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Problem: Let $V$ be a finite dimensional Vector Space over a field $\mathbb{F}$ and $F,G \in \text{End}(V) $Show that $F \circ G$ and $G \circ F$ have the same Eigenvalues $\lambda$

**My approach**: Let $\sigma_F:= \lbrace \lambda_i \in \mathbb{F} \mid \lambda_i \text{ is a Eigenvalue of } F\rbrace $

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Thus I have to show that $\sigma_{F \circ G} = \sigma_{ G \circ F} $ and because of the symmetry it is sufficient to show that for example $\sigma_{ F \circ G} \subset \sigma_{G \circ F} $

Suppose $\lambda$ is a Eigenvalue of $F \circ G$ and $v \in V \setminus \lbrace 0 \rbrace$ the related Eigenvector, it follows that for **Case 1)** $G(v) \neq 0$: $$ F(G(v))=\lambda v \implies (G\circ F)(Gv)=G(F(Gv))=G(\lambda v)=\lambda G(v) \\ \implies G(v) \text{ is a Eigenvector to $G \circ F$ with Eigenvalue $\lambda$}$$

I am sure the second part is just as easy, but I have been dealing with this problem for too long on my own without consulting help, so I guess I suffer from *not seeing the forest for the trees*.

**Case 2)** Let $G(v) = 0$, $v$ as above it follows that: $$ F(G(v))=\lambda v = F(0)=0 \implies \lambda =0 \text{ because } v \neq 0 $$ It is clear to me that I cannot use $G(v)= 0$ as an Eigenvector of $G \circ F$ because $G(v)$ is now the zero vector and therefore, by definition, not a Eigenvector. Therefore I have to come up with a better statement $$\ker (G \circ F) \neq \lbrace 0 \rbrace \iff \lambda =0 \text{ is a Eigenvalue of } G \circ F \tag{*} $$

I have several problems now with the above statement (*), focussing on the $\implies$ direction

Can I use the same Vector $v \in V \setminus \lbrace 0 \rbrace $ for this statement? It is not really clear to me if and why I can do this, I believe that I have to, because all I know for this case is that $G(v)=0 \implies v \in \ker \lbrace G \rbrace $ and applying $F$ to both sides would give $ F(G(v))=F(0)=0 \implies v \in \ker \lbrace F \circ G \rbrace $.

If I for example just say let $ k \in \ker (G \circ F ) \neq \lbrace 0 \rbrace $ such that $k$ is an Eigenvector of $G \circ F$ with Eigenvalue $\lambda$ it would follow that $$G(F(k))= \lambda k = 0 \implies \lambda =0 $$

which seems very artificial (or forced up) to me, furthermore I have not even made use of my premise that $G(v) = 0$, so the above procedure must be wrong.

On the other hand if I choose $v \in \ker \lbrace G \circ F\rbrace \neq \lbrace 0 \rbrace $ (which I don’t know if I am allowed to do) it would follow that: $$G(F(v))=\lambda v=0 \implies \lambda =0 $$

which looks better to me, but just as artificial and again I have not made use of the premise.

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Here’s what you’re missing:

You are correct in stating $0 \in \sigma_A \iff \ker(A) \neq \{0\}$.

If $G(v) = 0$ for some $v \neq 0$, then clearly $\ker(FG)\neq \{0\}$. To complete our proof, it suffices to show that $\ker(GF) \neq \{0\}$.

Now, suppose there is some vector $u\neq 0$ such that $F(u) = v$. Then clearly $G(F(v)) = 0$, which means that $\ker(GF) \neq \{0\}$, as desired.

Suppose to the contrary that there is no such vector. Then $F \in \text{End}(V)$ *is not onto*. What can we deduce about the kernel of $F$?

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