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Prompt

Provide me stepwise solution of question 1.49.

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## Problem 1.49

**Question:** Let \( A \in \mathbb{C}^{m \times n} \) and \( B \in \mathbb{C}^{n \times m} \). Show that \( \det(I_m + AB) = \det(I_n + BA) \).

**Solution:**

1. **Step 1:** Recall Theorem 1.32 which states that for matrices \( A \in \mathbb{C}^{m \times n} \) and \( B \in \mathbb{C}^{n \times m} \), the matrices \( AB \) and \( BA \) have the same nonzero eigenvalues.
2. **Step 2:** If \( \lambda \) is an eigenvalue of \( AB \), then \( 1 + \lambda \) is an eigenvalue of \( I_m + AB \). Similarly, if \( \lambda \) is an eigenvalue of \( BA \), then \( 1 + \lambda \) is an eigenvalue of \( I_n + BA \).
3. **Step 3:** Since \( AB \) and \( BA \) have the same nonzero eigenvalues, it follows that \( I_m + AB \) and \( I_n + BA \) have the same eigenvalues that are not equal to 1.
4. **Step 4:** The determinant of a matrix is the product of its eigenvalues. Since \( I_m + AB \) and \( I_n + BA \) have the same set of non-unit eigenvalues (from Theorem 1.32), and any extra eigenvalues equal to 1 multiply to 1, we have:
\[
\det(I_m + AB) = \det(I_n + BA)
\]
5. **Step 5 (Alternative proof using matrix identity):** Consider the matrix identity:
\[
\begin{bmatrix}
I_m & -A \\
0 & I_n
\end{bmatrix}
\begin{bmatrix}
I_m & A \\
B & I_n
\end{bmatrix}
\begin{bmatrix}
I_m & 0 \\
-B & I_n
\end{bmatrix}
=
\begin{bmatrix}
I_m - AB & 0 \\
0 & I_n
\end{bmatrix}
\]
6. **Step 6:** Taking determinants of both sides:
\[
\det\left(\begin{bmatrix}
I_m & -A \\
0 & I_n
\end{bmatrix}\right)
\det\left(\begin{bmatrix}
I_m & A \\
B & I_n
\end{bmatrix}\right)
\det\left(\begin{bmatrix}
I_m & 0 \\
-B & I_n
\end{bmatrix}\right)
=
\det\left(\begin{bmatrix}
I_m - AB & 0 \\
0 & I_n
\end{bmatrix}\right)
\]
7. **Step 7:** Evaluate each determinant separately:

For block triangular matrices:
\[
\det\left(\begin{bmatrix}
I_m & -A \\
0 & I_n
\end{bmatrix}\right) = \det(I_m)\det(I_n) = 1
\]

\[
\det\left(\begin{bmatrix}
I_m & 0 \\
-B & I_n
\end{bmatrix}\right) = \det(I_m)\det(I_n) = 1
\]

For the middle matrix, using the block matrix determinant formula:
\[
\det\left(\begin{bmatrix}
I_m & A \\
B & I_n
\end{bmatrix}\right) = \det(I_n + BA)
\]

For the right-hand side:
\[
\det\left(\begin{bmatrix}
I_m - AB & 0 \\
0 & I_n
\end{bmatrix}\right) = \det(I_m - AB)\det(I_n) = \det(I_m - AB)
\]
8. **Step 8:** Substitute these results into the equation from Step 6:
\[
1 \times \det(I_n + BA) \times 1 = \det(I_m - AB)
\]
\[
\det(I_n + BA) = \det(I_m - AB)
\]
9. **Step 9:** Replace \( B \) with \( -B \) to get the final result:
\[
\det(I_n + BA) = \det(I_m + AB)
\]

**Conclusion:** We have shown using two different methods that \( \det(I_m + AB) = \det(I_n + BA) \) for any matrices \( A \in \mathbb{C}^{m \times n} \) and \( B \in \mathbb{C}^{n \times m} \).

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Problem #11