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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

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The most concise answer I can think of does actually involve traces, though for contradiction. Note $$\mathrm{tr}(B^{-1}AB - A) = \mathrm{tr}I \implies 0=n$$

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    $\begingroup$ (+1) nice answer. what's more cool is the same proof works for field with finite characteristic $p$ as long as $p \not|\;n$. $\endgroup$ – achille hui May 8 '13 at 17:36
  • $\begingroup$ I had the same thought, but it seemed a shame to add another line to a short post! Are there counterexamples if $p=n$ say? Probably easy to find one for $p=n=2$ but hey, I'm lazy! $\endgroup$ – Sharkos May 8 '13 at 17:37
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    $\begingroup$ There are 12 pairs of counterexamples for $p = n = 2$. For example, $$A = \begin{pmatrix}0&0\\0&1\end{pmatrix}\quad\text{ and }\quad B = \begin{pmatrix}0&1\\1&0\end{pmatrix}$$ $\endgroup$ – achille hui May 8 '13 at 18:51
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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.

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  • $\begingroup$ This isn't quite complete; $A$ and $A-1$ might have no eigenvalues in $\mathbb R$. You'd have to consider the matrices' action on complex vectors. $\endgroup$ – Sharkos May 8 '13 at 17:13
  • $\begingroup$ Where did I say they have eigenvalues in $\mathbb{R}$? It doesn't matter. It suffices e.g. to take the trace of $A-1$ and $A$, or to consider the eigenvalue of $A$ with the smallest real part. $\endgroup$ – Start wearing purple May 8 '13 at 17:15
  • $\begingroup$ I meant it's possible that "They don't have eigenvalues PERIOD" considered as matrices acting on $\mathbb R^n$. In general, eigen-stuff isn't really relevant here. $\endgroup$ – Sharkos May 8 '13 at 17:18
  • $\begingroup$ It's true that we can do without eigenvalues, and I agree that in this sense your solution is nicer. $\endgroup$ – Start wearing purple May 8 '13 at 17:20
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    $\begingroup$ ah ok now I finally see the source of confusion. By eigenvalues I meant (complex) roots of the characteristic polynomial, not implying the existence of eigenvectors. $\endgroup$ – Start wearing purple May 8 '13 at 17:30
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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 $\det(B)=0$.

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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 $Bx=0$.

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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

$B$ is nilpotent $\iff$ $\text{tr}(B^k)=0$ for any $1 \leq k \leq n.$

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(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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  • $\begingroup$ @user1551, Oh, I forgot raising the power to the $B$ inside the operator norm. Thank you, I fixed it. $\endgroup$ – Sangchul Lee Sep 3 '13 at 15:40
  • $\begingroup$ @user1551, you are right. I muse have overlooked some delicate aspect of the matrix power. I can modify my argument to only show that $AB - BA = B$ implies $\det B = 0$. $\endgroup$ – Sangchul Lee Sep 3 '13 at 16:02

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