I was unable to solve this question of linear algebra and hence I am posting it here.
Let A and B be n$\times $ n matrices over $\mathbb{C}$.Then which of the following are true:
A AB and BA have same set of Eigenvalues.
B If AB and BA have same set of eigenvalues then BA= AB.
C If $A^{-1}$ exists then AB and BA are similar
D The rank of AB is always equal as rank of BA.
I have done (C) . Tried some matrices for D, but couldn't contradict it. Fr A,B I don't have any idea so, your help is required.
Please guide me!
Best Answer
$AB$ and $BA$ have same set of Eigenvalues
True.
See here
If $AB$ and $BA$ have same set of eigenvalues then $BA$=$AB$.
False.
From (A), the if part in (B) is trivially true for any two matrices. So you just need to pick any two non-commuting matrices as counter-example. Google Image has a good number of those. (But funny you have to search examples of commutative matrices.)
The rank of $AB$ is always equal as rank of $BA$.
False.
The easiest way to spot why this is false is to consider matrices as linear operators. If $A$ has non-trivial nullspace, and $B$ happens to map into that, yet A still maps out of the nullspace of $B$...
For example, consider
$$ A=\begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 1 & 0 & 0 \end{bmatrix}, B=\begin{bmatrix} 0 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix} $$
In $\mathbb{R}^3, A$ only keeps 1st coordinate and throws it to 3rd. $B$ keeps 2nd & 3rd coordinate. Going through $A$ first and then $B$, you will end with 1st coordinate on 3rd slot. In other words, rank 1. Going through $B$ first and then $A$, since you lose 1st coordinate right away, you end with zero. In other words, rank 0.