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Let $V$ be a vector space with a basis $\{\alpha_1,\alpha_2,.....,\alpha_n\}. $

Let $(\ | \ ) $ be an inner product on $V$. If $c_1,c_2,...,c_n $ are any scalars in the field. Then show that there is exactly one vector $\alpha \ \text{in} \ V \ \text{such that} (\alpha_i|\alpha) =c_i, \ i=1 \ldots n. $

I have been stuck on this for almost two days now. Looking for hints.

Paul Aljabar
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Johnathan G
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2 Answers2

1

I'm going to use the notation $<\cdot, \cdot>$ to denote the inner product.

We have a basis so each vector in the space $V$ can be expressed as a linear combination of the basis vectors $\{\alpha_{1}, \ldots , \alpha_{n}\}$. The coordinates of a vector determine the weights to use in the linear combination.

Let $v$ be a vector with coordinates $r = (r_{1}, \ldots, r_{n})^{T} \in \mathbb{R}^{n}$, i.e. we have

$$ v = r_{1} \alpha_{1} + \ldots + r_{n} \alpha_{n} $$

We can represent the inner products $<\alpha_{i}, v>$ using the coordinate representation of $v$ because the inner product is linear:

$$ <\alpha_{i}, v> = r_{1} <\alpha_{i}, \alpha_{1}> + \ldots + r_{n} <\alpha_{i}, \alpha_{n}> $$

A vector containing all the $<\alpha_{i}, v>$, for $1 \le i \le n$, can be obtained by pre-multiplying $r$ by a matrix $A$ where $A_{ij} = <\alpha_{i}, \alpha_{j}>$.

The $n\times n$ matrix $A$ is the Gramian matrix for the set of vectors $\{\alpha_{1}, \ldots, \alpha_{n}\}$. It is non-singular because the vectors in the set are linearly independent (link) .

If we collect the values of $c_{i}$ described in the question into a vector, $c = (c_{1}, \ldots, c_{n})^{T}$, we can ask whether we can solve uniquely for $r$ in the equation $Ar = c$. The answer is yes because $A$ is invertible.

Given that we can obtain a unique set of coordinates $r$, we can deduce that there is a unique vector in $V$ corresponding to these coordinates. This is due to the isomorphism between $V$ and the space of its coordinates $\mathbb{R}^{n}$.

Paul Aljabar
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  • How did you get $<\alpha_i, v>=c_i$? – Johnathan G May 11 '17 at 19:16
  • The vector you write as $\alpha$, I am calling $v$. It is just a vector that satisfies the property given in the question. – Paul Aljabar May 11 '17 at 19:18
  • But you've not demonstrated that such a vector exists. – Johnathan G May 11 '17 at 19:19
  • If you can show that $A$ is full rank, then you can solve $Ar = c$ - then you have the vector $v$. – Paul Aljabar May 11 '17 at 19:20
  • Oh yes. After this the uniqueness follows because the basis vectors are independent and the system $AX=0$ has only the trivial solution. Thanks a lot. I should've thought of it as a system of equations. – Johnathan G May 11 '17 at 19:23
  • This is weird: the answer says "we're are given that $;\langle \alpha_i,,v\rangle=c_i;$ ..." , yet this is exactly what must be proved ! Nevertheless, the OP accepted this answer. Oh, well... – DonAntonio May 11 '17 at 20:28
  • My bad, have proved something slightly different. Namely that it there is a vector with the property​ (gives the $c_i$ when hit with the $\alpha_i$) then there isn't another. I should have phrased it as "can we solve uniquely for $r$ in $A r = c$?" The answer is yes, and a unique $r$ provides a unique vector in the original vector space, because of the isomorphism between a vector space and the $R^n$ space of its coordinates. – Paul Aljabar May 11 '17 at 21:43
  • Okay, hopefully have fixed now ... – Paul Aljabar May 12 '17 at 11:22
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Fill in details

First, there exists a unique linear functional $\;f\in V^*\;$ s.t. $\;f(\alpha_i)=c_i\;$ .

Next, Riesz Representation Theorem tells us there exists a unique vector $\;\alpha\in V\;$ s.t.

$$\;f(\alpha_i)=\langle\alpha_i,\,\alpha\rangle\;$$

DonAntonio
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