Given $T^2 = T$ find all eigenvalues of $T$...

Question

Given a linear operator $T$ on a finite-dimensional vector space $V$, satisfying $T^2 = T$ answer the following:

(a) Using the dimension theorem, show that $N(T) \bigoplus R(T) = V$;

(b) Identify all eigenvalues of $T$ and the corresponding eigenspaces and show that $T$ is diagonalizable.

I have already proven part (a). Basically define $Tv = w$ and the rest follows.

Part (b) however I am honestly confused on where to start.

Answer 1

By definition: $$Tv = \lambda v$$ Multiply both sides by $T$: $$T^2 v = \lambda T v.$$ Recall that $T^2 = T$: $$Tv = \lambda Tv.$$ Since $Tv = \lambda v$, then: $$\lambda v = \lambda \cdot \lambda v = \lambda^2 v.$$ That is:

$$\lambda = \lambda^2.$$

This equation has only two solutions: $\lambda = 0$ and $\lambda = 1$.

For $\lambda = 0$, the eigenspace corresponds to the null space of $T$. For $\lambda = 1$, the eigenspace is given by all vectors $v$ such that:

$$(T-I)v = 0,$$

where $I$ is the identity matrix.

Answer 2

Hints:

1. If $v$ is an eigenvector, then $\lambda v=Tv=T^2v=\lambda^2v$, so $\lambda=\lambda^2$.
2. $R(T)$ coincides with the eigenspace of $\lambda=1$, then use part (a).