# Every subspace of $\mathbb{R}^{n}$ has an orthonormal basis

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

• 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. Dec 8, 2013 at 6:53
• yes. i agree with donantonio.
– Haha
Dec 8, 2013 at 6:54

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 $$W$$.
"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.