1
$\begingroup$

Hey guys, I need to prove or refute that once given an eigenvalue t of a matrix AB and B is invertible,

so t is also eigenvalue of A.

I believe it's not true, but sadly beliefs are not enough in math : )

Thank you.

  • 5
    If you believe it's not true, you should try to come up with a counterexample. Here's a hint: try the simplest thing that could possibly work—let $A = [a]$ and $B = [b]$ be $1\times 1$ matrices.2011-03-22

1 Answers 1

1

Let $A=B^{-1}$, then $t=1$. Clearly there are invertible matrices $B$ which don't have $1$ as an eigenvalue.

  • 0
    I see, and from this we get that $1$ cannot be a eigenvalue of A also. Thanks.2011-03-22