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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!

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    What does it mean for the rank of a matrix to be similar to the rank of another? – mi.f.zh Nov 20 '20 at 10:21
  • @nhmwhhxx It was equal ,sorry! –  Nov 20 '20 at 10:53
  • I think the main tool that you're supposed to use here is the characteristic polynomial. For example, your first question follows from the fact that $AB$ and $BA$ have the same polynomial and hence the same roots (eigenvalues). Moreover, your second question is probably false due to the same reason, just havce to think of a 2x2 example. – Ariel Serranoni Nov 20 '20 at 11:03

1 Answers1

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$AB$ 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.

Argyll
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  • How do you get from $\det(AB) = (\det A)(\det B)$ to the fact that $AB$ and $BA$ have the same characteristic polynomials? There is more to be shown there. – Joppy Nov 20 '20 at 12:27
  • @Joppy: I may just delete the determinant part and keep the link. I'll give it some thought. As for your second statement, it is not true. You may be missing multiplicity somewhere. – Argyll Nov 20 '20 at 12:33
  • Yes sorry about that, I was thinking only about diagonalisable matrices. – Joppy Nov 20 '20 at 12:35