ST and TS have the same eigenvalue

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1. Mar 4, 2016

maNoFchangE

1. The problem statement, all variables and given/known data
Prove that, if $T,S\in \mathcal{L}(V)$ then $TS$ and $ST$ have the same eigenvalues.

2. Relevant equations

3. The attempt at a solution
Suppose $T$ is written in a basis in which its matrix is upper triangular, and so is $S$ (these bases may be of different list of vectors in $V$). Since both $T$ and $S$ are upper triangular, $TS$ and $ST$ are also upper triangular. Now, the diagonal element of $TS$ is
$$\sum_i T_{pi}S_{ip}$$
which is the same as the diagonal element of $ST$
$$\sum_i S_{pi}T_{ip}$$
Therefore, $ST$ and $TS$ have the same eigenvalues.
What bothers me is my starting assumption; taking $T$ and $S$ in their own upper triangular bases. Since these bases can be a different list of vectors, is it alright then to multiply their matrices?

2. Mar 4, 2016

Samy_A

Assume $\lambda$ is an eigenvalue of $TS$.
Then, by definition of an eigenvalue, there exists an $x\in V,\ x\neq \vec0$ such that $TSx=\lambda x$.
Now try to prove that $\lambda$ is an eigenvalue of $ST$.

(You don't need a matrix representation of the operators.)

3. Mar 4, 2016

maNoFchangE

Multiplying
$$TSx=\lambda x$$
with $S$, I get
$$ST(Sx) = \lambda (Sx)$$
Alright, that looks good. But how do I prove that $Sx$ is not zero given $x \neq 0$?

4. Mar 4, 2016

Samy_A

You can't. There is no reason why $Tx$ couldn't be the 0 vector.
Hint: multiply from the other side.

5. Mar 4, 2016

maNoFchangE

I have edited my post.
Why can't I have $Sx=0$? What if $x\in \textrm{null }S$?

6. Mar 4, 2016

Samy_A

Assume $Sx=\vec 0$. What does that imply, given that $TSx=\lambda x$?

7. Mar 4, 2016

maNoFchangE

That means $\lambda x=0$, since $x \neq 0$ then $\lambda=0$. But what does it say about $Sx$ being not 0?

8. Mar 4, 2016

Samy_A

Take the case $\lambda \neq 0$ first. What have you then proven?

9. Mar 4, 2016

maNoFchangE

In that case $x=0$, which disproves the starting hypothesis that $x\neq 0$. Therefore in this case, $Sx$ cannot be zero.
But in the case $\lambda=0$, what next?
It does not seem to violate any hypothesis we have assumed.

10. Mar 4, 2016

Samy_A

Ok, so you have proven that $TS$ and $ST$ have the same non-zero eigenvalues.

Now for the tricky case $\lambda =0$.
Still assuming $x \neq \vec 0$, you have $TSx=\vec0$, and thus $ST(Sx)=\vec 0$.
If $Sx \neq \vec 0$, then we are done.

If $Sx =\vec 0$, $S$ is not injective.
At this point, you need to take into account that $V$ is finite dimensional (you didn't specifically state this, but since you used matrices, I assume that it is the case).
If $S$ is not injective, it is not surjective. What does that tell you about $ST$?

11. Mar 4, 2016

maNoFchangE

If $S$ is not surjective, then $ST$ is not surjective either.
Sorry I just can't see where this is leading to. If $ST$ is not surjective, is it inconsistent with one of our assumptions?

12. Mar 4, 2016

Samy_A

Correct, $ST$ is not surjective.
Can it be injective? If not, what does that imply as far as eigenvalues of $ST$ are concerned?

13. Mar 4, 2016

maNoFchangE

Since $ST$ is an operator in a complex vector space, if it's not surjective, it will not be injective nor it be invertible, am I right?
Since $ST$ is not invertible, there must be at least one of its eigenvalues equal to $0$. But isn't it what we were starting from? In post #10, we have set $\lambda=0$, aren't we just circulating here?

14. Mar 4, 2016

Samy_A

No, in post #10 we started with an eigenvalue of $TS$. You now have proved that if 0 is an eigenvalue of $TS$, it also is an eigenvalue of $ST$.
We are not circulating, we went from $TS$ to $ST$.

Notice that for the case of eigenvalue 0, $V$ has to have finite dimension. If $V$ is an infinite dimensional vector space, it is possible for $TS$ to have 0 as eigenvalue, while $ST$ is injective.

15. Mar 4, 2016

maNoFchangE

I see! I was not aware that we have been switching to calculating the eigenvalue of $ST$. So, $\lambda=0$ is also an eigevalue of $ST$, and the corresponding eigenvector is zero, is that right?

16. Mar 4, 2016

Samy_A

Yes, that is the whole purpose of this exercise.
It is an eigenvalue, but we can't say anything about the eigenvector.
Certainly the eigenvector is not the zero vector. An eigenvector is, by definition, never the zero vector.
For a simple reason: $T\vec 0=\lambda \vec 0$ is true for any linear operator $T$ and any scalar $\lambda$.

17. Mar 4, 2016

maNoFchangE

Ok many thanks for the help.
Just one more thing, my original question was actually is it allowed to multiply two square matrices which are represented in different bases?

18. Mar 4, 2016

Samy_A

No, you can't just do that. You can of course multiply the matrices, but the resulting matrix is not necessarily the matrix representing the composition of the two operators (for what basis should that be?)

19. Mar 4, 2016

maNoFchangE

Ah, that makes sense.