# Positive operator, self adjoint and orthonormal basis

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?

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).
• $\langle Te_i,Te_j \rangle$ 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. Nov 15, 2018 at 6:28