Let $K$ be a field and $f: K^{2} \to K^{2}$ be a $K$-linear map with the multiplication of $ \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} $.
Prove that there is only one f-invariant subspace $W$ whereby $0 \subsetneq W \subsetneq K^{2}$.
I believe I have the $0 \subsetneq W$ down well: Since $\chi_{f}= (x-1)^{2}$, we have eigenvalue $1$ and eigenvector $ \begin{pmatrix} 1 \\ 0 \end{pmatrix} $.
The Eigenspace $V_{1}$ is f-invariant. Proof:
Let $v \in V_{1}$, then $\exists \alpha \in K$: $v=\alpha \begin{pmatrix} 1 \\ 0 \end{pmatrix}$.
$f(v)=f(\alpha \begin{pmatrix} 1 \\ 0 \end{pmatrix})=\alpha f(\begin{pmatrix} 1 \\ 0 \end{pmatrix})=\alpha \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 \\ 0 \end{pmatrix}=\alpha\begin{pmatrix} 1 \\ 0 \end{pmatrix} \in V_{1}$
Therefore $V_{1}$ is f-invariant subspace and since $\begin{pmatrix} 1 \\ 0 \end{pmatrix} \neq 0$ we have $0 \subsetneq V_{1}$.
Now for $K^{2}$ we have $dim_{K}V_{1}=1\neq 2 = dim_{K}K^{2}$. So, it automatically follows that $V_{1} \subsetneq K^{2}$.
I have tried proving that there is only one invariant subspace by finding two invariant subspaces$V, V'$ and then attempting to prove that $V=V'$, but I do not semm to get further.
Hint: If $0 \subsetneq W \subsetneq K^{2}$ then $\dim_K W=1$, hence any non-zero vector of $W$ has to be an eigenvector since $W$ is $f$-invariant.