0
$\begingroup$

Let $V$ be a finite dimensional inner product space over $k$ with basis $\{v_1,\dots,v_n\}$ and inner product $\langle \cdot,\cdot\rangle$. For any $\alpha_1,\dots,\alpha_n\in k$, there exists a unique $v\in V$ such that $\langle v,v_i\rangle = \alpha_i$ for each $i$.

I read this and didn't think much of it at first, but I am having difficulty proving existence. Showing uniqueness is easy. For existence, I tried using the matrix associated to the inner product, say $Q$, and writing $$ v^tQv_i = \alpha_i $$ and computing $Qv_i$ to see how I might choose $v$, but this doesn't seem to give me anything. How does existence work in this claim?

$\endgroup$
  • 1
    $\begingroup$ One possible hint: do Gram-Schmidt on $\{v_i\}$. $\endgroup$ – Willie Wong May 23 '14 at 11:21
1
$\begingroup$

The linear mapping $$\Phi:V\to k^n,v\mapsto (\langle v,v_1\rangle,\ldots,\langle v,v_n\rangle)$$ is one to one because its kernel is reduced to $\{0\}$, this is the uniqueness part of your statement. But since $\dim V=\dim k^n$, this is sufficient to prove that $\Phi$ is also surjective, and this is the existence part.

$\endgroup$
  • $\begingroup$ This is very clean! $\endgroup$ – user59083 May 23 '14 at 11:54
1
$\begingroup$

Assume that the basis is orthonormal. Then the solution is trivial: $$ v = \sum_i \alpha_i v_i, $$ since $\langle v,v_i \rangle = \langle \alpha_i v_i,v_i \rangle = \alpha_i$. Therefore, as suggested by Willie Wong in the comments, a good idea is to use the given basis to produce an orthonormal basis, and this procedure is well known.

$\endgroup$

Your Answer

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