Show that the set is a Jordan basis of T

Question

Let $V$ be a vector space over a field $K$. Let $T:V \rightarrow V$ be a linear map. Furthermore, let $T$ satisfy $T^2=0$. If $T(x) \neq 0$ and dim $V = 2$ show that the set $\{T(x),x\}$ is a Jordan basis for $T$.

Answer 1

Since $T^2=0$ then $\rm{im}(T)\subset \ker T$ and by the rank nullity theorem and that $T\ne0$ we find $$\rm{rank}(T)=\dim\ker(T)=1$$

Now it's easy to see that $(T(x),x)$ is basis of $V$ since the two vectors are linearly independent: In fact if $a,b\in\Bbb R$ such that $$aT(x)+bx=0$$ then applying $T$ to the above equality we find $b=0$ and then $a=0$.

Finally the matrix of $T$ in this basis is $$J=\begin{pmatrix}0&1\\0&0\end{pmatrix}$$