Let's assume that $V$ and $W$ are vector spaces over a field $\mathbb{K}$, $\lambda\in\mathbb{K}$, $\lambda\neq0$.

$S: V\rightarrow W$ and $T: W\rightarrow V$ are linear maps. Prove, that

$\lambda$ is an eigenvalue of $TS\iff\lambda$ is an eigenvalue of $ST$

What can be stated about the eigenvalues of the maps $TS$ and $ST$? Would it also be correct if $\lambda=0$?

That's how far I've come: I have to prove, that

  1. $\lambda$ is an eigenvalue of $TS\Rightarrow\lambda$ is an eigenvalue of $ST$
  2. $\lambda$ is an eigenvalue of $ST\Rightarrow\lambda$ is an eigenvalue of $TS$

Assuming $V=\mathbb{K}^n$ and $W=\mathbb{K}^m$, such that $S:\mathbb{K}^n\rightarrow\mathbb{K}^m$ and $T:\mathbb{K}^m\rightarrow\mathbb{K}^n$. Hence, $TS: \mathbb{K}^n\rightarrow\mathbb{K}^n$ and $TS: \mathbb{K}^m\rightarrow\mathbb{K}^m$. $TS$ and $ST$ are both endomorphisms. Since the eigenvalue is not zero, the matrices must be invertible and the determinant of both matrices is not zero. Let's say $A$ is the transformation matrix of $TS$ and $B$ the transformation matrix of $ST$

That's where I'm stuck right now. How do I go on from here?

Do I have to prove, that $det(A-2*I_3)=0=det(B-2*I_2)$?


If $\lambda$ is an eigenvalue for $TS$ then there is a $v\neq 0$ such that $$TSv=\lambda v$$

I claim that $Sv$ will be an eigenvector for $ST$ with eigenvalue $\lambda$:

$$ST(Sv)=S(TSv)=S\lambda v=\lambda Sv$$

The argument wouldn't hold if $\lambda=0$ because then $Sv=\lambda v=0$, but we require eigenvectors to be non-zero.

And, just to let you know, your work so far isn't quite right. Just because a matrix has a non-zero eigenvalue, that does not necessarily make it invertible. Consider for example

$$M=\begin{pmatrix}1&0\\0&0\end{pmatrix}$$ which has eigenvalue $1$, but is not invertible.

  • $\begingroup$ What I meant was, as long a matrix doesn't have 0 as an eigenvalue it is invertible. Your matrix M has the eigenvalues 1 and 0. $\endgroup$ – Arthur May 3 '15 at 19:57
  • $\begingroup$ @Arthur, yes that is correct. But the point is that we cannot assume in the proof that the matrix is invertible since we only know that one eigenvalue is non-zero. $\endgroup$ – Tyr Curtis May 3 '15 at 21:53

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