To prove that $I_n - AB$ is invertible if and only if $A^{-1} - B$ is invertible, we can proceed as follows:

Forward Direction: If $I_n - AB$ is invertible, then $A^{-1} - B$ is invertible.

Assume that $I_n - AB$ is invertible. We need to show that $A^{-1} - B$ is also invertible.

1. Start with the assumption: $(I_n - AB) \text{ is invertible.}$

2. Consider the matrix product $A(I_n - AB)$: $A(I_n - AB) = A - AAB = A - AB.$

3. Rewrite the expression $A(I_n - AB)$: $A(I_n - AB) = A - AB = A - A(B).$

   Notice that the matrix on the right-hand side can be factored: $A(I_n - AB) = A(I_n - B).$

4. Since $I_n - AB$ is invertible, there exists a matrix $C$ such that: $C(I_n - AB) = I_n.$

5. Multiply both sides by $A$: $AC(I_n - AB) = A.$

6. From the expression above, we see that $AC = (A^{-1} - B)^{-1}$: $(A^{-1} - B)^{-1} = AC.$ Since $AC$ is invertible (as a product of invertible matrices), $A^{-1} - B$ must also be invertible.

Reverse Direction: If $A^{-1} - B$ is invertible, then $I_n - AB$ is invertible.

Now assume that $A^{-1} - B$ is invertible. We need to show that $I_n - AB$ is also invertible.

1. Start with the assumption: $A^{-1} - B \text{ is invertible.}$

2. Multiply both sides of the equation $A^{-1} - B$ by $A$: $A(A^{-1} - B) = AA^{-1} - AB = I_n - AB.$

3. Since $A(A^{-1} - B) = I_n - AB$ and $A^{-1} - B$ is invertible, then $I_n - AB$ is a product of invertible matrices $A$ and $A^{-1} - B$, hence $I_n - AB$ is also invertible.

Conclusion:

Thus, we have shown that $I_n - AB$ is invertible if and only if $A^{-1} - B$ is invertible.

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Do you want more details or have any questions?

Here are some related questions you might find interesting:

1. What properties do invertible matrices satisfy?
2. How can we use determinant properties to determine the invertibility of a matrix?
3. Can you generalize the result to block matrices?
4. How does the proof change if $A$ and $B$ are not square matrices?
5. What are some applications of this result in linear algebra?

Tip: When working with matrix proofs, remember that the invertibility of a matrix product depends on the invertibility of the individual matrices involved.