$n\times n$ matrix with all eigenvalues equal to $1$ or $0$. Does a conjugated matrix with only $1$'s and $0$'s exist?

Question

Let $A$ be an $n\times n$ matrix with all eigenvalues equal to $1$ or $0$. Is there a conjugated matrix $B = XAX^{-1}$ for some $X$ such that all the elements equal either $1$ or $0$?

My thoughts so far: $A$ is not necessarily positive semi-definite..

Given the assumption that $A$ has only eigenvalues that are $0$ or $1$ doesn't really imply anything obvious that we can use. I don't see why there would be a conjugated matrix with all the elements equal either to $1$ or $0$.

So I'm currently trying to come up with a counterexample...

Edit: Actually, a counterexample might be the wrong path to take. The question is just asking for the existence of such a matrix...

Thanks in advance,

Answer 1

Hint: Jordan normal form. Note that since every field contains $1$ and $0$, the Jordan form of your $A$ always exists, even if the underlying field is not algebraically closed.