Remove old directions before adding a new one

Try

Start with v₁=(1,1,0) and v₂=(1,0,1). They are independent but not orthogonal. Removing v₂’s component along v₁ gives u₂=(1/2,−1/2,1), whose dot product with v₁ is zero.

Predict

Predict why you must subtract every earlier direction, not only the most recent. Gram–Schmidt builds uⱼ=vⱼ−Σprojᵤᵢ(vⱼ) over i<j. The nonzero uᵢ are orthogonal. Dividing each by its positive norm produces orthonormal qᵢ.

Experiment

Calculate v₂·v₁=1 and v₁·v₁=2, then subtract half v₁. Verify u₂·v₁=0 and ||u₂||²=3/2. Reconstruct v₂=u₂+(1/2)v₁ to show the span has not changed. No square-root decimal is needed for this unnormalized check.

Understand

Worked three-direction example: start with (1,0,0),(1,1,0),(1,1,1). The first direction stays fixed. Removing its projection from the second yields (0,1,0). Removing both earlier projections from the third yields (0,0,1). The resulting basis is orthonormal and spans the same R³.

In general each subtraction uses previously constructed orthogonal directions, so it removes exactly those components. Reversible reconstruction shows the span of each initial segment is preserved. For an independent input list no residual is zero; a zero residual instead reveals dependence and must not be normalized.

For a full-column-rank matrix A, put qᵢ in Q and the reconstruction coefficients in upper-triangular R. Then A=QR and QᵀQ=I. Q may be rectangular: this is thin QR.

Apply

Check dot products after each step and retain exact fractions. In a written QR answer, give both the orthogonality check and the reconstruction. Merely obtaining vectors of length one does not prove they are orthogonal.

Continue

Depth: classical Gram–Schmidt can lose orthogonality in floating-point arithmetic for nearly dependent inputs. Modified Gram–Schmidt and Householder QR improve numerical behavior. Exact zero and a small computed residual need different interpretations; production numerical algorithms belong to the later numerical sequence.