Questions: Gram-Schmidt Orthogonalization Process

Question 1

You apply Gram-Schmidt to a basis {v₁, v₂, v₃}. After computing e₁ (normalized v₁), you form u₂ = v₂ − ⟨v₂, e₁⟩e₁. What is guaranteed about u₂ before normalization?

• A. u₂ has unit length
• B. u₂ is perpendicular to e₁
• C. u₂ lies in a different subspace than v₂
• D. u₂ equals v₂ rotated exactly 90 degrees around the origin

Answer: (B)

The step u₂ = v₂ − ⟨v₂, e₁⟩e₁ subtracts from v₂ exactly its component in the e₁ direction (the projection). Computing ⟨u₂, e₁⟩ = ⟨v₂, e₁⟩ − ⟨v₂, e₁⟩⟨e₁, e₁⟩ = ⟨v₂, e₁⟩ − ⟨v₂, e₁⟩ = 0 confirms orthogonality. u₂ is not yet normalized (that comes next), does not 'rotate' v₂ (it changes direction by subtraction), and stays in the same span as v₁ and v₂.

Question 2

After applying Gram-Schmidt to three basis vectors of a plane in ℝ⁴, a student claims the resulting orthonormal vectors now describe a different plane because they point in different directions than the originals. Is this correct?

• A. Yes — Gram-Schmidt rotates the basis to align with coordinate axes, changing which plane is described
• B. No — Gram-Schmidt only changes how the plane is described (the basis), not the plane itself; the orthonormal vectors span exactly the same subspace as the originals
• C. Yes — normalization rescales the vectors, which alters the geometry of what they span
• D. No — but only if the original basis vectors were already mutually orthogonal

Answer: (B)

This is the crucial invariant of Gram-Schmidt: the subspace spanned does not change. At each step, uₖ = vₖ − (projections onto previous vectors) is a linear combination of vₖ and the earlier basis vectors — so it lies in the span of {v₁, …, vₖ}. The orthonormal set {e₁, …, eₖ} therefore spans the same k-dimensional subspace as {v₁, …, vₖ}. You are changing the *representation* of the space, not the space itself.

Question 3

Gram-Schmidt changes the subspace being described because the resulting orthonormal vectors are not parallel to the original basis vectors.

• T. True
• F. False

Answer: False

The direction of individual basis vectors changes, but the subspace spanned — the set of all linear combinations — does not. Because each new orthonormal vector is a linear combination of the original vectors (and vice versa), the two sets span exactly the same subspace. Gram-Schmidt is a change of basis within a fixed subspace, not a change of subspace.

Question 4

If two of the original basis vectors are already orthogonal to each other, the Gram-Schmidt projection step between them will produce a zero projection, leaving those vectors unchanged up to normalization.

• T. True
• F. False

Answer: True

If v₂ is already perpendicular to e₁, then ⟨v₂, e₁⟩ = 0, so the subtraction step u₂ = v₂ − 0·e₁ = v₂ leaves v₂ unchanged. Gram-Schmidt 'discovers' that no projection needs to be removed. This is correct behavior — the algorithm handles the already-orthogonal case gracefully and the only remaining step is normalization.

Question 5

Explain the geometric meaning of 'subtracting the projection' in a Gram-Schmidt step. Why does subtracting the projection of vₖ onto e₁ guarantee that the result is perpendicular to e₁?

Model answer: The projection ⟨vₖ, e₁⟩·e₁ is the component of vₖ in the direction of e₁ — its 'shadow' onto e₁. When you subtract this shadow, you remove everything vₖ shares with the e₁ direction, leaving only the part that is perpendicular to e₁. Algebraically: ⟨vₖ − ⟨vₖ, e₁⟩e₁, e₁⟩ = ⟨vₖ, e₁⟩ − ⟨vₖ, e₁⟩·⟨e₁, e₁⟩ = ⟨vₖ, e₁⟩ − ⟨vₖ, e₁⟩ = 0. The dot product is zero, confirming orthogonality.

This geometric intuition — project, then subtract the projection — is why Gram-Schmidt works. The projection captures exactly how much of vₖ 'points in the e₁ direction'; subtracting it leaves the remainder pointing purely perpendicular. Repeating for each previous direction strips out all overlap with the established basis, guaranteeing the new vector is orthogonal to all of them.