Gram Schmidt Orthonormalization Process With an Example

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In linear algebra and numerical analysis, Gram–Schmidt orthonormalization process is a process of transforming two or more vectors that are perpendicular to each other.

  • It is a case of inner product space, which is characterized by a set of linearly independent vectors.
  • The Gram-Schmidt orthonormalization process is also known as the orthogonalization or the Gram-Schmidt algorithm.
  • It is a set of linearly independent vectors S = {v1, ..., vk} for k ≤ n and generates an orthogonal set S′ = {u1, ..., uk}.
  • The process is useful for computing an orthonormal basis in an inner product space.
  • The Gram-Schmidt orthonormalization process, a fundamental tool in mathematical analysis, finds its applications in various real-world scenarios. 
  • It is used in motion planning, sensor calibration, and robot localization tasks, demonstrating its practical significance beyond theoretical understanding.

Key Terms: Gram-Schmidt Orthonormalization Process, Vectors, Linear Algebra, Euclidean Space, Variables


What is Gram-Schmidt Orthonormalization Process?

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The Gram-Schmidt orthonormalization process comprises a series of steps that transform a set of linearly independent vectors into a set of orthonormal vectors that cover the same space as the original set.

  • It is most commonly used in the Euclidean space Rn, which is equipped with the standard inner product.
  • The Gram-Schmidt orthonormalization process was discovered by Jørgen Pedersen Gram and Erhard Schmidt.
  • The process takes into account those vectors that generate the same number of orthonormal vectors.
  • The method was generalized by the Iwasawa decomposition using the theory of lie group decompositions.
  • It is used in matrices where column vectors of a rank matrix yield an orthogonal and a triangular matrix.

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State and Prove Gram-Schmidt Orthonormalization Process

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Statement: Let (v1,...,vm) be a set of linearly independent vectors in V lying in the k-dimensional subspace of Rn then (e1,...,em) be an orthonormal list.

Proof: We will begin with any basis for vector V to determine the orthonormal basis for vector V.

= Consider {v1,…,vk} to be a non-orthonormal matrix for V. 

= We will form a set of {u1,…,uk} repeatedly until {u1,…,up} will form an orthonormal basis for the vector {v1,…,vp}.

⇒ For p = 1 Let u1=1/ ∥v∥ 

= Suppose {u1,…,up-1} is an orthonormal basis for the span of a vector set {v1,…,vp}.

= Now determine the vector d such that ‖d‖=1,ui⋅d=0 for i=1,…,p−1.

= vp can be represented as a linear combination of u1,…,up−1,d.

⇒ For time being eliminate the condition ‖d‖=1 and determine the value for d in the following system:

u1. d = 0

u2.d = 0

up-1.d = 0

= λ1u1⋯⋅⋯λp−1up−1+αd = vp

= λ1,…,λp−1 and d1,…,dn represent the set of variables here. However, the final equation used in the system is not linear. 

Since vp does not lie inside the range of u1,…,up−1, we cannot have any solution for value α=0. 

= Now consider that α≠0 and solve the system of equations by substituting the value of u=αd.

(1/α)u1⋅u=0

(1/α)u2⋅u=0

(1/α)up-1⋅u=0

λ1u1⋯⋅⋯λp−1up−1+u = vp

= In other words it can be represented as:

u1⋅u = 0

u2⋅u = 0

up-1⋅u = 0

= λ1u1⋯⋅⋯λp−1up−1+u = vp

= Now determine u = vp−(λ1u1+⋯+λp−1up−1) using the last equation of the system. 

= Substituting the ith value in the equation i∈{1,…,p−1}, we will get

ui ⋅ (vp−(λ1u1+⋯+λp−1up−1))=0

= This is equivalent to ui⋅vpi

= The required value of u is represented as: u = vp−((u1⋅vp)u1 + (u2⋅vp)u2 + ⋯ +(up−1⋅vp)up−1) where u≠0. 

= Now consider up=(1 / ∥u∥)u.

