The Gram–Schmidt orthonormalization process is a procedure for orthonormalizing a set of vectors in an inner product space, most often the Euclidean space Rⁿ provided with the standard inner product, in mathematics, notably linear algebra and numerical analysis. Let us explore the Gram Schmidt orthonormalization process with a solved example in this article.

What is Gram Schmidt Orthonormalization Process?

Let V be a k-dimensional subspace of Rⁿ. Begin with any basis for V, we look at how to get an orthonormal basis for V.

Allow {v₁,…,v_k} to be a non-orthonormal basis for V. We’ll build {u₁,…,u_k} repeatedly until {u₁,…,u_p} is an orthonormal basis for the span of {v₁,…,v_p}.

We just use u₁=1/ ∥v₁∥ for p=1.

u₁,…,u_p-1 is assumed to be an orthonormal basis for the span of v₁,…,v_p.

Note that for i=1,…,p₁, we require a vector d such that ∥d∥=1, u_i⋅d=0, and that v_p may be expressed as a linear combination of u₁,…,u_p-1, d. For the time being, we’ll ignore the requirement ∥d∥=1 and work for d in the following system:

u₁. d = 0

u₂.d = 0

…

…

…

u_p-1.d = 0

λ₁u₁⋯⋅⋯λ_p−1u_p−1+αd = v_p

λ₁,…,λ_p−1 and d₁,…,d_n are the variables here. However, the system’s final equation isn’t linear. However, since v_p is not inside the range of u₁,…,u_p−1, we cannot have α=0 for any solution. Considering that α≠0, we can solve for the system by substituting u=αd.

(1/α)u₁⋅u=0

(1/α)u₂⋅u=0

…

…

…

(1/α)u_p-1⋅u=0

λ₁u₁⋯⋅⋯λ_p−1u_p−1+u = v_p

or, in other words:

u₁⋅u = 0

u₂⋅u = 0

u_p-1⋅u = 0

λ₁u₁⋯⋅⋯λ_p−1u_p−1+u = v_p

We can construct u = v_p−(λ₁u₁+⋯+λ_p−1u_p−1) using the last equation. By plugging this into the ith equation i∈{1,…,p−1}, we get

u_i ⋅ (v_p−(λ₁u₁+⋯+λ_p−1u_p−1))=0

which is equivalent to u_i⋅v_p=λ_i. As a result, we must

u = v_p−((u₁⋅v_p)u₁ + (u₂⋅v_p)u₂ + ⋯ +(u_p−1⋅v_p)u_p−1).

It’s worth noting that u≠0. As a result, we can take u_p=(1 / ∥u∥)u.

The following is a summary of the procedure:

u₁=(1/||v₁||). v₁

u₂ = (1/||e₂||). e₂ where e₂=v₂−(u₁⋅v₂)u₁

u₃ = (1/||e₃||). e₃ where e₃=v₃−((u₁⋅v₃)u₁+(u₂⋅v₃)u₂)

…

…

u_i = (1/||e_i||). e_i, where e_i=v_i−((u₁⋅v_i)u₁+(u₂⋅v_i)u₂+⋯+(u_i−1⋅v_i)u_i−1))

…

…

…

u_k = (1/||e_k||). e_k where e_k=v_k−((u₁⋅v_k)u₁+(u₂⋅v_k)u₂+⋯+(u_k−1⋅v_k)u_k−1))

Gram Schmidt Orthonormalization Process Example

Example:

Assume that

Apply the Gram Schmidt process to {v1, v2, v3}.

Solution:

As we know,

Now,

Hence,

v₂ and u₁ are already orthogonal at this point. By simply normalizing v₂, u₂ can be obtained.

Now,

As, ||e₃|| = 1 and now we have

As a result, for the span ({v₁, v₂, v₃}), {u₁, u₂, u₃} is an orthonormal basis.

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