Final answer:
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.