How do I find an orthonormal basis of the column space of A?

For any matrix A, I would do this by (1) Finding a basis (any basis) for the column space of A, then (2) Applying the Gram-Schmidt process to the basis produced in (1). The standard way of doing (1) is to apply row operations to A until it is in row echelon form, then taking the columns of A that correspond to columns that have pivots in them in the row echelon form. If you apply that here, though, you find that the row echelon form of A has a pivot in every column. This implies that the column space of A is all of R^3. There is an obvious orthonormal basis to use for R^3, namely (1,0,0), (0,1,0), and (0,0,1), so you could just do that. More generally, whenever A is an nxn matrix whose row echelon form has a pivot in every column, you can just take the standard basis of R^n as an orthonormal basis for the column space of A. You really only need to resort to the process (1) and (2) above when the column space is not all of R^n but some proper subspace of it.

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I will use C(V) for the orthogonal complement of V suppose x is in C(V) then <x,v> = 0 for all v in V. But U subspce of V means that this is true for all u in U. Therefore x is in C(U). therefore C(V) is subset of C(U). 2. let v be in V. Now for all x in C(V), <v,x> = 0. Therefore v is in C(C(V)). Therefore V i subset of C(C(V)). 4. let x be in C(U+V). then for all y in (U+V), <x,y> = 0. in particular, let u be in U. Then <x,u> = 0, and x is in C(U) analagously, let c be in V. Then <x,v> = 0, and x is in C(V), therefore C(U+V) subset of C(U) intersect C(V) now let y be in C(U) intersect C(V). let z be in U+V. then z can be written as au + bv for scalars a,b, u in U, and v in V <y,z> = a<y,u> + b<y,z>. By hypothesis, both dot products are 0, and therefore <y,z> = 0. So C(U) intersect C(V) is a subset of C(U+V). qed