Finding a basis of the vector space of $\mathbb{C}G$ homomorphisms. Let $G=C_4=\langle g\mid g^4=1\rangle$ and let $V_1,V_2$ be two $\mathbb{C}G$-modules with bases $\mathcal{B}_1=\{v_1,v_2\}$ and $\mathcal{B}_2=\{w_1,w_2\}$ respectively. Define the action of $G$ on $V_1,V_2$ by $[g]_{V_1}=\left(\begin{matrix}0 & 1\\ -1 & 0\end{matrix}\right)$ and $[g]_{V_2}=\left(\begin{matrix}-3 & 8\\ -1 & 3\end{matrix}\right)$.
Find a basis of $\operatorname{Hom}_{\mathbb{C}G}(V_1,V_2)$.
Attempt:
If $A\in $Hom$_{\mathbb{C}G}(V_1,V_2)$ then we must have $A[g]_{V_1}=[g]_{V_2}A$ so if $A=\left(\begin{matrix}a & b\\ c & d\end{matrix}\right)$ then;
$\left(\begin{matrix}-b & a\\ -d & c\end{matrix}\right)=\left(\begin{matrix}-3a+8c & -3b+8d\\ -a+3c & -b+3d\end{matrix}\right)$.
Then we have 4 equations:
$-3a+8c+b=0$
$-3b+8d-a=0$
$-a+3c+d=0$
$-b+3d-c=0$
But these equations result in $a=b=c=d=0$, so does this mean that the set of homomorphisms is just the zero homomorphism, or have I gone wrong?
 A: The solution is indeed that the space of homomorphisms is trivial, and your approach is completely correct.

An alternative perspective.  Let $P$ be a matrix such that $P^{-1}[g]_{V_1}P$ is diagonal.  We then have
$$
A[g]_{V_1} = [g]_{V_2}A \iff\\
(P^{-1}AP)(P^{-1}[g]_{V_1}P) = (P^{-1}[g]_{V_2}P)(P^{-1}AP).
$$
Because $(P^{-1}[g]_{V_1}P)$ is diagonal, the above will hold if and only if the columns of $P^{-1}AP$ are either zero-vectors or eigenvectors of $P^{-1}[g]_{V_2}P$ associated with the eigenvalues $\pm i$ of $[g]_{V_1}$.
However, the eigenvalues of $[g]_{V_2}$ are $\pm 1$, which means that the only matrix satisfying the above criteria is the zero matrix.
A: Your attempt is correct, you have not gone wrong. An easier way to reach the same conclusion is to note that $A$ must also commute with $g^2$; both $[g]_{V_1}^2$ and $[g]_{V_2}^2$ are diagonal matrices, making the computations a lot easier while still reaching the conclusion that $A=0$.
In fact, having computed the squares of both matrices, there is no more need for matrices at all:
Computing the squares of the two matrices shows that for all $v\in V_1$ and $v'\in V_2$ you have
$$g^2v=-v\qquad\text{ and }\qquad g^2v'=v'.$$
It follows that for all $f\in\operatorname{Hom}_{\Bbb{C}G}(V_1,V_2)$ and all $v\in V_1$ you have
$$f(v)=g^2f(v)=f(g^2v)=f(-v)=-f(v),$$
so $f(v)=0$. This shows that indeed $\operatorname{Hom}_{\Bbb{C}G}(V_1,V_2)=0$, and the empty set is a basis.