⇒ The summary of the procedure is as follows:

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)

ui = (1/||ei||). ei, where ei=vi−((u1⋅vi)u1+(u2⋅vi)u2+⋯+(ui−1⋅vi)ui−1))

uk = (1/||ek||). ek where ek=vk−((u1⋅vk)u1+(u2⋅vk)u2+⋯+(uk−1⋅vk)uk−1))


Gram-Schmidt Orthonormalization Process Example

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The example of gram-schmidt orthonormalization process is as follows:

Example: Consider the following set of values:

\(V_1 = \begin {bmatrix} 1\\1 \\1 \end{bmatrix} V_2 = \begin {bmatrix} 1\\2 \\0 \end{bmatrix} \; and\; V_3 = \begin {bmatrix} 2\\0 \\1 \end{bmatrix}\)

\(\text{solution let }u_1=v_1= = \begin {bmatrix} 1\\1 \\1 \end{bmatrix}and \; similarly\)

\(\Rightarrow u_2=v_2-proj_{u_1}.{v_2} = \begin {bmatrix} 1\\2 \\0 \end{bmatrix}-\frac{3}{3}\begin {bmatrix} 1\\1 \\1 \end{bmatrix}= \begin {bmatrix} 0\\1 \\-1 \end{bmatrix}\)

\(\Rightarrow u_3=\begin {bmatrix} 2\\0 \\1 \end{bmatrix}-\frac{3}{3}\begin {bmatrix} 1\\1 \\1 \end{bmatrix}+\frac{1}{2}\begin {bmatrix} 0\\1 \\-1 \end{bmatrix}= \begin {bmatrix} 1\\-0.5 \\0.5 \end{bmatrix}\)

\(\text{Now determine the orthogonal basis: }\)

\(W_1= \frac{u_1}{ {|u_1|}}=\frac{1}{\sqrt 3}\begin {bmatrix} 1\\1 \\1 \end{bmatrix}=W_1=\begin {bmatrix} 0.57\\0.57 \\0.57 \end{bmatrix}\)

\(Similarly \;W_2 =\frac{1}{\sqrt2}\begin {bmatrix} 0\\1 \\-1 \end{bmatrix}= \begin {bmatrix} 0\\0.707 \\-0.707 \end{bmatrix}\)

\(W_3=\frac{\sqrt 6}{2}\begin {bmatrix} 0\\0.5 \\-0.5 \end{bmatrix}= \begin {bmatrix} 0\\0.4082 \\-0.4082 \end{bmatrix}\)


Things to Remember

  • Technically, the Gram–Schmidt orthonormalization process is a method of constructing an orthonormal basis from a set of vectors in an inner product space.
  • Orthogonalization refers to the construction of a set of orthogonal vectors formed from a particular subspace.
  • Orthonormalization is a type of orthogonalization that forms a unit vector.
  • Before the term was coined, Pierre-Simon Laplace analyzed it.
  • The Gram-Schmidt orthonormalization process can be stabilized using a technique called modified Gram-Schmidt or MGS.

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Sample Questions

Ques: What is Gram-Schmidt orthonormalization process? (2 marks)

Ans: The Gram-Schmidt orthonormalization process transforms a set of vectors into an equivalent set of orthonormal vectors. It aims to convert a set of linearly independent vectors into a set of orthonormal vectors covering the same space as the original set.

Ques: What is the application of the Gram-Schmidt orthonormalization process? (3 marks)

Ans: The Gram–Schmidt orthogonalization process is a method for ortho normalizing a set of vectors in an inner product space. The QR decomposition is achieved by employing the Gram–Schmidt process on the column vectors of a full column rank matrix. It is used to orthonormalize vectors in an inner product space, typically the Euclidean space Rn with the standard inner product.

Ques: What is the primary purpose of the Gram-Schmidt process? (2 marks)

Ans: The primary purpose of the Gram-Schmidt process is to convert a set of linearly independent vectors into a set of orthonormal vectors covering the same space as the original set.

