# What's the connection between rank of matrix and $0$ eigenvalue?

My matrix B is nxn and know nothing about if diagonalizble, but I know that rank B = 1. Therefore the geometric multiplicity of λ=0 as an eigenvalue is n-1. But by knowing the rank is 1, can I say that the algebraic multiplicity is n-1 too or could it be more?

• It can't be more, because if it was more, it would be the null matrix. – Git Gud Jun 21 '14 at 21:05

I'm assuming you're asking about whether the algebraic multiplicity could be less than $n-1$. Here's a hint. What are the eigenvalues and eigenvectors of
$$\left[\begin{array}{cccccc}1 & 0&0&0&0 &\\0&0&1&0&0&\\0&0&0&1&0&\\0&0&0&0&1&\\0&0&0&0&0&\\&&&&&\ddots\end{array}\right]?$$