showing the direct sum of an endomorphism $f$ where $f$ is a projection $f^2=f$

Question

Endomorphism $f$ of vector space $V$ is a projection such that $f^2=f$. Show that $f=f_0 \oplus f_1$ where $f_0$ is the zero-map on $E_0$ and $f_1$ is the identity map on $E_1$.

Attempt thus far:

I know that first I have to show that $V$ is the direct sum of $E_0$ and $E_1$.

$E_1=ker(f-id_v)=im(f)$ and $E_0=ker(f)$

To prove $V=im(f)\oplus ker(f)$ I have to show that:

1. $im(f)+ker(f)=V$
2. $im(f) \cap ker(f)=\{0\}$

I know how to do all this but where I get stuck is understand how $f=f_0 \oplus f_1$

If anyone can explain this in a clear way, I'd much appreciate this.

Answer 1

You need to do the following (in order of increasing difficulty):

1. Show that $f|_{E_0} = f_0$
2. Show that $f|_{E_1} = f_1$
3. Show that $E_0 \cap E_1 = 0$. In other words, if $(f - \operatorname{id}_V)x = 0$ and $fx = 0$, then it must hold that $x = 0$.
4. Show that $V = E_0 + E_1$. That is, show that for every $x \in V$, there are vectors $x_0 \in E_0$ and $x_1 \in E_1$ for which $x = x_0 + x_1$. Hint: take $x_1 = f(x)$.

Facts 3 and 4 imply that $V = E_0 \oplus E_1$. From there, the fact that $f|_{E_0} = f_0$ and $f|_{E_1} = f_1$ allow you to conclude that (by definition of the direct sum of two linear maps) $f = f_0 \oplus f_1$.