Question

Let A and B be n x n matrices.

(i) Let 1+0. Show that is an eigenvalue of AB if and only if it is also an eigenvalue of BA SEMESTER 1 MAIN, 2019 PAGE 19 OF 21
(ii) Show that I + AB is invertible if and only if I, + BA is invertible, where I. is the identity n x n matrix.

Answer 1

To show that 1+0 is an eigenvalue of AB if and only if it is also an eigenvalue of BA, we can use the fact that the determinant of a matrix and its transpose are the same.

Step-by-step explanation:

To show that 1+0 is an eigenvalue of AB if and only if it is also an eigenvalue of BA, we can use the fact that the determinant of a matrix and its transpose are the same. Let's assume that 1+0 is an eigenvalue of AB. This means that there exists a nonzero vector v such that ABv = (1+0)v. We can rearrange this equation to get BA(Bv) = (1+0)(Bv), which shows that 1+0 is an eigenvalue of BA.

Conversely, let's assume that 1+0 is an eigenvalue of BA. This means that there exists a nonzero vector w such that BAw = (1+0)w. We can rearrange this equation to get AB(Aw) = (1+0)(Aw), which shows that 1+0 is an eigenvalue of AB.

Therefore, 1+0 is an eigenvalue of AB if and only if it is also an eigenvalue of BA.