Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?
share|improve this question

3 Answers 3

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.

share|improve this answer
    
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
1  
@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

$$\pmatrix{1&1&0&0\cr0&1&0&0\cr0&0&1&1\cr0&0&0&1\cr}$$

share|improve this answer
  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 $x-1$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.