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

1 Answer

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.