Vectors shared by two different eigenspaces

A matrix $B$ of size $n\times n$ has two eigenvalues $\lambda_1\ne\lambda_2$, with eigenspaces $E_{\lambda_1}$ and $E_{\lambda_2}$. How many vectors lie in $E_{\lambda_1}\cap E_{\lambda_2}$? Answer $-1$ if there are infinitely many.

$1$. A common vector $v$ satisfies $\lambda_1v=Bv=\lambda_2v$, so $(\lambda_1-\lambda_2)v=0$ and therefore $v=0$. The intersection is just the zero vector.