So I have this exercise where I want to check my solutions. Can someone help me?
Let $A = \begin{pmatrix} a &b \\ 0& d \end{pmatrix} ∈ Mat_{2,2}(K)$ with $a, b, d ∈ K$, where $a ≠ 0$ and $d ≠ 0$. Let $l_A: K^2 → K^2$ be the corresponding $K$-linear map. A $K$-subspace $U$ of $K^2$ is said to be $l_A$-invariant if $l_A(U) ⊂ U$.
a) Show that if $a = d$ and $b ≠ 0$, then there exists exactly one $l_A$-invariant $K$-subspace of dimension $1$ in $K^2$.
So I have read that the condition for existence of a 1-dimensional invariant subspace is expressed as: $A\mathbf {v} =\lambda \mathbf {v}$, where $\lambda$ is a scalar (in the base field of the vector space.
This is basically an eigevalue problem.
So I have for a) $\det (\begin{pmatrix} a-\lambda &b \\ 0& a-\lambda \end{pmatrix})=(a-\lambda)^2=0 \Rightarrow \lambda=a$
We have therefore $\begin{pmatrix} 0 &b \\ 0& 0 \end{pmatrix}$ and this takes us to $v=\begin{pmatrix}1\\0\end{pmatrix}$ which was what we had to prove.
Am I Right or am I doing something wrong?
Another question: If the problem was with another matrix and was to prove the existence of exactly one $l_A$-invariant $K$-subspace of dimension $2$, should we have proved that there are two independent eigenvectors?
Thanks for the help :)