TITLE: 0 as the unique eigenvalue of a matrix
QUESTION [6 upvotes]: I want to prove the following statement :
Let $B$ be a matrix, such that $B$ has the eigenvalue $0$ and no other eigenvalue. Then $B^2=0$.
In the context of the statement, $B$ is of size $2$. Is this hypothesis necessary for the statement to hold ?
How to prove the statement ?

REPLY [15 votes]: The characteristic polynomial $f$ of $B$ has degree 2 and since 0 is an eigenvalue of B it has 0 as a root, so $f$ splits into linear factors: $f(X) = X(X-a)$ for some $a$. Since 0 is the only eigenvalue we must have $a=0$, hence $f(X) = X^2$. By the Cayley-Hamilton theorem, $0 = f(B) = B^2$.