A matrix satisfying $AB-BA=B$
Andrew Henderson 
If $A$ and $B$ are two matrices of $\mathcal{M}_n(\mathbb{R}$) such that $$AB-BA=B$$ how can we prove that $B$ isn't invertible?
my attempt: I found that $\mathrm{tr}(B)=0$ but I know that this is insufficient. Thanks
$\endgroup$6 Answers
$\begingroup$The most concise answer I can think of does actually involve traces, though for contradiction. Note $$\mathrm{tr}(B^{-1}AB - A) = \mathrm{tr}I \implies 0=n$$
$\endgroup$ 3 $\begingroup$Suppose that $B$ is invertible, then $A-1=BAB^{-1}$. Hence $A-1$ and $A$ should have the same set of eigenvalues, which is impossible.
$\endgroup$ 7 $\begingroup$Here is a proof, not by contradiction. Let $C=A+kI$. The given assumption implies that $$CB = B(C+I).\tag{1}$$ When $k$ is sufficiently large, $\det(C)\neq\det(C+I)$ and both determinants are nonzero. Hence $(1)$ implies that $\det(B)=0$ must be zero. Alternatively, pick a (perhaps complex) number $k$ such that $C$ is singular but $C+I$ is not. Again, $(1)$ implies that $\det(B)=0$.
$\endgroup$ $\begingroup$Here is another proof, which is rewritten from O.L.'s. It does not only prove that $B$ is singular, but it also gives a nontrivial solution to $Bx=0$.
The equation $AB-BA = B$ implies that $$AB = B(A+I).\tag{1}$$ Now, if $(\lambda,v)$ is an eigenpair (which is perhaps complex) of $A$ and $Bv\neq0$, then it follows from $(1)$ that $(\lambda+1,\,Bv)$ is also an eigenpair of $A$. Continue in this manner, we get a chain of eigenpairs $(\lambda,v),(\lambda+1,\,Bv),(\lambda+2,\,B^2v),\ldots$ of $A$. Since $A$ (as a complex matrix) has only finitely many eigenvalues, we must have $B^kv=0$ for some positive integer $k$. Pick the smallest such $k$ and let $x$ be the real or imaginary part (whichever is nonzero) of $B^{k-1}v$, we get a nontrivial solution to $Bx=0$.
$\endgroup$ $\begingroup$The following works over any algebraically closed field.
Show by (strong) induction that that $AB^k-B^kA=m_kB^k$ for any $k \in \mathbb{N}$ and some $m_k\in \mathbb{N},$ and then use the fact that
$\endgroup$ $\begingroup$$B$ is nilpotent $\iff$ $\text{tr}(B^k)=0$ for any $1 \leq k \leq n.$
(As I found a silly mistake, I made amend to my argument.)
It is quite a late answer: By induction, it is easy to check that
$$AB^{n} = B^{n}(A + nI) \quad \text{for }n = 1, 2, 3, \cdots.$$
Thus if $\det B \neq 0$, then
$$ \det(A) = \frac{\det(A)\det(B^{m})}{\det(B^{m})} = \det (A + nI). $$
This implies that the characteristic polynomial of $A$ has infinitely many zeros, which is clearly absurd.
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