Possible Duplicate:
Show that $V = \mbox{ker}(f) \oplus \mbox{im}(f)$ for a linear map with $f \circ f = f$
I think I need to use the fact that if $v \in V$, then $v = (v - \pi(v)) + \pi(v)$
Possible Duplicate:
Show that $V = \mbox{ker}(f) \oplus \mbox{im}(f)$ for a linear map with $f \circ f = f$
I think I need to use the fact that if $v \in V$, then $v = (v - \pi(v)) + \pi(v)$
Take $v\in V$. You need to show that it can be expressed as a sum of a vector in the kernel of $\pi$, and a vector in the image of $\pi$, and in exactly one way.
You have actually already shown that $v$ decomposes in this way : $v=\big(v-\pi(v)\big)+\pi(v)$ and $v-\pi(v)\in\mathrm{Ker}(\pi)$, since $\pi\big(v-\pi(v)\big)=\pi(v)-\pi\circ\pi(v)=\pi(v)-\pi(v)=0~$ because $\pi$ is a projection, while $\pi(v)\in\mathrm{Im}(\pi)$. So $v$ is the sum of a vector in the kernel of $\pi$ and one in one its image.
Thus far we know that $V=\mathrm{Im}(\pi)+\mathrm{Ker}(\pi)$, and what remains to be shown is that $\mathrm{Im}(\pi)\oplus\mathrm{Ker}(\pi)$ i.e. $\mathrm{Im}(\pi)\cap\mathrm{Ker}(\pi)=\lbrace 0\rbrace$. So consider $w\in\mathrm{Im}(\pi)\cap\mathrm{Ker}(\pi)$ : by definition, there is a vector $v\in V$ with $w=\pi(v)$, and $\pi(w)=0$, but then $0=\pi(w)=\pi(\pi(v))=\pi\circ\pi(v)=\pi(v)=w,$ so $w=0$ and $\mathrm{Im}(\pi)\cap\mathrm{Ker}(\pi)=\lbrace 0\rbrace$, so finally $V=\mathrm{Im}(\pi)\oplus\mathrm{Ker}(\pi).$
Let v be in V. Since π is a projection, π^2 = π. Thus if v is in V, π(v - π(v)) = π(v) - π^2(v) = π(v) - π(v) = 0, i.e., v - π(v) is in Ker(π). So v = (v - π(v)) + π(v) is in Ker(π) + Im(π). Since v is arbitary, this shows that V = Ker(π) + Im(π). To see that the sum is direct, suppose w is an element of the intersection of Ker(π) and Im(π). Then π(w) = 0 and w = π(u) for some u in V. Hence w = π(u) = π^2(u) = π(π(u)) = π(w) = 0. Hence V = Ker(π) ⊕ Im(π).
Let's call $\pi$ the projection. So: $\pi^2=\pi \Rightarrow \pi(\pi-Id)=0 \Rightarrow t(t-1)$ is divided by minimim polynomial that nullify $\pi; $ so $ \pi$ is diagonalizzable, so you did it.
$V=W_0 \oplus W_1$ (since Generalized Eigenspace Decomposition Theorem of V)$ = V_0 \oplus V_1$ (since diagonazation)
For further info check: http://www.dm.unipi.it/~benedett/JORDAN.pdf.