I need to find the Jordan normal form of a linear operator $T$ that acts on a complex vector space $V$, such that there is a unique one–dimensional $T$–invariant vector subspace of $V$. I have been given a hint that in such a setting there is only one eigenvalue. Could someone tell me why is this the case?
Also, assuming this I would like you guys to check my argument to finally construct the Jordan matrix. Say that the associated matrix of the operator is $A$ and that the unique one dimensional $T$ invariant subspace is $W$. Becuase such a subspace exists we know there is some $w \in W$ and some $\lambda \in \mathbb{C}$ such that $Aw= \lambda w$ . This is true because in this case $T$ acts on $w$ by giving another vector that is in the same space. But then this means that $w$ is an eigenvector and $\lambda$ an (the unique) eigenvalue. If $v$ is another eigenvector, then by definition $Av= \lambda v$. By uniqueness of the subspace $W$ I think it is meant that the matrix only acts as a scalar multiplication on elements of $W$. With it we can conclude that if $v$ is any other eigenvector then it is a linear combination of $w$.
So there is only one linearly independent eigenvector associated with $\lambda$. This means that the the Kernel of $(A -\lambda I)$ is one dimensional, or that the geometric multiplicity of $\lambda$ is one. We conclude that the Jordan matrix is the following:
$$J=\begin{pmatrix} \lambda & 1 & 0 & \cdots & 0 \\ 0 & \lambda & 1 &\cdots& 0 \\ 0 & 0 & \lambda &\cdots& 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots& \lambda \end{pmatrix}$$
Thanks in advance!