Finding invariant subspaces of matrix operator $\mathbb{R}^2\to\mathbb{R}^2$ $$A =\begin{bmatrix}0&0\\1&0\end{bmatrix}  $$
So I need to find the 3 invariant subspaces and prove why they are the only one
I found only one so far which I think is the only one which is \begin{bmatrix}0\\1\end{bmatrix} which is going to   \begin{bmatrix}0\\0\end{bmatrix}
and maybe  \begin{bmatrix}1\\0\end{bmatrix} because after $f^2$ it would go to  \begin{bmatrix}0\\0\end{bmatrix}
and for the same reason  \begin{bmatrix}1\\1\end{bmatrix} would go also to \begin{bmatrix}0\\0\end{bmatrix}
is it right or am I totally wrong? thanks
 A: You have to find a subspace $U$ of $R^2$ such that for all $u\in U$, we have that $Au\in U$. Note that $$\begin{bmatrix}0 & 0\\ 1 & 0 \end{bmatrix}\begin{bmatrix}0\\ 1 \end{bmatrix}=\begin{bmatrix}0\\ 0 \end{bmatrix}$$ Thus, $v=\begin{bmatrix}0\\ 1 \end{bmatrix}$ is an eigenvector of $A$ with eigenvalue $0$. Define $$U_1=\{\lambda v : \lambda\in R\}=\text{span}(v)$$ Then you should prove that $U$ is an invariant subspace under $A$.
Now for the second one. Consider the subspace $$U_2=\{\begin{bmatrix}0\\ 0 \end{bmatrix}\}$$ You should prove that this is an invariant subspace under $A$.
For the third one, let $$U_3=R^2$$ Then you should prove that $U_3$ is invariant under $A$.

As for proving that these are the only ones. Note that any two dimensional subspace of $R^2$ is $R^2$ itself. Thus, there is only one $2$ dimensional invariant subspace of $A$.
The only $1$ dimensional invariant subspace of $A$ is the one which is the span of its eigenvectors. As you should verify that $A$ has only one eigenvector $v=\begin{bmatrix}0 \\ 1 \end{bmatrix}$. Thus, there is only one $1$ dimensional invariant subspace of $A$.
The only $0$ dimensional invariant subspace of $A$ is $U_3$.
