Question

Prove that if (i - ab) is invertible, then i - ba is invertible

Answer 1

Answer:

$(I-BA)$ is invertible and its inverse is given by $(B(I-AB)^{-1}A+I)$.

Step-by-step explanation:

Suppose $(I-AB)$ is invertible, that is, there exists $(I-AB)^{-1}$.

Then $I+B(I-AB)^{-1}A+I$ exists either. We will show that $I+B(I-AB)^{-1}A+I$ is the inverse of $I-BA$ by computing the product:

$(I-BA)(B(I-AB)^{-1}A+I) = B(I-AB)^{-1}A+I-BAB(I-AB)^{-1}A-BA$

$B((I-AB)^{-1}-AB(I-AB)^{-1})A+I-BA = B((I-AB)(I-AB)^{-1})A+I-BA$

Since (I-AB)(I-AB)^{-1} = I,

$B((I-AB)(I-AB)^{-1})A+I-BA = BA+I-BA = I$

Then $(I-BA)(B(I-AB)^{-1}A+I) = I$

It is completely analogous to prove that

$(B(I-AB)^{-1}A+I)(I-BA) = I$

Therefore, we explicit the inverse.

That is, $(B(I-AB)^{-1}A+I) = (I-BA)^{-1}$ and $(I-BA)$ is invertible.