0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.