Invariant subspaces

Well-defined. If $x\in F$, then $f(x)\in f(F)\subseteq F$, so $f|_F$ really maps $F$ into $F$.

Linearity. For $x,y\in F$ and $\lambda\in\mathbb{R}$: $f|_F(x+\lambda y)=f(x+\lambda y)=f(x)+\lambda f(y)=f|_F(x)+\lambda f|_F(y)$.

Eigenspaces. Let $x\in E_\lambda=\ker(f-\lambda\,\mathrm{id})$, so $f(x)=\lambda x$. Then $f(x)=\lambda x\in E_\lambda$ because $E_\lambda$ is a subspace. In particular, the line spanned by an eigenvector is $f$-invariant.

Conclusion. $f|_F$ is a well-defined linear endomorphism of $F$, and eigenspaces are invariant.

Remark (block form). If $(b_1,\ldots,b_k)$ is a basis of $F$ completed to a basis of $E$, then $f(b_j)\in F$ for $j\le k$, so the matrix of $f$ has the block form $\begin{pmatrix} A' & * \\ 0 & A'' \end{pmatrix}$, where $A'$ is the matrix of $f|_F$.