2
$\begingroup$

Is there a non-constructive proof of this statement, i.e., one that avoids Gram-Schmidt?

$\endgroup$
2
  • 2
    $\begingroup$ I don't know but I think there is not, as GS is precisely the theorem (the only one I know of) that assures us the existence of orthonormal basis. $\endgroup$
    – DonAntonio
    Dec 8, 2013 at 6:53
  • $\begingroup$ yes. i agree with donantonio. $\endgroup$
    – Haha
    Dec 8, 2013 at 6:54

1 Answer 1

2
$\begingroup$

Here's a non-constructive Gram-Schmidt, I'm not sure its what you want though. All 1 dimensional subspaces have an orthonormal basis. Now for a $k$-dimensional subspace, $W$, with $k\geq2$, pick any $k-1$ dimensional of $V\subset W$. By induction $V$ has an orthonormal basis $\{x_1,\dots, x_{k-1}\}$. We'll let $V^{\perp}$ denote the orthogonal complement to $V$ in $W$. By dimension considerations $V^{\perp}$, has a non-zero vector, $x_k$. Then $\{x_1,\dots,x_{k-1},x_k\}$ is an orthonormal basis for $W$.

Remark:

"dimension considerations" are essentially using the fact that $V\oplus V^{\perp}=W$, you will want to make sure that you are not using Graham-Schmidt in order to prove this to avoid circularity. This can be done, $V\oplus V^{\perp}=W$ is just rank nullity applied to the projection operator.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .