Question

both A,B in square matrix space
statement1: All eigenvalues of AB are Zero
statement2: All eigenvalues of BA are Zero
Does statement1 imply statement2? If yes, prove it. If no, give a counterexample
Anyone helps me, please

Answer 1

let there be a "non-zero" eigen value \(\lambda_!\) for AB
then and only then there exists a corresponding eigen vector \(v\) such that
\[(AB)v=\lambda_1v\\
\text{pre-multiplying with B}\\
BABv=\lambda_1Bv\\
(BA)w=\lambda_1w
\]
hence, \(\lambda_1\) is an eigen value of BA with a corresponding eigen vector "w".

but since out \(\lambda_1=0\), the eigen vector \(v\) does not exist and hence neither does \(w\).

Answer 2

and hence BA will also have the same eigen values (0)

Answer 3

let \(\lambda \) be the eigenvalue of B and \(\mu \) of A, then:
\(ABv=A(Bv)=A(\lambda v)=\lambda(Av)=\lambda\mu v\)
I could put first in the paréntesis A with v, insted of B with v like I did, so AB and BA have same eigenvalues

Answer 4

got it?

Answer 5

the first step goes to the second one seems not stable

Answer 6

?

Answer 7

@myko don't get

Answer 8

using characteristic eq,
\[
|AB-\lambda I|=0\\
|AB|=0\implies|A||B|=0=|B||A|\\
\implies|BA|=0\implies|BA-0I|=0
\]

Answer 9

@electrokid how can you know |AB| =0

Answer 10

oh. missed a step
\[|AB-\lambda I|=|AB-0I|=|AB|\]

Answer 11

says nothing

Answer 12

using matrices, I simply set
\[w=Bv\]

Answer 13

so, in general, the eigen values for AB and BA are same

Answer 14

makes sense?

Answer 15

you mean AB =BA?

Answer 16

in general,
BA does not equal AB

but we are talking baout eigen values of (AB) and (BA)

Answer 17

sorry young guys. I need time to understand. I will let you know if I have more question or I got it.

Answer 18

ok, got it. how old and how slow am I. sorry guys.

Answer 19

@Hoa you have been enlightened. you are one of us now! lol

Answer 20

hihihi.... give you medal already. good friend