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If we have $A_{n \times n}$ and $AX-XA=A$, how do we show that $\det(A)=0$?

I've tried taking the trace on both sides to get

$$tr(AX)-tr(XA)=tr(A)$$ $$tr(AX)-tr(AX)=tr(A)$$ $$0=tr(A)$$

However, that doesn't necessarily show that $\det(A)=0$

sai-kartik
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John Rawls
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    I'm not sure whether to mark this as a duplicate. Nilpotent matrices necessarily have determinant 0. $A$ commutes with its commutator, hence is nilpotent -- this is known as Jacobson's Lemma. See e.g. here https://math.stackexchange.com/questions/227984/if-a-and-ab-ba-commute-show-that-ab-ba-is-nilpotent – user8675309 Jun 25 '20 at 01:25
  • The answers posted so far seem to assume---very reasonably, given the context---that the matrices have entries in a field of characteristic $0$. In characteristic $p$, (1) if $p | n$ then the $n \times n$ identity matrix has trace $0$ (and so there is not contradiction in Andreas' and River's answers), and (2) it's possible for all of $\lambda, \lambda + 1, \lambda + 2, \ldots$ to be eigenvalues of a matrix, since there are only $p$ distinct numbers in that list. I'm curious whether the original statement is even true in positive characteristic. – Travis Willse Jun 25 '20 at 01:37
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    In fact it is not true; in characteristic $2$, the matrices $A = \pmatrix{1&1\\cdot&1}$, $X = \pmatrix{\cdot&1\1&1}$ satisfy $A X - X A = A$. – Travis Willse Jun 25 '20 at 01:50

3 Answers3

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Rewrite the equation as $A (X-I) = XA$. If $A$ is nonsingular, that implies $X - I = A^{-1} X A$, i.e. $X$ is similar to $X-I$. That says $\lambda$ is an eigenvalue of $X$ if and only if $\lambda$ is an eigenvalue of $X-I$, i.e. $\lambda+1$ is an eigenvalue of $X$. But this is impossible, as it would mean $\lambda+n$ is an eigenvalue of $X$ for every integer $n$, but $X$ has only finitely many eigenvalues.

Robert Israel
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Similar to Robert Israel's answer, but without explicitly using eigenvalues: If $A$ were invertible, we could multiply the given equation on the right by $A^{-1}$ to get $AXA^{-1}-X=I$. Taking the trace of both sides and remembering that $AXA^{-1}$ has the same trace as $X$ (and that trace is linear), we get that the trace of the identity matrix is $0$, a contradiction.

[Pedantic technicality: That final "contradiction" assumes that you don't allow $0\times0$ matrices. If you do allow them, the result is false, as the hypothesis holds when both $A$ and $X$ are the empty matrix, but the determinant of the empty matrix is $1$.]

Andreas Blass
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Suppose $\det A \ne 0$.

Left-multiplying $A^{-1}$ on both sides of $AX - XA = A$, we have $X - A^{-1}XA = I$.

Right-multiplying $A^{-1}$ on both sides of $AX - XA = A$, we have $AXA^{-1} - X = I$.

Add them up to get $AXA^{-1} - A^{-1}XA = 2I$.

Note that $\mathrm{Tr}(AXA^{-1}) - \mathrm{Tr}(A^{-1}XA) = \mathrm{Tr}(X) - \mathrm{Tr}(X) = 0$, and $\mathrm{Tr}(2I) \ne 0$.

Contradiction. (Q. E. D.)

River Li
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