What is the Gram-Schmidt basis?

How it works

The Gram-Schmidt process takes linearly independent vectors v₁, v₂, …, vₖ and builds orthonormal vectors e₁, e₂, …, eₖ that span the same subspace:

1. Start with the first vector: u₁ = v₁, and e₁ = u₁ ÷ |u₁|.
2. For each next vector vᵢ, subtract its projection onto every earlier unit vector: uᵢ = vᵢ − (vᵢ·e₁)e₁ − … − (vᵢ·eᵢ₋₁)eᵢ₋₁.
3. Normalize: eᵢ = uᵢ ÷ |uᵢ|, where |u| = √(u·u).

The vectors u₁, u₂, … are orthogonal; the vectors e₁, e₂, … are orthonormal. The calculator takes the vectors in the order you enter them (rows of the grid).

The page shows:

• Orthonormal basis: e₁, e₂, … as rows separated by semicolons, for example [0, 0.6, 0.8; …].
• u1, u2, …: the orthogonal vectors.
• Working: each uᵢ with its projection coefficients vᵢ·eⱼ, its length, and eᵢ. The same text is listed step by step under "How it is worked out".

Numbers are rounded to 10 significant figures. A component within 10⁻¹² of 0 (relative to the length of the vector it comes from) is shown as 0. Negative numbers use a plain hyphen, such as -0.48. The projection coefficients in the working follow the same zero rule. Numbers of 10²¹ or more, or under 10⁻⁶ (other than 0), are written in e-notation, for example 1.4e+22 or 1.5e-7.

Assumptions

• Up to 10 vectors, each with up to 10 real components.
• The vectors must be linearly independent. When the part of vᵢ outside the span of the earlier vectors is at most 10⁻¹⁰ of |vᵢ| (for example a zero vector, or v₂ = 2v₁), there is no answer. More vectors than components are always dependent.
• The page uses a numerically stable form of the same process (modified Gram-Schmidt, run twice), which gives the same vectors as the formulas above.
• Old links from the previous version of this page (vectorCount, dimension, and cells v00 to v99) open the same vectors. A missing cell counts as 0.

Worked examples by hand

The default vectors v₁ = (0, 3, 4), v₂ = (1, 0, 1), v₃ = (1, 1, 3).

• u₁ = (0, 3, 4), |u₁| = 5, so e₁ = (0, 0.6, 0.8).
• v₂·e₁ = 0 + 0 + 0.8 = 0.8. u₂ = (1, 0, 1) − 0.8(0, 0.6, 0.8) = (1, −0.48, 0.36). |u₂| = √1.36 = 1.1661904, so e₂ = (0.8574929257, −0.4115966043, 0.3086974533).
• v₃·e₁ = 0.6 + 2.4 = 3; v₃·e₂ = (1 − 0.48 + 1.08) ÷ 1.1661904 = 1.6 ÷ 1.1661904. u₃ = v₃ − 3e₁ − (1.6 ÷ 1.36)u₂ = (1, 1, 3) − (0, 1.8, 2.4) − (1.1764706, −0.5647059, 0.4235294) = (−0.1764705882, −0.2352941176, 0.1764705882), which is (−3, −4, 3)/17. |u₃| = √34/17, so e₃ = (−3, −4, 3)/√34 = (−0.5144957554, −0.6859943406, 0.5144957554).

Two vectors in the plane v₁ = (1, 1), v₂ = (1, 0). e₁ = (1, 1)/√2 = (0.7071067812, 0.7071067812). v₂·e₁ = 1/√2, so u₂ = (1, 0) − ½(1, 1) = (0.5, −0.5) and e₂ = (0.7071067812, −0.7071067812).

Vectors in 4D v₁ = (1, 2, 3, 4), v₂ = (4, 3, 2, 1). v₂·v₁ = 4 + 6 + 6 + 4 = 20 and v₁·v₁ = 30, so u₂ = v₂ − (20/30)v₁ = (4 − 2/3, 3 − 4/3, 2 − 2, 1 − 8/3) = (3.333333333, 1.666666667, 0, −1.666666667).

Each vector adds one axis v₁ = (2, 0, 0), v₂ = (3, 4, 0), v₃ = (5, 6, 7). e₁ = (1, 0, 0); u₂ = (3, 4, 0) − 3e₁ = (0, 4, 0), e₂ = (0, 1, 0); u₃ = (5, 6, 7) − 5e₁ − 6e₂ = (0, 0, 7), e₃ = (0, 0, 1).

One vector v₁ = (3, 4, 0): u₁ = (3, 4, 0), |u₁| = 5, e₁ = (0.6, 0.8, 0).