Help with Linear Algebra proof

[evilpostingmong]

Homework Statement

Suppose S, T \in the set of linear transformations
from V to V. Prove that ST and TS have the same eigenvalues.

Homework Equations

T=\lambdaI

The Attempt at a Solution

let v\inV.
For TS
T(Sv)=T(I\lambdav)=\lambdaT(Iv)
=\lambda'\lambdaIv.
For ST
S(Tv)=S(\lambda'v)=\lambda'S(Iv)=
\lambda'IS(v)=\lambda'I\lambdaIv
=\lambda'\lambdaIv.

 

[Mark44]

Your "relevant equation" T = \lambdaI 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 = \lambdav.

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.

 

[evilpostingmong]

For ST
Knowing that Tv=\lambdav, STv=S\lambdav
=\lambdaSv=\lambda \lambda'v
\lambda \lambda'=\lambda1

For TS
knowing that Sv=\lambda'v, TSv=T\lambda'v
=\lambda'Tv since we know from ST that Tv=\lambdav
we get \lambda' Tv=\lambda'\lambdav=\lambda \lambda'v
\lambda\lambda'=\lambda2

 

[Mark44]

Looks OK, if a little rough around the edges.

 

[Dick]

For ST
Knowing that Tv=\lambdav, STv=S\lambdav
=\lambdaSv=\lambda \lambda'v
\lambda \lambda'=\lambda1

For TS
knowing that Sv=\lambda'v, TSv=T\lambda'v
=\lambda'Tv since we know from ST that Tv=\lambdav
we get \lambda' Tv=\lambda'\lambdav=\lambda \lambda'v
\lambda\lambda'=\lambda2

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?

 

[evilpostingmong]

STv=\lambdav
TSw=\lambda2w
STv-\lambdav=0
TSw-\lambda2w=0
(ST-\lambda)v=0
(TS-\lambda2)w=0
ST-\lambda=0
TS-\lambda2=0
ST=\lambda
TS=\lambda2
T=S^-1\lambda2
T=S^-1\lambda
plugging in S^-1\lambda2 to ST-\lambda=0 for T
S(S^-1\lambda2)-\lambda=0
I\lambda2-\lambda=0
\lambda2-\lambda=0
\lambda2=\lambda
since ST=\lambda and TS=\lambda2
and \lambda=\lambda2
ST and TS have the same eigenvalues.

 

[kof9595995]

(ST-\lambda)v=0
(TS-\lambda2)w=0
do not imply
ST-\lambda=0
TS-\lambda2=0

 

[kof9595995]

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

 

[evilpostingmong]

still don't get it

 

[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).

 

[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.

 

[evilpostingmong]

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).

TS=\lambda2
ST=\lambda1
TS=S^(-1)\lambda1S
TS=\lambda1I
TS=\lambda1
Since \lambda1=TS=\lambda2,
\lambda1=\lambda2
ST=T^(-1)TST
ST=T^(-1)\lambda2T
ST=I\lambda2

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=\lambdav, STv=S\lambdav
=\lambdaSv=\lambda \lambda'v
\lambda \lambda'=\lambda1For TS
knowing that Sv=\lambda'w, TSv=T\lambda'w
=\lambda'Tw since we know from ST that Tv=\lambdav
we get \lambda' Tw=\lambda'\lambdav=\lambda \lambda'w
\lambda\lambda'=\lambda2

 

[kof9595995]

Check your latex codes,
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.

 

[Dick]

TS=\lambda2
ST=\lambda1
TS=S^(-1)\lambda1S
TS=\lambda1I
TS=\lambda1
Since \lambda1=TS=\lambda2,
\lambda1=\lambda2
ST=T^(-1)TST
ST=T^(-1)\lambda2T
ST=I\lambda2

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=\lambdav, STv=S\lambdav
=\lambdaSv=\lambda \lambda'v
\lambda \lambda'=\lambda1


For TS
knowing that Sv=\lambda'w, TSv=T\lambda'w
=\lambda'Tw since we know from ST that Tv=\lambdav
we get \lambda' Tw=\lambda'\lambdav=\lambda \lambda'w
\lambda\lambda'=\lambda2

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?

 

[evilpostingmong]

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