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: $\phi (v) = <v, u>$ 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?

  • $\begingroup$ 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 $\mathbb C$. $\endgroup$ – Lee Mosher May 22 at 16:38
  • $\begingroup$ @LeeMosher The field is C or R, the question should work for both. $\endgroup$ – Omer May 22 at 16:39
  • $\begingroup$ 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. $\endgroup$ – Lee Mosher May 22 at 16:41
  • $\begingroup$ @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? $\endgroup$ – 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).

  • $\begingroup$ 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 $\phi (u) = ||u||^2$. $\endgroup$ – Omer May 26 at 16:41
  • $\begingroup$ 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 $\ker\phi$. $\endgroup$ – user10354138 May 26 at 17:29
  • $\begingroup$ 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 $d(v, ker) + d(w, ker) = d(v+w, ker)$ $\endgroup$ – Omer May 26 at 18:16

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