In each of the following given are an inner product space $(V,\cdot)$ and a linear operator $T : V → V$.
Is $T$ symmetric? If the operator is symmetric, find an orthonormal basis of eigenvectors for $T$.
(a) For $M_{2\times 2}$ with $A \cdot B = \textrm{trace}(A^T B)$, let $T(A)=A^T$.
I know that an operator T is symmetric if $u \cdot T(v) = T(u) \cdot v$ for all $u$ and $v$. This is also equivalent to this equality holding for all vectors $u$ and $v$ in some orthonormal basis, and also to $M_B(T)$ being symmetric for some orthonormal basis. But I am not really sure what after.
It's easy to check that $$ T(A)\bullet B=\operatorname{trace}(AB) $$ and that $$ A\bullet T(B)=\operatorname{trace}(A^TB^T)=\operatorname{trace}((BA)^T) $$ Can you end the check?
Hints: $\operatorname{trace}(C)=\operatorname{trace}(C^T)$; $\operatorname{trace}(AB)=\operatorname{trace}(BA)$.
If $A$ is an eigenvector for $T$, then $A^T=\lambda A$, for some $\lambda$. Thus $a_{ji}=\lambda a_{ij}$ for all $i,j$ (with I hope obvious notation).
In particular $a_{ij}=\lambda a_{ji}=\lambda^2 a_{ij}$, hence $$ (1-\lambda^2)a_{ij}=0 $$ for all $i,j$. Since $A$ is nonzero, this implies $\lambda=\pm1$.
Can you go on?
Hint: an eigenvector is either a symmetric or an antisymmetric matrix.