for the following exercise i have some questions, i appreciate some help or hints for this exercise.
Let be V an euclidean or unitary vectorspace and $ p:V \to V$ a self-adjoint Projection.
$(a)$ Show that $ker(p) \perp im(P)$ and $V=ker(p) \oplus im(p)$.
$(b)$ Show that $1-p$ is a self-adjoint projection on $ker(p)$.
for (a) i collected following informations:
because $p$ is a self-adjoint projection,
$$\langle p(p(v)), w \rangle = \langle v, p(p(w)) \rangle, \ \ \ \ \forall v,w \in V$$ is valid.when look at the notation $ker(p) \perp im(p)$ the following things coming thru my mind:
$ker(p) \perp im(p)$ called the orthogonal sum of the kernel and image of $p$, if following properties are fulfilled:
- Let $U:= ker(p) \oplus im(p)$, than $\forall u \in U : u=x+y, \ \ \ with \ \ \ x\in ker(p) and \ \ \ y \in im(p)$
- $x+y=0 \Rightarrow x=y=0$
- $ker(p) \cap im(p)=0$
This three points are equivalent and if one is fulfilled, than all are fulfilled and we call U the inner direct sum of $$ker(p) \ \ \ and \ \ \ im(p).$$
In addition the following must be fulfilled, so that one can speak of an orthogonal sum:
- $ x \perp y, \ \ \ \forall y \in im(p) \ \ \ \ and \ \ \ \ x \in ker(p).$
My first question would be, how can I show that U is a inner dricet sum or how should I start?
For the part that $x \perp y$ , I have to show that $\langle x,y \rangle = 0$. My intuition tells me that for $x \in ker(p)$, x:=p(v)= o. Because p ist a projection, it holds p(p(v)) = 0 $\Rightarrow$ p(v)=0 $\Rightarrow$ v = 0. . Therefore applies $\langle x,y \rangle = 0$, for all $p(w) \in im(p).$
Is this the way to show that the elements of ker(p) and im(p) are orthogonal?
For the item (b) I dont know what is to shown... can someone give me a hint for this?
kind regards,
WomBud