Ques: Give an example of the Gram-Schmidt orthogonalization process? (3 marks)

Ans: An example of the Gram-Schmidt orthogonalization process involves considering the vectors (2,1) and (3,2) in the (x,y) plane, which form a basis. These vectors are neither perpendicular nor of unit length. By applying the Gram-Schmidt process, the vectors (1,0) and (0,1) are obtained, which form an orthonormal basis for the (x,y) plane.

Ques: What is the utility of the Gram-Schmidt orthogonalization process? (2 marks)

Ans: The Gram-Schmidt orthogonalization process is utilized to transform a set of linearly independent vectors into an orthonormal set, thereby creating an orthonormal basis. This process facilitates the determination of vector linear independence.

Ques: How is the Gram-Schmidt process implemented? (3 marks)

Ans: The Gram-Schmidt process can be executed manually or through computer algorithms. Programming languages like MATLAB or Python offer efficient implementations of the Gram-Schmidt process. By hand, the algorithm involves iteratively computing vector projections and normalizations.

Ques: What is the role of the Gram-Schmidt process? (2 marks)

Ans: The Gram-Schmidt process is employed to orthonormalize vectors, ensuring orthogonality and unit magnitude for each vector. It is commonly used to construct orthonormal bases in inner product spaces.

Ques: What is the Gram-Schmidt orthogonalization process in the field of digital communication? (3 marks)

Ans: In digital communication, binary bits are used as input and a digital modulator transforms them into waveforms and symbols. 

  • The waveforms ought to be easily distinguishable from one other so that we can determine which bit or symbol is being transmitted. 
  • To make them unique, we apply the Gram-Schmidt Orthogonalization procedure.
  • The waveforms can be easily recognized on the demodulation side and do not overlap when they are orthogonalized.

Ques: What is the difference between the Gram-Schmidt orthogonalization process and the modified Gram-Schmidt process?  (3 marks)

Ans: The computational procedures involved in modified Gram-Schmidt are identical to those in classical Gram-Schmidt. It does so, albeit, in a somewhat different order. In conventional Gram-Schmidt, a total involving all previously computed vectors is calculated in each iteration. The updated version allows you to fix mistakes at every stage.

Ques: Consider the following set of values: \(V_1=\begin {bmatrix} 1\\1 \\1 \end{bmatrix}, V_2=\begin {bmatrix} 1\\1 \\-1\\-1 \end{bmatrix} and\; V_3= \begin {bmatrix} 0\\-1 \\2\\1 \end{bmatrix}\) (4 marks)

Ans: \(\text{As we know}\; u_1=\frac{1}{||v_1||}=\frac{1}{2}\begin {bmatrix} 1\\1 \\1\\1 \end{bmatrix}\)

\(\Rightarrow e_2=v_2-(u_1.v_2)u_1=\begin {bmatrix} 1\\1 \\-1\\-1 \end{bmatrix}-(0)(u_1)\)

\(=e_2=\begin {bmatrix} 1\\1 \\-1\\-1 \end{bmatrix}\)

\(\Rightarrow similarly\;u_2=\frac{1}{2}\begin {bmatrix} 1\\1 \\-1\\-1 \end{bmatrix}\)

\(= \text{Since v_2 and u, are orthogonal}\)

\(\Rightarrow e_3=v_3-(u_1.v_3)u_1-(u_2.v_3)u_2\)

\(\Rightarrow e_3=\begin {bmatrix} 0\\-1 \\2\\1 \end{bmatrix}-\frac{1}{2}\begin {bmatrix} 1\\1 \\1\\1 \end{bmatrix}+\begin {bmatrix} 1\\1 \\-1\\-1 \end{bmatrix}\)

\(\Rightarrow e_3=\begin {bmatrix} \frac{1}{2}\\-\frac{1}{2} \\\frac{1}{2}\\-\frac{1}{2} \end{bmatrix}and\;u_3=\begin {bmatrix} \frac{1}{2}\\-\frac{1}{2} \\\frac{1}{2}\\-\frac{1}{2} \end{bmatrix}\)

\(\therefore \{V_1,V_2,V_3\}and\{u,u_2,u_3\}are \;orthornormal\)

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