Prove that the only eigenvalue of a unipotent matrix is 1.

Question

An $n\times n$ matrix is $\textit{unipotent}$ if $(M-I)^k=0$ for some $k\in\mathbb{N}$, where $I$ is the $n\times n$ identity matrix. Show that the only eigenvalue of a unipotent matrix is $1$.

I found the following proof:

$\lambda$ is an eigenvalue of $M-I$ iff $1+\lambda$ is an eigenvalue of $M$. Then, because $C_{M-I}(x)=\prod_{i=1}^n(x-\lambda_i)$, we have $C_M(x)=\prod_{i=1}^n(x-(1+\lambda_i))$. Since $M-I$ is nilpotent, its only eigenvalue is $0$, and it follows that $C_M(x)=(x-1)^n$, as desired.

My question is specifically about the first statement. How are we able to decompose the spectrum of the difference between $M$ and $I$? Also, is there a simpler approach to this problem?

Answer 1

Another approach: $M(x)=cx$ implies that $(M-I)(x)=cx-x=(c-1)x$.

Suppose that $(M-I)^p(x)=(c-1)^p(x)$, $(M-I)^{p+1}(x)=(M-I)(c-1)^px=(c-1)^pM(x)-(c-1)^px=c(c-1)^px-(c-1)^px=(c(c-1)^p-(c-1)^p)x=(c-1)^{p+1}(x)$.

You deduce that $(M-I)^k(x)=(c-1)^kx=0$ implies that $c=1$.

$\lambda$ is an eigenvalue of $M-I$ i.e there exists a non zero $x$ such that $(M-I)(x)=M(x)-x=\lambda x$, this is equivalent to saying that $M(x)=x+\lambda x=(1+\lambda )x$ i.e that $1+\lambda$ is an eigenvalue of $M$.

Answer 2

As we know, eigenvalues of a nilpotent matrix can only be $0$. Because $A$ is a unipotent matrix iff $A-I$ is a nilpotent matrix, and $$\det(0I-(A-I)) = \det(I-A) = 0,$$ it's obvious that the only eigenvalue of a unipotent matrix is 1.