Let V be a vector subspace with a inner product. Let U be a vector subspace with finite dimensions of V and ${u_{1},…,u_{n}}# be a base of U. Let x be an element of V.

(I) If x = u + v, with $u space epsilon space U $ and $v space epsilonspace U_{orthogonal}$, so u and v are unique.

(II) The element of U nearest to x is

(III) If U is a subspace of V such that $V = U oplus U’$, so $U’ = U_{orthogonal}$.

…

I was able to prove the first one (I), but the main problems are with II and III.

I can’t see why II is wrong! I mean, basically what we are saying here is that the nearest element of U to a vector x of V is $proj_{“vec u’sspace space space”} space space vec x$. Isn’t this right? In anallogy, seeing this as a 3d space, the shortest distance of an element in space to a plane is measured using a line that is orthogonal to the plane, and the point in the plane that is cutted by the line is the nearest point of the plane to the vector. So is it here, so why is it wrong?

And also there is a problem about the III! $U’$ is such that $U’ cap U = {0} $. But or we have a vector that is orthogonal to the linear combination of u, or that isn’t, there is no other definition. So, since U’ has no vectors in U, all its vectors need to be orthogonal to U!! So why isn’t “U’=U”??

OBS: the answer apparenttly is “the only true is I”