- #1

joypav

- 151

- 0

**Problem:**

Suppose $V$ is a complex vector space of dimension $n$, where $n > 0$, and suppose that $T$ is a linear map from $V$ to $V$. Suppose that if $\lambda$ is any eigenvalue of $T$, then $ker(T−\lambda I)^2 = ker(T−\lambda I)$. Prove that $T$ is diagonalizable.

**Here's what I think I need to use:**

if $ker(T−\lambda I)^2 = ker(T−\lambda I)$ then $ker(T−\lambda I) \cap Range(T-\lambda I) = \left\{0\right\}$

and

$T$ is diagonalizable if $V = ker(T-\lambda_1 I) \bigoplus ker(T-\lambda_2 I) \bigoplus ... \bigoplus ker(T-\lambda_n I)$

$\lambda_1, \lambda_2, ... ,\lambda_n$ distinct eigenvalues of $T$.If someone could help me out I would appreciate it!