Show that if a self-adjoint matrix is triangular, then it must be diagonal. I want to show that if a self-adjoint matrix is triangular, then it must be diagonal.
Here's my thoughts so far:
Self adjoint is simply $\langle A \pmb x, \pmb y \rangle = \langle \pmb x, A \pmb y\rangle$
If $A$ is lower triangular then $\langle \pmb b, \pmb y \rangle = \langle \pmb x , \pmb c\rangle$ where
$$
\pmb b = \begin{bmatrix}a_{11}x_1 \\ a_{21} x_2 + a_{22} x_2 \\ \vdots \\ \sum_{i=1}^n a_n x_i\end{bmatrix}
$$
We don't really know anything about $\pmb c$ beyond that $A \pmb y = \pmb c$ and the equality shown above.
Intuitively, I want to think of the inner product as being characterized by the identity matrix because then I'd have a straightforward system of equations that would show that $A$ must only have entries along the diagonal...
Another thought is to somehow show that $A$ must be both upper and lower triangular to be self-adjoint which would imply it's diagonal.
I'm not sure how to advance from here.

Hint: Take $x$ to be the standard basic vector $e_i$ and $y$ to be the standard basic vector $e_j$. So $\langle e_i, Ae_j\rangle=\langle Ae_i, e_j\rangle$. This should give you some info about the matrix $A$.
