I have been working on homework for a CS course and it is rife with linear algebra. On one of the questions, I came to brood about the following conundrum (not really):
If a matrix has only eigenvalues of 1 and 0, is it always the case that the characteristic polynomial will be of the form $p(x)=(1-x)^kx^l$, $s.t.:k,l\in \mathbb R$?
The eigenvalues of a matrix are the roots of the characteristic polynomial $p(x)$. If the only eigenvalues of the matrix are 0 and 1, the fundamental theorem of algebra tells us $p(x)=a(x-1)^kx^l$ for $a\in\mathbb{R}$ and $k,l\in\mathbb{N}$. But the characteristic polynomial is always monic, so we must have $a=1$.