A linear transformation $T:\mathbb{V}$ $\rightarrow$ $\mathbb{V}$ on an inner product space is called self-adjoint if: $\langle Tx,y \rangle = \langle x,Ty \rangle$ for all $x,y \in V$. Prove that $T:\mathbb{R}^n\rightarrow\mathbb{R}^n$ is self-adjoint if and only if its associated matrix (in the standard basis) is symmetric.
I'm not too sure how to tackle this, I'm guessing using the properties of a matrix being symmetric in inner product terms is a starting point. i.e. $\langle$$Ax,y$$\rangle$ = $\langle$$x,Ay$$\rangle$ where $A\in{M}_n(\mathbb{R})$.
Best Answer
Hint: the standard inner product on $\mathbb{R}^n$ is just matrix multiplication, by $$\langle u,v\rangle=u^Tv.$$