## Diagonalizable matrix with only one eigenivalue is a scalar matrix

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

My proof so far:

Suppose that $[T]_{a}$ 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?