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The question asks how does the identity $A(I+BA)=(I+AB)A$ connect the inverses of $I+BA$ and $I+AB$, and I am able to express $(I+BA)^{-1}$ as $A^{-1}(I+AB)^{-1}A$. However, I fail to see how this tells that $I+AB$ and $I+BA$ are either both invertible or both singular. Can anyone give me a hint? I really appreciate it!

lulu
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Xiangnong Wu
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    Are you assuming that $A$ is invertible? If not, what do you mean by $A^{-1}$? – lulu May 21 '24 at 21:09
  • Yeah I think I assumed that A is invertible but the problem doesn't say so. – Xiangnong Wu May 21 '24 at 21:13
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    Remark: Many books are called Introduction to Linear Algebra. Give the author name as well. – Sean Roberson May 21 '24 at 21:13
  • Sorry! The author is Gilbert Strang. – Xiangnong Wu May 21 '24 at 21:15
  • Take the determinant on both sides. If $A$ is invertible, this will tell you that either both $I+AB$ and $I+BA$ are invertible or they are both singular. In fact, if $A$ is invertible, this tells you that both $I+AB$ and $I+BA$ have the same determinant. – AnCar May 21 '24 at 21:16
  • More generally, you can think of this equation as saying that $I+AB$ and $I+BA$ are conjugates of each other if $A$ is invertible, since you can write $I+BA=A^{-1}(I+AB)A$. You will see later on, when you look at the properties of the characteristic polynomial, that conjugated matrices share the same characteristic polynomial. In particular the free term in the characteristic polynomial of a matrix $M$ is, up to a sign, the determinant of $M$. – AnCar May 21 '24 at 21:22
  • You may take a look at https://math.stackexchange.com/a/4079591/316749 where you also find many links. – Hanno May 21 '24 at 21:25
  • Suppose $x \in \ker (I+BA)$ and $(I+AB)$ is invertible $\implies x \in \ker A$. Then $Ax = \mathbf 0 = (I+BA)x = Ix +B(Ax) = x+\mathbf 0$, i.e. $\dim \ker (I+BA) = 0$. Then when $(I+BA)$ is invertible suppose $x^T \in \text{left nullspace}(I+AB)$... – user8675309 May 22 '24 at 04:07
  • with more algebra, https://math.stackexchange.com/questions/17831/how-can-we-prove-sylvesters-determinant-identity/17837 and https://math.stackexchange.com/questions/311342/do-matrices-ab-and-ba-have-the-same-minimal-and-characteristic-polynomia/ also give the answer – user8675309 May 22 '24 at 04:13

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