# Linear algebra and vector space

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 $$v \space \epsilon\space U_{orthogonal}$$, 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 $$proj_{"\vec u's\space \space space"} \space \space \vec x$$. Isn't this right? In analogy, 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 cut 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'=Uort"?

OBS: the answer apparently is "the only true is I".

The assertion II is indeed false. It would be true if the $$u_k$$'s were orthogonal, but you are not assuming that. Take, for instance, $$V=\Bbb R^3$$, take the usual inner product, take $$U=\{(x,y,0)\mid x,y\in\Bbb R\}$$, take $$u_1=(1,0,0)$$, take $$u_2=(1,1,0)$$, and take $$x=(1,1,1)$$. Then the element of of $$U$$ which is closest to $$x$$ is $$(1,1,0)$$, but$$\frac{\langle x,u_1\rangle}{\|u_1\|^2}u_1+\frac{\langle x,u_2\rangle}{\|u_2\|^2}u_2=(2,1,0).$$
Concerning III, consider the same $$V$$ and $$U$$ as above, and let $$U'=\{(x,x,x)\mid x\in\Bbb R\}$$. Then $$V=U\oplus U'$$, but$$U^\perp=\{(0,0,x)\mid x\in\Bbb R\}\ne U'.$$
• Use Gramm-Schmidt to find an orthonormal basis of $U$ starting from $u_1,\ldots,u_n$ and then use the formula from II with that basis. Jul 20, 2021 at 21:48