# The question is incorrect?

Given an Inner product space $$V$$ over a field $$F = \mathbb{R}$$ or $$F = \mathbb{C}$$ (the question should work for both) and $$\phi : V \to F$$ is a functional, and also $$< \cdot, \cdot>$$, and an orthonormal basis $$v_1,...v_n$$ we define $$u = \sum_{i=1}^{n} \phi(v_i)v_i.$$

(a) Show that for all $$v \in V$$ the following equality holds: $$\phi (v) = $$ and $$u$$ is the only vector in $$V$$ that satisfies that. I think this includes a mistake and that $$\phi (v_i)$$ should be the complex conjugate of $$\phi (v_i)$$ in the sum.
(b) Show that $$d(v, Ker \phi) = \frac{\langle v, u \rangle}{||u||}$$
Here is what I tried for (b) which simply doesn't look correct to me.
Use Gram-Schmidt to find an orthonormal basis for the Kernel, say $$e_1,...,e_k$$, and complete it to an orthonormal basis of $$V$$: $$e_1,...,e_k,...,e_n$$, and write $$v = \sum_{i=1}^{n}a_iv_i$$.
Define $$u' = \sum_{i=1}^{n} \overline \phi(e_i) e_i$$, so $$u'$$ satisfies (a) and therefore $$u'=u=\sum_{i=1}^{n}\overline \phi (e_i)e_i$$.
Now, on the one hand we have: $$d(v, Ker \phi) = ||v-P_{Ker\phi}(v)|| = ||v-\sum_{i=1}^k \langle v,e_i \rangle e_i||=||v-\sum_{i=1}^{k}a_ie_i|| = ||\sum_{i=k+1}^{n}a_ie_i||$$.
On the other hand, we have: $$\frac{\langle v,u \rangle}{||u||} = \frac{\langle \sum_{i=1}^{n}a_ie_i, \sum_{i=1}^{n}\overline \phi(e_i)e_i\rangle}{||u||} = \frac{\sum_{i=1}^{n}a_i \phi(e_i)}{||u||} = \frac{\sum_{i=k+1}^{n}a_i \phi(e_i)}{||u||}$$
And we have $$||u||^2 = \langle u,u \rangle = \langle\sum_{i=1}^n\overline \phi(e_i)e_i, \sum_{i=1}^n\overline \phi(e_i)e_i \rangle = \sum_{i=1}^{n}\overline \phi(e_i) \phi(e_i) = \sum_{i=1}^n |\phi(e_i)|^2 = \sum_{i=k+1}^{n} \phi (e_i)^2$$

And that means $$||u|| = \sqrt{\sum_{i=k+1}^{n}\phi(e_i)^2}$$

We also have $$||\sum_{i=k+1}^{n} a_ie_i||=\sqrt{\sum_{i=k+1}^na_i^2}$$.
Therefore the claim is: $$\sqrt{\sum_{i=k+1}^na_i^2} = \frac{\sum_{i=k+1}^na_i\phi(e_i)}{\sqrt{\sum_{i=k+1}^n \phi(e_i)^2}}$$ And I don't see any reason for this to be true. is (b) false as well?

• What is the field of scalars for this problem? The language of (a) suggests that it is $\mathbb R$, in which case (a) is correct as stated. But your own language and notation suggests instead $\mathbb C$. – Lee Mosher May 22 at 16:38
• @LeeMosher The field is C or R, the question should work for both. – Omer May 22 at 16:39
• Well, as I said, (a) is correct for $\mathbb R$ but needs the fix you suggested to work for $\mathbb C$. You should edit your post to clearly state the field of scalars. – Lee Mosher May 22 at 16:41
• @Lee Mosher I've edited, I am pretty sure (a) is incorrect for $\mathbb{C}$ but actually my main problem is (b). Is it even correct? – Omer May 22 at 16:43

Assuming $$\phi\neq 0$$ (so the RHS is well-defined), (b) is correct for $$\mathbb{R}$$ except it needs an absolute sign (unless you want to consider signed distance for some reason). Over $$\mathbb{C}$$ you need to correct the $$u$$ too (as in (a)).
The reason is that $$\phi$$ induces a linear map $$V/\ker\phi\to\mathbb{R}$$ that is $$0$$ on $$\ker\phi$$, so up to some scalar multiple this must be the (signed) distance from $$\ker\phi$$. Since we know $$u$$ is orthogonal to $$\ker\phi$$, our function needs to give $$1$$ for the unit vector $$u/\lVert u\rVert$$, which leads to $$\lvert \phi(v)\rvert/\lVert u\rVert$$. Now use (a).
• I didn't really understand, obviously $\phi$ is $0$ on its kernel, why does it mean that the distance from some vector $v$ to the kernel is some multiple of this? Also, how is the fact that $u$ is orthogonal to the kernel means that $\phi (u/||u||) = 1$? it doesnt even look correct to me because from (a) we have $\phi (u) = ||u||^2$. – Omer May 26 at 16:41
• Combine the following three facts: (1) $\phi\colon V\to\mathbb{R}$ is a linear map that vanish on $\ker\phi$. (2) the space of all linear maps $V\to\mathbb{R}$ which vanish on $\ker\phi$ has dimension 1 unless $\ker\phi=V$ (in which case it is 0) (3) signed distance is a linear map $V\to\mathbb{R}$ that vanish on $\ker\phi$. – user10354138 May 26 at 17:29
• why is (2) true, and for (3), I think you meant that if we denote by $P(v)$ the closest vector in the kernel to v, then P is a linear map, correct? because I don't see why $d(v, ker) + d(w, ker) = d(v+w, ker)$ – Omer May 26 at 18:16