On linear transformations that have only one eigenvalue Let $V$ be a $n$-dimensional complex vector space and $f:V\rightarrow V$ be an endomorphism on $V$.  Suppose $f$ has only one eigenvalue $\lambda \in \mathbb{C}$.


*

*$f$ can be  an endomorphism whose matrix presentation is a Jordan block or a scalar matrix.  What else can $f$ be?

*Suppose $\lambda = 1$ and $f^2 = 1_V$.  Then $f$ is necessarily either $1_V$ or $-1_V$.  How do you prove this?

 A: $$\pmatrix{1&1&0&0\cr0&1&0&0\cr0&0&1&1\cr0&0&0&1\cr}$$
A: Not sure what you mean by "Jordan block matrix", but given that you are separating out scalar matrices, I expect you mean a matrix consisting of a single Jordan block associated to $\lambda$. In fact, it can be a matrix similar to a matrix with any number of Jordan blocks associated to $\lambda$, of any sizes that add up to $n$; the scalar matrix is a special case, in which you have $n$ $1\times 1$ Jordan blocks. 
If $f^2=1_n$, then $f$ satisfies the polynomial $x^2-1$, hence the minimal polynomial of $f$ divides $x^2-1 = (x-1)(x+1)$. Since $\lambda$ is an eigenvalue if and only if $x-\lambda$ divides the minimal polynomial, the fact that there is a single eigenvalue says that the minimal polynomial is either $x-1$ or $x+1$, hence either $f=1_n$ or $f=-1_n$. However, since you say that the only eigenvalue is $\lambda=1$, that means that we must have $f=1_n$, and $f=-1_n$ is impossible.
A: *

*$f$ has a Jordan form that is composed of Jordan blocks that all have the same eigenvalue, and there is no further restriction.  Any matrix similar to a Jordan block would be an example, but you can also have direct sums of such, like $\begin{bmatrix}2&1&0&0\\0&2&0&0\\0&0&2&0\\0&0&0&2\end{bmatrix}$.

*If $f^2=I_n$, then the minimal polynomial of $f$ divides $x^2-1=(x+1)(x-1)$.  If $f$ has only one eigenvalue, this means that the minimal polynomial of $f$ is $x+1$ or $x-1$.  
