Q: Let $f:V\rightarrow V$ be a linear operator on real vector space $V$. Prove if dim$V\geq2$, then there exist invariant subspace of $V$, with dimension $2$.
My approach:
Induction: Basis - if dim$V=2$, we can take V be that invariant subspace, because if $v\in V$ then it is obvious that $f(v)\in V$.
Assumption - We assume there exist invariant subspace for every vector space such that dim$V=n-1$.
Proof
Let $V_{n-1}$ be a vector space with dimension of n-1, and $W_{n-1}$, be the mentioned invariant space. Now:
let $g:V_{n-1}\rightarrow \mathbb{R}^{n-1}$ be isomorphism, such that if $v\in V_{n-1}$ then $g(v)=\begin{pmatrix}a_1\\a_2\\ \vdots \\ a_{n-1}\end{pmatrix}\in\mathbb{R}^{n-1}$. Now, let's define $g^*:\mathbb{R}^{n-1}\rightarrow \mathbb{R}^{n-1}$ recall that $f:V_{n-1}\rightarrow V_{n-1}$ have the invariant subspace $W_{n-1}$ such that if $w\in W_{n-1}$ then $f(w)\in W_{n-1}$.
Now, because of the isomorphism of $g$, if $w\in W_{n-1}$ then we will have some subspace $W^*_{n-1}$ of $\mathbb{R}^{n-1}$, such that if $g(f(w))\in W^*_{n-1}$ then $g^*(g(f(w)))\in W^*_{n-1}$. Now, let $h:\mathbb{R}^{n-1}\rightarrow \mathbb{R}^n$ be defined by $h(\begin{pmatrix}a_1\\a_2\\ \vdots \\ a_{n-1}\end{pmatrix}\in\mathbb{R}^{n-1})=\begin{pmatrix}a_1\\a_2\\ \vdots \\ a_{n-1}\\a_n\end{pmatrix}\in\mathbb{R}^{n}$, for example: $h(\begin{pmatrix}4\\6\\1\end{pmatrix})=(\begin{pmatrix}4\\6\\1\\-12\end{pmatrix})$. Now, what i'm not sure, is if we can say that the linear opator $h^*:\mathbb{R}^n\rightarrow \mathbb{R}^n$ defined by $h^*\Bigg(h(g^*(g(f(w))))\in \mathbb{R}^{n}\Bigg)=(\begin{pmatrix}g^*(g(f(w)))\\a_n\end{pmatrix}\in\mathbb{R}^n)$ have an invariant subspace $W^*_n$, such that dim$W^*=2$. I think that it can be true, because even though I am defining $h$ by some specific $g^*$, We know(by the assumption), that every $g^*$ have the invariant subspace $W^*_{n-1}$. From here, let's define $i^*:\mathbb{R}^n\rightarrow V^n$ be isomorphism, such that if $\begin{pmatrix}g^*(g(f(w)))\\a_n\end{pmatrix}\in\mathbb{R}^n$ then $i^*(\begin{pmatrix}g^*(g(f(w)))\\a_n\end{pmatrix})\in V^n$. For the last time, let's define $i:V^n\rightarrow V^n$ be a linear operator, with some subspace $W_n$, such that, if $i^*(\begin{pmatrix}g^*(g(f(w)))\\a_n\end{pmatrix})\in W_n$ then $i(i^*(\begin{pmatrix}g^*(g(f(w)))\\a_n\end{pmatrix}))\in W_n$, and because dim$W^*_n=2$, we can conclude that for every vector space $V_n$ there exist some invariant subspace $W_n$ with dimesion $2$.$\square$
Note that I know that there are already proofs of this question(or similar ones) on this site, I just want to make sure, that my proof is good!