Linear Algebra · step 10 of 13

Orthogonality, projections & Gram–Schmidt

Two vectors are orthogonal when their dot product is zero. The projection of v onto u is (v·u/u·u)u — the component of v along u — and subtracting it leaves the part orthogonal to u. Gram–Schmidt turns any basis into an orthogonal one by repeatedly subtracting projections onto everything already produced.

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Method: how to approach it

The order below is what examiners expect to see, and each step carries its own marks.

Project onto a single vectorproj_u(v) = (v·u / u·u)·u. Normalising u first simplifies the denominator to 1.
Subtract to get the orthogonal partv − proj_u(v) is orthogonal to u. Verify with a dot product — it is a free check.
Run Gram–Schmidtu₁ = v₁; then uₖ = vₖ minus its projections onto every earlier uᵢ.
Normalise lastDivide each uᵢ by its length only at the end, to keep the arithmetic in fractions rather than surds.

Worked example

Apply Gram–Schmidt to v₁ = (1, 1, 0) and v₂ = (1, 0, 1).

u₁ = v₁ = (1,1,0), with u₁·u₁ = 2.
v₂·u₁ = 1·1 + 0·1 + 1·0 = 1.
proj = (1/2)(1,1,0) = (1/2, 1/2, 0).
u₂ = (1,0,1) − (1/2,1/2,0) = (1/2, −1/2, 1).

Answer. Orthogonal basis {(1,1,0), (1/2,−1/2,1)}. Check: 1·(1/2) + 1·(−1/2) + 0 = 0 ✓.

Where marks get dropped

These are the specific errors that cost credit on orthogonality, projections & gram–schmidt questions — QED's rubric penalises each of them separately.

Forgetting to divide by u·u in the projection formula. Omitting it is only correct when u is a unit vector.
Projecting onto the ORIGINAL vectors instead of the already-orthogonalised ones during Gram–Schmidt.
Normalising too early, which fills the remaining arithmetic with square roots for no benefit.

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Orthogonality, projections & Gram–Schmidt — frequently asked questions

Why is an orthogonal basis convenient?

Coordinates become simple dot products: v = Σ(v·uᵢ/uᵢ·uᵢ)uᵢ with no system to solve. For orthonormal bases the denominators vanish entirely.

What is the QR decomposition?

Gram–Schmidt applied to the columns of A, written as A = QR with Q orthonormal and R upper triangular. It is the numerically stable way to solve least-squares problems.

How does projection relate to least squares?

The least-squares solution of Ax = b projects b onto the column space of A, making the residual orthogonal to it — which gives the normal equations AᵀAx = Aᵀb.

The rest of Linear Algebra

Matrices, systems, determinants, eigenvalues. Each subtopic below has its own method, worked example and mark-losing traps.

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