Linear transformations and eigen values

Question

Let $V$ be a finite dimensional vector-space over $\mathbb{K}$ and $\phi:V\to V$ a linear transformation with $\phi \circ \phi = \phi$

1. Show that $\phi$ can have only the eigenvalues $0$ and $1$.
2. Describe all endomorphisms $\phi : V \to V$ with $\phi \circ \phi = \phi$ that have only the eigenvalue $0$ and prove your claim.
3. Describe all endomorphisms $\phi : V \to V$ with $\phi \circ \phi = \phi$ that have only the eigenvalue $1$ and prove your claim.
4. Specify a vector space $V$ and a linear transformation $\phi : V \to V$ with $\phi \circ \phi = \phi$ that has eigenvalues $0$ and $1$.

1.) Let $v\in V$ and $v\neq 0$ with $ \phi(v)=\lambda v$ and $ \lambda \in \mathbb{K}$

Since $\phi \circ \phi = \phi$, we have $(\phi \circ \phi)(v) = \lambda\lambda v\iff \phi(v) = \lambda^2v$

How do I show, that this will only hold for eigenvalues $1$ and $0$?

2.) With $\ker\phi$ we only map those vectors that are zero, that's what happens, then $\lambda = 0$

(How do I prove this?)

3.) Not really sure, (maybe the image of the linear transformation?)

4.)

Could I just use the given general vector-space and linear transformation as an example?

Answer 1

Hints:

1. If $\phi(v) = \lambda v$ then $$\lambda v =\phi(v) = (\phi\circ\phi)(v) = \phi(\lambda v) = \lambda^2 v$$ so $\lambda^2 = \lambda$ if $v \ne 0$.

2. The polynomial $x^2-x = x(x-1)$ annihilates $\phi$ so $\phi$ is diagonalizable. The only eigenvalue is $0$ so $\phi$ diagonalizes to the zero matrix.

3. The same reasoning as above implies $\phi$ diagonalizes to the identity matrix.
4. Consider $\phi : \mathbb{R}^2 \to \mathbb{R}^2$, $\phi(x,y) = (x,0)$.