What can one deduce from $\text{rk}(\text{id}_V-s)=1$?

Question

Let $V$ be a vectorspace over a characteristic $0$ field $k$ of dimension $n$. By $\text{rk}$ I mean the rank of the matrix.

If $s\in\text{End}(V)$ and $\text{rk}(\text{id}_V-s)=1$ then what can we say about $s$?

So the rank being as it is means that $I-s$ has a one dimensional image. So there is some $v\in V$ such that $(I-s)(\lambda v)=\lambda w$ for all $\lambda \in k$. $$\lambda(v-s(v))=\lambda w\implies v-s(v)=w.$$

Now any other vector than $v$ is annihilated, so that means that any other vector than $v$ is an eigenvector with eigenvalue $1$, so there is an $n-1$ dimensional eigenspace corresponding to eigenvalue $1$.

What more can I deduce from this condition?

Answer 1

The characteristic polynomial $p_s(x)$ has the form $(x - 1)^{n-1} (x - \lambda)$ for some $\lambda \in \mathbb{F}$. If $\lambda \neq 1$ then $s$ is diagonalizable and with respect to basis of eigenvalues it can be represented as $$\operatorname{diag}(\underbrace{1,\dots,1}_{n - 1 \text{ times}}, \lambda). $$

If $\lambda = 1$ then $n \geq 2$ and $s$ must be non-diagonalizable (because otherwise $s = \operatorname{id}_V$ and then $\operatorname{rk}(\operatorname{id}_V - s) = 0$). In this case, $s$ can be represented by the block diagonal matrix

$$ \begin{pmatrix} A_{2 \times 2} & 0 \\ 0 & B_{(n - 2) \times (n - 2)} \end{pmatrix} $$

with

$$ A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, B = \operatorname{diag}(\underbrace{1, \dots, 1}_{n - 2 \text{ times}}). $$