\}$. The condition you want is that the diagonal elements are non-negative but the matrix is not positive definite. $\endgroup$ Nov 15, 2018 at 6:28

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