Given a finite dimensional space $V$, with the map $A:V \rightarrow V$, identified as a regular and symmetric matrix $A$, and a subspace $U \subset V$. So $v=u+u^\perp$ holds for every $v\in V$, with $<u,Au^\perp>=0$, that is $u\perp_Au^\perp$ for all $u \in U$,$u^\perp \in U^\perp$. We define the "A-orhogonal projection" $P_A:V \rightarrow U$ by $P_A(v):=u$ I've got to show (i) $<P_Av,w>_A=<v,w>_A$, for all $w\in U$ and
(ii)$||P_Av-v||_A=min_{w\in U}||w-v||_A$
I don't know how to handle with the Matrix-notation $A$ in the inner product and in the norm. Can anyone explain it or give me a hint? How can I use the symmetric property here? I've tried $<P_Av,w>_A=(P_Av)^TAw=v^TP_A^TAw=v^TAw=<v,w>_A$, but in the exericise the semi positive property is not given and I doubt whether $P_A^TA=IA$ is right....
The importance of the symmetry of $A$ is that $\langle Av,w\rangle = \langle v,Aw\rangle$. To start with your first problem, write $v=u+u_\perp$. Then we can get
$$\langle v,w\rangle_A = \langle u,Aw\rangle+\langle u_\perp,Aw\rangle = \langle P_A,w\rangle_A + \langle w,A u_\perp\rangle=\langle P_A,w\rangle_A,$$
where we used the fact that $w\in U$ to cancel the second term. See if you can use this to get the second one.