Show that $f(E_{\lambda}(g)) \subseteq E_{\lambda}(g)$ if $f \circ g = g \circ f$

Question

We have 2 linear functions.

$f: V \to V$ and $g: V \to V$

Show that for each Eivenvalue $\lambda$ of g we have for the corresponding Eigenspace $E_{\lambda}(g)$: $$f(E_{\lambda}(g)) \subseteq E_{\lambda}(g)$$

I started off like this:

$$g(E_{\lambda}(g)) \subseteq E_{\lambda}(g)$$ $$f(g(E_{\lambda}(g))) \subseteq f(E_{\lambda}(g))$$ $$\implies g(f(E_{\lambda}(g))) \subseteq f(E_{\lambda}(g))$$

Now I'm stuck, how can I continue? Did I even start this proof correctly?

Answer 1

Okay I was able to solve it.

Let $x \in E_{\lambda}(g)$

$$g(x) = \lambda x$$ $$f(g(x)) = f(\lambda x) = \lambda f(x)$$ $$g(f(x)) = \lambda f(x)$$ $$\implies f(x) \in E_{\lambda}(g)$$