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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 $(AB)^{-1}$

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


$I=B^{-1} +I$ 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!

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$(A+BA)$ does not equal $A(I+B)$. You also assume $B$ is invertible without justification. – Chris Eagle Mar 27 '12 at 19:21
Even if you assume that $A$ has an inverse, how do you conclude that $AB$ has an inverse? – Arturo Magidin Mar 27 '12 at 19:22
What you do have is $ABA^{-1} = I + B$... – Dustan Levenstein Mar 27 '12 at 19:23
up vote 7 down vote accepted

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 $(I+B)v = (1+\lambda) v$.

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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. $\det A=0$.

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Thanks a lot. it is very clear now – Christine Mar 27 '12 at 19:53

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 $Ax_k=0$?

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