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In this question, I denote $$A=\begin{bmatrix}a&b\\c&d\end{bmatrix},$$ $$B=\begin{bmatrix}e&f\\g&h\end{bmatrix}.$$ So, from $|I_2 + AB| = 0$, I'll have: $$(ae+fc+1)(gb+hd+1) = (ga+hc)(eb+fd).$$

If $|I_2 + BA| = 0$ (which we have to prove here), then $$(ae+bg+1)(cf+hd+1) = (af+bh)(ce+dg).$$ From both equations, I can see that what we're going to prove here is $cf=bg$, more precisely, $g=f$ and $b=c$.

Is my thinking path right or wrong? I still have not figured out the solution yet, I appreciate any help.

Gary
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Thu Nguyễn
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    Have you studied eigenvalues? If you do, try to think of what $\det(I_2+AB)=0$ means in terms of eigenvalues. – Marcos Dec 15 '21 at 00:15
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    See https://mathoverflow.net/questions/31595/how-would-you-solve-this-tantalizing-halmos-problem and https://math.stackexchange.com/questions/1078781/proof-if-ab-i-invertible-then-ba-i-invertible – lhf Dec 15 '21 at 00:16
  • Regarding your approach so far, there is no reason to believe that we have $g = f$ or $b = c$. As an example, if we take $$ A = B = \pmatrix{0&-1\1&0}, $$ then it is true that $|I + AB| = 0$, but it is not true that $g = f$ or $b = c$. – Ben Grossmann Dec 15 '21 at 15:28
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    Proof on wiki: https://en.wikipedia.org/wiki/Weinstein%E2%80%93Aronszajn_identity – ComptonScattering Dec 15 '21 at 15:35
  • Order 2 like $A^{2} = I$? – Elmex80s Dec 15 '21 at 16:25

3 Answers3

1

We have $$ |I_2 + AB| = a e (d h + 1) - b c f g + d h + 1 $$ which is clearly invariant under swapping $a\leftrightarrow e$, $b\leftrightarrow f$, $c\leftrightarrow g$, $d\leftrightarrow h$. Hence $$ |I_2 + AB| = |I_2 + BA| $$

lhf
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Note that $|I + AB| = 0$ if and only if $I + AB$ is non-invertible, which is true if and only if the equation $(I + AB) x = 0$ has a non-zero solution $x$.

Suppose that $x \neq 0$ is such that $(I + AB)x = 0$. We see that $$ Ix + ABx = 0 \implies ABx = -x. $$ Note that because $A(Bx) = -x \neq 0$, it must be the case that $Bx \neq 0$. Using our observations so far, note that $$ BA(Bx) = B(AB)x = B(-x) = -Bx. $$ In other words, $Bx \neq 0$, and we have $$ BA(Bx) = -Bx \implies IBx + BABx = 0 \implies (I + BA)Bx = 0. $$ So, the equation $I + BA y = 0$ has a non-zero solution. It follows that $|I + BA| \neq 0$, which was what we wanted.

Ben Grossmann
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It is just a corollary (taking $n = m = 2$) of the result below:

For $A \in F^{m \times n}, B \in F^{n \times m}$, it follows that \begin{equation} \det(I_{(m)} + AB) = \det(I_{(n)} + BA). \tag{$*$} \end{equation}

To prove identity $(*)$, consider the block matrix: \begin{align} \Delta = \begin{pmatrix} I_{(m)} & A \\ -B & I_{(n)} \end{pmatrix}, \end{align} and verify decompositions below: \begin{align*} & \begin{pmatrix} I_{(m)} & A \\ -B & I_{(n)} \end{pmatrix} \begin{pmatrix} I_{(m)} & 0 \\ B & I_{(n)} \end{pmatrix} = \begin{pmatrix} I_{(m)} + AB & A \\ 0 & I_{(n)} \end{pmatrix}, \tag{1} \\ & \begin{pmatrix} I_{(m)} & A \\ -B & I_{(n)} \end{pmatrix} \begin{pmatrix} I_{(m)} & -A \\ 0 & I_{(n)} \end{pmatrix} = \begin{pmatrix} I_{(m)} & 0 \\ -B & I_{(n)} + BA \end{pmatrix} \tag{2} \end{align*}

Taking determinants on both sides of $(1)$ and $(2)$ then yields $$\det(\Delta) = \det(I_{(m)} + AB) = \det(I_{(n)} + BA).$$

Zhanxiong
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