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Let $\mathbb{V}$ be a finite dimensional inner product space and $\alpha : \mathbb{V} \rightarrow \mathbb{R}$ a linear functional. Prove that there is a unique vector $\overrightarrow v_{0} \in \mathbb{V}$ such that $\alpha(\overrightarrow v)=\langle\overrightarrow v,\overrightarrow v_{0}\rangle$ for all $\overrightarrow v \in \mathbb{V}$.

My approach:

I suppose that there is exists another vector $\overrightarrow w_{0} \in \mathbb{V}$ that satisfies the same property. We get $\langle\overrightarrow v,\overrightarrow v_{0}-\overrightarrow w_{0}\rangle=0$ and I need to show that $\overrightarrow v_{0}=\overrightarrow w_{0}$ somehow. Any tips on how to do that? I tried taking an orthonormal basis for $\mathbb{V}$ but that didn't help in the end.

Guy Fsone
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Omrane
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  • What if you set $v=\vec{v}_0-\vec{w}_0$? – fred Apr 11 '16 at 20:02
  • This will show uniqueness, but you still have show existence. That will actually follow from uniqueness and finite-dimensionality. – fred Apr 11 '16 at 20:04
  • @fred, according to the answer given below, proving uniqueness seems to be a simple implication. Any tips on how to show existence? I did not see that part of the question for some reason. – Omrane Apr 11 '16 at 20:08
  • @fred, Any tips on how to show existence? – Omrane Apr 11 '16 at 20:22
  • @fred, attempt of mine: I can create a basis for V with $n<\infty$ dimensions. Therefore I can use the linear extension theorem: that $\exists$ a linear transformation $\alpha$ and we define $\alpha(v)=<v,v_{0}$ which is possible since inner products pertain to the vector space \mathbb{R}? – Omrane Apr 11 '16 at 20:28

1 Answers1

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You get

$$\langle v,\,v_0-w_0\rangle=0\;\;\color{red}{\forall\,v\in V}\iff v_0-w_0=0$$

as zero is the only vector which is orthogonal to the whole space.

Existence Choose an orthonormal basis $\;\{u_1,...,u_n\}\;$ of $\;V\;$ , and let

$$v_0:=\sum_{k=1}^n\alpha(u_k)u_k\in V\;\;\implies\;\;\forall\,v=\sum_{i=1}^n a_iu_i\in V:$$

$$\langle v_0,v\rangle=\sum_{i,k=1}^n\alpha(u_k)a_i\langle u_k,u_i\rangle=\sum_{k=1}^n\alpha(u_k)a_k\stackrel{\text{linearity of}\;\alpha}=\alpha\left(\sum_{k=1}^n a_ku_k\right)=\alpha(v)$$

DonAntonio
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    I've actually never seen that property. Can there be no other vector orthogonal to the whole space? If $\mathbb{V}$ is a plane we can always find a vector orthogonal to the whole plane in that case, so doesn't that disprove what you just wrote? – Omrane Apr 11 '16 at 20:10
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    @Sadem It follows at once from $;\langle v,v\rangle=0\iff v=0;$, which is one of the axioms defining inner product in a linear space. And pay attention to what I wrote: the only vector orthogonal to the whole space is the zero vector. Of course subspaces can have non-zero orthogonal vectors, and in fact they're pretty important. – DonAntonio Apr 11 '16 at 20:11
  • but $\overrightarrow v\not = \overrightarrow v_{0}-\overrightarrow w_{0}$, so you can't apply that property can you? – Omrane Apr 11 '16 at 20:13
  • @Sadem Choose $;v=v_0-w_0;$ !! You can because the equality is true for all $;v\in V;$ ? – DonAntonio Apr 11 '16 at 20:14
  • I'm sorry I am kind of lost bear with me. Wouldn't choosing $v=v_{0}-w_{0}$ contradict the fact that we want the equality to be true for all $v$?

    Edit: I understand now, the uniqueness part should come after the existence part and the existence part says that the equation is true for all $v\in \mathbb{V}$. Thanks for your help. Any tips on how to show existence?

     – Omrane Apr 11 '16 at 20:16
  • @Sadem Look at the addition to my answer now. – DonAntonio Apr 11 '16 at 20:27
  • @Joanpemo I think you might have a typo in the first line in the existence part. – fred Apr 11 '16 at 20:44
  • @fred Thank you, but I can't see it. Would you be so kind to tell what exactly? – DonAntonio Apr 11 '16 at 20:46
  • @Joanpemo, I guess I don't understand it. $v_0$ is the thing we are trying to find, right? How does $v_0\in V$ imply the thing on the other side of the arrow? I think you might need to change the duality pairing on the next line to $\langle v_0,w\rangle$ from $\langle v,w\rangle$. I just couldn't follow it. Thanks. – fred Apr 11 '16 at 20:58
  • @fred Thank you, now I see the typo but in the second line: it should be $;\langle v_0,v\rangle;$ , not $;\langle v,w\rangle;$ . – DonAntonio Apr 11 '16 at 21:00