Let A and B be invertible n × n matrices. Prove that In − AB is also invertible if
and only if A−1 − B is invertible.

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.

Let A and B be invertible n × n matrices. Prove that In − AB is also invertible if and only if A−1 − B is invertible.

Explore the proof that for invertible n × n matrices A and B, In − AB is invertible if and only if A−1 − B is invertible. This solution covers essential linear algebra concepts including matrix invertibility and product of matrices. Suitable for undergraduate students in linear algebra.

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