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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 $\lambda \in \mathbb{C}$.

  1. $f$ can be an endomorphism whose matrix presentation is a Jordan block or a scalar matrix. What else can $f$ be?
  2. Suppose $\lambda = 1$ and $f^2 = 1_V$. Then $f$ is necessarily either $1_V$ or $-1_V$. How do you prove this?
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up vote 3 down vote accepted

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.

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On the question 1, is "the matrix presentation of $f$ is a matrix consisting of Jordan blocks associated to $\lambda$" the necessary and sufficient condition? – Pteromys Feb 12 '12 at 5:17
@Pteromys: "Necessary and sufficient condition" for what? For $f$ having a single eigenvalue? Yes: over $\mathbb{C}$, a linear transformation has a single eigenvalue if and only if its coordinate matrix is similar to a diagonal block matrix, with all diagonal blocks a Jordan block associated to the same $\lambda$. – Arturo Magidin Feb 12 '12 at 5:19


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  1. $f$ 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 $\begin{bmatrix}2&1&0&0\\0&2&0&0\\0&0&2&0\\0&0&0&2\end{bmatrix}$.

  2. 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 $x-1$.

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