Does every bilinear (symmetric) form admit a self-adjoint map representation? Let $Q(x,y)$ be a bilinear (symmetric) form on a finite dimensional vector space $V$, $Q: V \times V \to \mathbb{R}$, is it true that it will admit a linear self-adjoint map $K$ such that
$$(K(x),y) = Q(x,y)$$?
This looks like some sort of Rieze representation, but I am not too sure.
I found this claim in a proof I am reading, but I am not sure if this is true in general.

Your space has to be Euclidean. If you fix $x$ in your bilinear form, you get a linear form on $y$. By Riesz representation, this linear form is given by the inner product with a given vector $h(x)$. It is easy to see that the map $x\to h(x)$ is linear.
No coordinate representation is needed here.
