Show that a diagonalizable matrix with only one eigenvalue is a scalar matrix.

My proof so far:

Suppose that $\displaystyle [T]_{a} $ is diagonal for some ordered basis a. Let W be the only eigenvalue of T, that is, T(v) = Wv for some vector v.

T(v) - Wv = 0

(T - W)v = 0

How should I follow?