# Help with Linear Algebra proof

1. Jun 1, 2009

### evilpostingmong

1. The problem statement, all variables and given/known data
Suppose S, T $$\in$$ the set of linear transformations
from V to V. Prove that ST and TS have the same eigenvalues.

2. Relevant equations
T=$$\lambda$$I

3. The attempt at a solution
let v$$\in$$V.
For TS
T(Sv)=T(I$$\lambda$$v)=$$\lambda$$T(Iv)
=$$\lambda$$'$$\lambda$$Iv.
For ST
S(Tv)=S($$\lambda$$'v)=$$\lambda$$'S(Iv)=
$$\lambda$$'IS(v)=$$\lambda$$'I$$\lambda$$Iv
=$$\lambda$$'$$\lambda$$Iv.

2. Jun 1, 2009

### Staff: Mentor

Your "relevant equation" T = $\lambda$I doesn't seem at all relevant, as far as I can see.

If $\lambda$ is an eigenvalue of T, then for some v $\neq$ 0,
Tv = $\lambda$v.

I would approach this by assuming that $\lambda_1$ and $\lambda_2$ are eigenvalues of ST and TS, then showing that the two eigenvalues are equal.

3. Jun 1, 2009

### evilpostingmong

For ST
Knowing that Tv=$$\lambda$$v, STv=S$$\lambda$$v
=$$\lambda$$Sv=$$\lambda$$ $$\lambda$$'v
$$\lambda$$ $$\lambda$$'=$$\lambda$$1

For TS
knowing that Sv=$$\lambda$$'v, TSv=T$$\lambda$$'v
=$$\lambda$$'Tv since we know from ST that Tv=$$\lambda$$v
we get $$\lambda$$' Tv=$$\lambda$$'$$\lambda$$v=$$\lambda$$ $$\lambda$$'v
$$\lambda$$$$\lambda$$'=$$\lambda$$2

Last edited: Jun 1, 2009
4. Jun 1, 2009

### Staff: Mentor

Looks OK, if a little rough around the edges.

5. Jun 1, 2009

### Dick

That's really not a very good proof at all. It only works if S and T have the same eigenvectors. The result is true even if they don't. You want to show ST and TS have the same characteristic polynomial. Start with the case where S is invertible. Can you show ST and TS are similar matrices?

6. Jun 1, 2009

### evilpostingmong

STv=$$\lambda$$v
TSw=$$\lambda$$2w
STv-$$\lambda$$v=0
TSw-$$\lambda$$2w=0
(ST-$$\lambda$$)v=0
(TS-$$\lambda$$2)w=0
ST-$$\lambda$$=0
TS-$$\lambda$$2=0
ST=$$\lambda$$
TS=$$\lambda$$2
T=S^-1$$\lambda$$2
T=S^-1$$\lambda$$
plugging in S^-1$$\lambda$$2 to ST-$$\lambda$$=0 for T
S(S^-1$$\lambda$$2)-$$\lambda$$=0
I$$\lambda$$2-$$\lambda$$=0
$$\lambda$$2-$$\lambda$$=0
$$\lambda$$2=$$\lambda$$
since ST=$$\lambda$$ and TS=$$\lambda$$2
and $$\lambda$$=$$\lambda$$2
ST and TS have the same eigenvalues.

Last edited: Jun 1, 2009
7. Jun 1, 2009

### kof9595995

(ST-$\lambda$)v=0
(TS-$\lambda$2)w=0
do not imply
ST-$\lambda$=0
TS-$\lambda$2=0

8. Jun 1, 2009

### kof9595995

Dick's suggestion is good, and if you start with a singular matrix S,you can substitute S-$\epsilon$I instead of S, and let $\epsilon$ tend to 0. It's a standard homotopy trick frequently used in linear algebra

9. Jun 1, 2009

### evilpostingmong

still dont get it

10. Jun 1, 2009

### Dick

Do you know similar matrices have the same eigenvalues? Can you show if S is nonsingular then ST and TS are similar? Do you know that nonsingular matrices are dense in the space of all matrices? Do you know that the coefficients of the characteristic polynomial are continuous functions of the elements of the matrices? You have to put all of these things together to get it. Big hint: TS=S^(-1)*(ST)*(S).

11. Jun 1, 2009

### evilpostingmong

Oh, my bad, I should've explained that I'm taking this off of Axler's book, which is my text, he says nothing of similar matrices,
so I'm not too familiar with those. Though I see your point in that there
can be an eigenvector (1,0) that undergoes $$\lambda$$(1,0)= (2*1, 2*0)=(2,0)
and another that undergoes $$\lambda$$(0, 1)=(2*0, 2*1)=(0,2)
the same transformations.

12. Jun 2, 2009

### evilpostingmong

TS=$$\lambda$$2
ST=$$\lambda$$1
TS=S^(-1)$$\lambda$$1S
TS=$$\lambda$$1I
TS=$$\lambda$$1
Since $$\lambda$$1=TS=$$\lambda$$2,
$$\lambda$$1=$$\lambda$$2
ST=T^(-1)TST
ST=T^(-1)$$\lambda$$2T
ST=I$$\lambda$$2

I still don't know what I just did. Does the one below work?
If not, then I'll stop trying to prove this in this manner.
w and v are either different or the same, as opposed to just using v.
I'm not trying to be the my proofs better than yours guy,
I just need convincing.

For ST
Knowing that Tv=$$\lambda$$v, STv=S$$\lambda$$v
=$$\lambda$$Sv=$$\lambda$$ $$\lambda$$'v
$$\lambda$$ $$\lambda$$'=$$\lambda$$1

For TS
knowing that Sv=$$\lambda$$'w, TSv=T$$\lambda$$'w
=$$\lambda$$'Tw since we know from ST that Tv=$$\lambda$$v
we get $$\lambda$$' Tw=$$\lambda$$'$$\lambda$$v=$$\lambda$$ $$\lambda$$'w
$$\lambda$$$$\lambda$$'=$$\lambda$$2

Last edited: Jun 2, 2009
13. Jun 2, 2009

### kof9595995

and I think Dick has given the proof:TS=S^(-1)*(ST)*(S), that's all of it, plus the homotopy trick I mentioned,then the proof will be finished.

14. Jun 2, 2009

### Dick

I really don't think you can prove the general case that way. The eigenvalues and eigenvectors of S and T don't have to be the same as those of ST and TS. Nor is it generally true that the product of the eigenvalues of S and T are the eigenvalues of ST and TS. Try some nontrivial examples. Say S=[[2,1],[1,2]] (eigenvalues 3 and 1) and T=[[1,1],[8,-1]] (eigenvalues -3 and 3). The eigenvalues of ST or TS are (9+/-3*sqrt(21))/2. Now do you believe me?

Last edited: Jun 2, 2009
15. Jun 2, 2009

### evilpostingmong

Oh, ok I see that. Yeah, thats my problem with these proofs,
I tend to neglect some cases. Thanks for your efforts.