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
- $\lambda$ is an eigenvalue of $TS\Rightarrow\lambda$ is an eigenvalue of $ST$
- $\lambda$ 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 $det(A-2*I_3)=0=det(B-2*I_2)$?
Best Answer
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.