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Let $A$ and $B$ be $n\times n$ matrices over the field $F$. Prove that if $I-AB$ is invertible, then $I-BA$ is invertible and $(I-BA)^{-1}=I+B(I-AB)^{-1}A$.

My attempt: Since $M_{n\times n}(F)$ is linear algebra over $F$, we have $$\begin{align*} (I-BA)\cdot (I+B(I-AB)^{-1}A) &=I-BA+(I-BA)\cdot (B(I-AB)^{-1}A)\\ &=I-BA+B(I-AB)^{-1}A-BAB(I-AB)^{-1}A\\ &=I-BA+B[(I-AB)^{-1}-AB(I-AB)^{-1}]A\\ &=I-BA +B[(I-AB)(I-AB)^{-1}]A\\ &=I-BA+BA \\&=I \end{align*}$$ Similarly $(I+B(I-AB)^{-1}A)\cdot (I-BA)=I$. Hence $I-BA$ is invertible and $(I-BA)^{-1}=I+B(I-AB)^{-1}A$. Is my proof correct?

Que: $I-BA$ is invertible$\iff$$\text{det}(I-BA)\neq 0_F$. Can we prove $\text{det}(I-BA)\neq 0_F$, given $\text{det}(I-AB)\neq 0_F$? If we can show that, then we don’t have to show $(I+B(I-AB)^{-1}A)\cdot (I-BA)=I$ in above attempt, because of following lemma: Let $P,Q,R\in M_n(F)$. If $QP=PR=I$, then $Q=R$.

user264745
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  • The proof cannot be accepted because we need to prove that $I-BA$ is invertible. Yet you begin with $(I-BA)\cdot(I-BA)^{-1}= \cdots$. This implicitly assumes that $(I-BA)^{-1}$ exists, which is the same as saying that $(I-BA)$ is invertible. We just cannot do that. – Li Kwok Keung Nov 03 '22 at 15:00
  • @LiKwokKeung yess.. I should have use different notation for $I+B(I-AB)^{-1}A$. I have edited my post – user264745 Nov 03 '22 at 15:06
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    I think simplify showing that $$(I-BA)\cdot\left( I+B(I-AB)^{-1}A\right)=I$$and the other way round is sufficient. – Li Kwok Keung Nov 03 '22 at 15:12
  • @LiKwokKeung Without knowing invertibility of $I-BA$, it is not sufficient to only show $(I-BA)\cdot (I+B(I-AB)^{-1}A)=I$, I think. – user264745 Nov 03 '22 at 15:13
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    The inverse of a matrix $X$ is defined to be the matrix $Y$ such that $XY=YX=I$.

    Once we have shown that$(I-BA)\cdot(I+(I-AB)^{-1}A)=I$ and $(I+(I-AB)^{-1}A)\cdot(I-BA)=I$. We can say that $(I+(I-AB)^{-1}A$ is the inverse of $I-BA$

     – Li Kwok Keung Nov 03 '22 at 15:17
  • @LiKwokKeung yeah you’re correct. I’m trying to make my proof more efficient by first proving $I-BA$ is invertible. Read Que part in post. There is different equivalent definition of invertible matrix. Like $\text{det}$ of that matrix is non zero. – user264745 Nov 03 '22 at 15:20
  • for your "que". Yes. You can even prove that $BA$ and $AB$ have the same characteristic polynomial. See e.g. the blocked argument under the bounty award here: https://math.stackexchange.com/questions/821934/eigenvalues-of-ab-and-ba-where-a-and-b-are-square-matrices . Alternatively depending on how much algebra you know, work over $\mathbb Z[\mathbf x]$, apply Newton's Identities and then do a substitution homomorphism into your field. – user8675309 Nov 03 '22 at 15:42
  • @user8675309 that is precisely next exercise in the textbook. That problem is generalization of Que part. I can’t even prove Que part $(\text{det}(I-AB)\neq 0_F\implies \text{det}(I-BA)\neq 0_F)$ by myself. I need hint. About alternative solution, I don’t know anything (for e.g. Newton’s identities). – user264745 Nov 03 '22 at 16:48
  • See https://math.stackexchange.com/q/2054109/121671 – Mittens Nov 03 '22 at 17:10
  • Your proof is fine. No need to prove that the determinant is non-zero. – azif00 Nov 03 '22 at 17:39
  • One more option: let $\det(I-AB)=0$. Suppose $(AB)$ has minimal polynomial $p(x)=q(x)(1-x)^k$ (i.e. $q$ has all roots of 1 factored out) and suppose for contradiction that $\det(I-BA)\neq0$ . Then $\mathbf 0=p(AB)=p(AB)\cdot A=A\cdot p(BA)=A \cdot q(BA)(I-BA)^k= A\cdot q(BA)$ $=q(AB)\cdot A=q(AB)\cdot (AB)$ which contradicts the uniqueness of the minimal polynomial of $AB$. (If you don't know minimal polynomials you can let $p$ be characteristic polynomial, $\mathbf 0=p(AB)$ by Cayley Hamilton and examine $q(AB)\cdot (AB)\cdot\mathbf v$ where $\mathbf v \in \ker(I-AB)$) – user8675309 Nov 04 '22 at 04:27
  • @user8675309 you have proved if $\text{det}(I-AB)=0$, then $\text{det}(I-BA)=0$. Which is same as if $\text{det}(I-BA)\neq 0$, then $\text{det}(I-AB)\neq 0$, but that is opposite of our desired result. I will learn about some notion (minimal polynomial & Cayley Hamilton) in next section of the textbook. – user264745 Nov 04 '22 at 09:11
  • @user264745 No. If you read it more carefully you'll see I proved the desired result not "the opposite". The underlying point is that the labeling of matrices is arbitrary, so what I have actually proven is $\det(I-AB)=0\implies \det(I-BA)=0$ and $\det(I-BA)=0\implies \det(I-AB)=0$ but you know $\det(I-AB)\neq 0$ thus $\det(I-BA)\neq 0$. – user8675309 Nov 04 '22 at 15:55
  • @user8675309 well that’s what essentially we’re trying to prove. “labeling of matrices is arbitrary” is same as saying $BA$ and $AB$ have same characteristic value. – user264745 Nov 04 '22 at 17:33
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    @user264745 No it is not. Stating this in very explicit language, re-label as follows: $A':= B$ and $B':=A$. My argument proves that $\det(I-A'B')=0\implies \det(I-B'A')=0$. These are very basic proof techniques. – user8675309 Nov 04 '22 at 18:03
  • @user8675309 ohh… I missed that, mainly because I don’t understand proof. I knew we have to change little bit in proof to make it work. – user264745 Nov 04 '22 at 18:25
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    @user264745 Fair enough. Perhaps it is worth revisiting these different proof approaches once you cover Cayley Hamilton. Proving the same thing multiple ways yields a lot of insight (sometimes). Good luck. – user8675309 Nov 04 '22 at 23:26

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