Derivations on matrix algebra

Question

Let $M=M_2(\mathbb{C})$ and let $\delta:M \to M$ be a $\mathbb{C}-$linear map such that $\forall a,b \in M$ we have $\delta(ab)=\delta(a)b+a\delta(b)$. Prove that $\delta$ is of the form $\delta(m)=am-ma$ for some $a \in M$.

I've tried to mess around with elementary matrices, but I did not advance much.

Any help? Thanks!

Answer 1

Denote $[A,B]:=AB-BA$.

Hints:

1. Calculate the following ones to recover the matrix $A$ of the desired equation $\delta(M)=[A,M]$: $$\left[\pmatrix{a&b\\c&d},\ \pmatrix{1&0\\0&0}\right],\quad\quad \left[\pmatrix{a&b\\c&d},\ \pmatrix{0&1\\0&0}\right],\quad \dots $$
2. Note that if $A$ is a solution (i.e. $\delta(M)=[A,M]$ for all $M$) then $A+\lambda I$ is also a solution for any $\lambda\in\Bbb R$.
3. Prove that trace of $\delta(A)$ must be zero for any $A$