Show the Linear Map is Diagonalizable

[joypav]

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!

 

[jvicens]

Hello there,

You are correct in identifying the key concepts that need to be used to prove that $T$ is diagonalizable. I will walk you through the proof step by step.

First, we know that if $\lambda$ is an eigenvalue of $T$, then $ker(T-\lambda I)^2 = ker(T-\lambda I)$. This means that the null space of $T-\lambda I$ does not change when we apply the map $T-\lambda I$ a second time. In other words, the null space of $T-\lambda I$ is invariant under $T-\lambda I$.

Now, let's consider the Jordan canonical form of $T$. This form tells us that $T$ can be written as a block diagonal matrix with Jordan blocks corresponding to each distinct eigenvalue of $T$. Each Jordan block has the form:
$\begin{pmatrix}
\lambda & 1 & 0 & \cdots & 0\\
0 & \lambda & 1 & \cdots & 0\\
\vdots & \vdots & \ddots & \ddots & \vdots\\
0 & 0 & \cdots & \lambda & 1\\
0 & 0 & \cdots & 0 & \lambda
\end{pmatrix}$

Notice that the null space of $T-\lambda I$ is the same as the null space of the Jordan block, which is just the span of the last column. This means that the null space of $T-\lambda I$ is invariant under $T-\lambda I$, just like we saw earlier. This is true for each distinct eigenvalue, so we can conclude that $ker(T-\lambda I)$ is invariant under $T$ for each eigenvalue $\lambda$.

Next, we need to show that the intersection of the null spaces of $T-\lambda I$ is trivial. This is where we use the fact that $ker(T-\lambda I)^2 = ker(T-\lambda I)$. Suppose there are two distinct eigenvalues $\lambda_1$ and $\lambda_2$ such that $ker(T-\lambda_1 I) \cap ker(T-\lambda_2 I) \neq \{0\}$. Then, there exists a non-zero vector $v$ such that $(T-\lambda_1 I)v = 0$ and $(T-\lambda_2 I)v