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I was just encountered with a rather tough problem as follows:

Suppose $A,B\in M_n(\mathbb R)$, prove:
$$\det(I_n+AB)\ne0\Rightarrow\det(I_n+BA)\ne0$$

Although at this moment I am still at a loss how to go about proving this, I seemed to have derived something that looks very stunning to me in my previous failed attemps:
First, I think this one is obvious: $$\det(I_n+AB)\ne0\Leftrightarrow-1\text{ is not an eigenvalue of }AB$$ And likewise, $$\det(I_n+BA)\ne0\Leftrightarrow-1\text{ is not an eigenvalue of }BA$$ So what I'm asked to prove is actually equivalent to showing $$-1\text{ is not an eigenvalue of }AB\Rightarrow-1\text{ is not an eigenvalue of }BA$$ Taking the converse-negative, that is to say $$-1\text{ is an eigenvalue of }BA\Rightarrow-1\text{ is an eigenvalue of }AB$$ I cannot find any flaw in my reasoning. But if what I'm about to prove is true (I'm certain to say, yes it's true, because months ago I solved it in an extremely tricky way, which didn't, of course, follow my current threads), then it means for two arbitrary matrices $A,B$ of the same size, even if they don't commute, $AB$ and $BA$ will share $-1$ as an eigenvalue!! I think it is VERY unlikely.
So could you please point out where the flaw of my reasoning lies? Or, could you help me prove this tough thing? Thanks in advance!

Cameron Williams
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Vim
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2 Answers2

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the nonzero eigenvalues of $AB$ and $BA$ are the same. the reason is $tr(AB) = tr(BA), tr((AB)^2) = tr((BA)^2), \cdots$ implying that the the sum of the powers of eigenvalues of $AB$ and $BA$ are the same. through the newtons formula, this implies that the leading coefficients of characteristic polynomials are the same.

abel
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Say $v\in \Bbb R^n$ is such that $ABv = -v$. Then I claim that $u = Bv$ is a vector such that $BAu = -u$. In fact: $$ BAu = BABv = B(ABv) = B(-v) = -Bv = -u $$ So if $AB$ has eigenvalue $-1$, then so does $BA$. The opposite is of course true as well, by an analoguous argument. This argument also works for any other eigenvalue, so $AB$ and $BA$ share all eigenvalues.

Small caveat: we actually have to check that $u \neq 0$. But as long as the eigenvalue in question is not $0$, we have $A(Bv) \neq 0$, so it follows that $Bv = u \neq 0$.

Arthur
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