Matrix - Show $\det(A) =0$ I am a little stuck on this Matrix problem.
Suppose that for complex square matrix A,B the following holds:  
$AB -BA = A$ Show that $\det(A)=0$
That would mean that A has no inverse. So I thought, let's suppose there exists an Inverse element and that would lead me to a contradiction later on.
$AB = A + BA \implies AB = A(I+B)$
multiply though with $(AB)^{-1}$
$I = B^{-1}A^{-1}A(I+B)$
$I=B^{-1}(I+B)$
$I=B^{-1} +I$ so $B^{-1}$ is equal to $0$. Contradiction in my view. Or what does this mean? Could you please help me interpret? :)
Thanks in advance!
 A: As I already noted, what you can derive from $AB = A+BA$ is $ABA^{-1} = I + B$. Conjugation must preserve eigenvalues, but adding $I$ clearly changes the set of eigenvalues, because if $B v = \lambda v$, then $(I+B)v = (1+\lambda) v$.
A: Hint #1: Assume that $x_1$ is an eigenvector of $B$ belonging to the eigenvalue $\lambda$. Show that $x_2=Ax_1$ is either an eigenvector of $B$ belonging to the eigenvalue $\lambda-1$, or  $x_2=0$. Look up ladder-operator.
Hint #2: Replace $x_1$ with $x_2$ in the former case, and study $x_3=Ax_2$. Repeat. Why must we eventually find a non-zero vector $x_k$ ($k$ some natural number) with the property that $Ax_k=0$?
A: Suppose that $AB-BA=A$. 
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. $\det A=0$.
