Zero determinant of a matrix

Suppose that for the complex square matrices $A, B$ it holds

(1) $\begin{equation*} AB - BA =A \end{equation*}$

Prove that $\det A =0$ .

Solution

If $\det A\ne0$ , then $A$ is invertible. So we get

$I=A^{-1}A=A^{-1}(AB-BA)=B-A^{-1}BA$

But this equality is impossible: taking trace on both sides, we get

$n=\text{Tr}(I)=\text{Tr}(B-A^{-1}BA)=\text{Tr}(B)-\text{Tr}(A^{-1}BA)=\text{Tr}(B)-\text{Tr}(B)=0$

We get a contradiction, which shows that $A$ is singular, i.e. $\det A=0$ .

Series of Bessel function