There is a linear operator $\mathcal{A}:\mathbb{R}^2\mapsto \mathbb{R}^2$, such that $\mathcal{A}(x,y)=(x,x+y)$. Find all $\mathcal{A}$-invariant subspaces of $\mathbb{R}^2$.
Let $S$ be some arbitrary subspace of $\mathbb{R}^2$ which is $\mathcal{A}$-invariant. Dimension of such space can be $0\lor 1\lor 2$. For $\dim S=0$ and $\dim S=2$, we have trivial solutions $S_0=\{(0,0)\}$ and $S_2=\mathbb{R}^2$ respectively. But what about the $\mathcal{A}$-invariant subspace $S$, such that $\dim S_1=1$.
It is obviously $S_1=\{(0,a)|a\in\mathbb{R}\}$, which can easily be proved. My question, though, is how can we find this value? Here it was pretty obvious, but then again it was pretty much trial and error, since I guessed what it would be and then just proved it, which obviously isn't efficient.
Take the canonical basis of $\mathbb{R}^2$, $\{e_1,e_2\}$. Tha matrix of the transformation in the given basis is $$\begin{pmatrix} 1&0\\1&1\end{pmatrix},$$ whose eigenvalues is $1$ with multiplicity $2$. The invariant subspaces are given by the span of the eigenvectors, which are the solutions of the system $x=0$, that means the only eigenvector is $(0,1)$ which gives us the unique $\mathcal{A}$-invariant subspace $$V_1=\{(0,\lambda):\lambda\in\mathbb{R}\}.$$ Is invariant since given $v\in V_1$, $v=(0,\lambda)$, which implies that $$\mathcal{A}v=\lambda\mathcal{A}e_2=\lambda e_2=v\in V_1.$$