Problem 13 - Advanced Engineering Mathematics

Question

The given vectors span a subspace \(W\) of \(R^{3}\). Use the Gram-Schmidt orthogonalization process to construct an orthonormal basis for the subspace. $$ \mathbf{u}_{1}=\langle 1,5,2\rangle, \mathbf{u}_{2}=\langle-2,1,1\rangle $$

Step-by-Step Solution

Answer

The orthonormal basis is \(\{ \mathbf{e}_1, \mathbf{e}_2 \}\).

Step 1: Start with Basis Vectors

The basis of the subspace \(W\) is given by the vectors \(\mathbf{u}_1 = \langle 1, 5, 2 \rangle\) and \(\mathbf{u}_2 = \langle -2, 1, 1 \rangle\). Our goal is to apply the Gram-Schmidt process to these vectors to obtain an orthonormal basis.

Step 2: Normalize the First Vector

First, calculate the norm of \(\mathbf{u}_1\): \[\| \mathbf{u}_1 \| = \sqrt{1^2 + 5^2 + 2^2} = \sqrt{1 + 25 + 4} = \sqrt{30}\]The normalized vector \(\mathbf{e}_1\) is:\[\mathbf{e}_1 = \frac{\mathbf{u}_1}{\| \mathbf{u}_1 \|} = \left\langle \frac{1}{\sqrt{30}}, \frac{5}{\sqrt{30}}, \frac{2}{\sqrt{30}} \right\rangle\]

Step 3: Make Second Vector Orthogonal to First

Project \(\mathbf{u}_2\) onto \(\mathbf{e}_1\):\[\text{proj}_{\mathbf{e}_1} \mathbf{u}_2 = \frac{\mathbf{u}_2 \cdot \mathbf{e}_1}{\mathbf{e}_1 \cdot \mathbf{e}_1} \mathbf{e}_1\]Calculate the dot product:\[\mathbf{u}_2 \cdot \mathbf{e}_1 = \left( -2, 1, 1 \right) \cdot \left( \frac{1}{\sqrt{30}}, \frac{5}{\sqrt{30}}, \frac{2}{\sqrt{30}} \right) = \frac{-2}{\sqrt{30}} + \frac{5}{\sqrt{30}} + \frac{2}{\sqrt{30}} = \frac{5}{\sqrt{30}}\]Thus, the projection is:\[\text{proj}_{\mathbf{e}_1} \mathbf{u}_2 = \frac{5}{30} \mathbf{e}_1 = \frac{1}{6} \mathbf{e}_1\]Now, calculate the orthogonal component, \(\mathbf{v}_2\):\[\mathbf{v}_2 = \mathbf{u}_2 - \text{proj}_{\mathbf{e}_1} \mathbf{u}_2 = \left\langle -2, 1, 1 \right\rangle - \frac{1}{6} \left\langle 1, 5, 2 \right\rangle = \left\langle -2 - \frac{1}{6}, 1 - \frac{5}{6}, 1 - \frac{2}{6} \right\rangle = \left\langle -\frac{13}{6}, \frac{1}{6}, \frac{2}{3} \right\rangle\]

Step 4: Normalize the Orthogonal Vector

Calculate the norm of \(\mathbf{v}_2\):\[\| \mathbf{v}_2 \| = \sqrt{\left( -\frac{13}{6} \right)^2 + \left( \frac{1}{6} \right)^2 + \left( \frac{2}{3} \right)^2} = \sqrt{\frac{169}{36} + \frac{1}{36} + \frac{4}{9}}\]\[= \sqrt{\frac{169 + 1 + 16}{36}} = \sqrt{\frac{186}{36}} = \sqrt{\frac{31}{6}} = \frac{\sqrt{31}}{6}\]The normalized vector \(\mathbf{e}_2\):\[\mathbf{e}_2 = \frac{\mathbf{v}_2}{\| \mathbf{v}_2 \|} = \left\langle -\frac{13}{\sqrt{31}}, \frac{1}{\sqrt{31}}, \frac{4}{\sqrt{31}} \right\rangle\]

Step 5: Conclusion

The orthonormal basis for the subspace \(W\) is \(\{ \mathbf{e}_1, \mathbf{e}_2 \}\). Specifically:\[\mathbf{e}_1 = \left\langle \frac{1}{\sqrt{30}}, \frac{5}{\sqrt{30}}, \frac{2}{\sqrt{30}} \right\rangle\]\[\mathbf{e}_2 = \left\langle -\frac{13}{\sqrt{31}}, \frac{1}{\sqrt{31}}, \frac{4}{\sqrt{31}} \right\rangle\]

Key Concepts

Gram-Schmidt process

The Gram-Schmidt process is a method used to transform a set of linearly independent vectors into an orthogonal or orthonormal set. This process helps in finding a basis where vectors are mutually perpendicular to each other. The process involves two main steps: making a vector orthogonal to previous vectors, and then normalizing it.

• Start with a set of vectors that span the subspace of interest.
• Take the first vector as it is, and for each subsequent vector, subtract the projection of the previous orthogonal vectors from it to make it orthogonal.
• Normalize each orthogonal vector to turn them into unit vectors, completing the creation of an orthonormal set.

The Gram-Schmidt process is an integral part of linear algebra, used to ensure a simpler representation of vector spaces and ease computations like finding coefficients in vector projections.
Recognizing its utility is vital for applications like QR factorization and signal processing.

Vector normalization

Vector normalization is a crucial step in the Gram-Schmidt process. It involves converting a vector to a unit vector, meaning a vector that retains its direction but has a length (or magnitude) of one.

• Calculate the norm of a vector, which is the square root of the sum of the squares of its components.
• Divide each component of the vector by its norm.

In the solution, vector normalization was applied to both the original vector (\(\mathbf{u}_1\)) and the orthogonalized vector, resulting in unit vectors \(\mathbf{e}_1\) and \(\mathbf{e}_2\). This conversion is necessary for creating an orthonormal basis, where all vectors have a length of one and are orthogonal to each other.
Normalization is widely utilized across various fields such as data science for preparing input data and machine learning models.

Orthogonal basis

An orthogonal basis for a vector space is a set of basis vectors that are all perpendicular, or orthogonal, to each other. This concept simplifies many computations in linear algebra, as orthogonal vectors allow for straightforward calculation of coefficients in linear combinations.

• Orthogonality makes it easier to handle projections and simplifies matrix operations due to the diagonal structure of the resulting Gram matrix.
• Each vector in the orthogonal basis can be expressed independently of the others.

In the context of the example, after using the Gram-Schmidt process on vectors \(\mathbf{u}_1\) and \(\mathbf{u}_2\), we obtained orthogonal vectors and then normalized them. This ensures that any vector in the subspace can be projected more simply, as each orthogonal vector aligns perpendicularly to the others.

Orthonormal basis

An orthonormal basis takes the concept of an orthogonal basis a step further by ensuring that all basis vectors are unit vectors. That means each vector not only is orthogonal to the others but also has a magnitude of one.

• Orthonormal bases simplify computations even more, allowing straightforward computation of vector lengths and making calculations like dot products particularly easy.
• In an orthonormal basis, the Gram matrix becomes the identity matrix, which is especially useful for solving equations.

In our example, the formation of the orthonormal basis using the vectors \(\mathbf{e}_1\) and \(\mathbf{e}_2\) completed the solution. Elements of this basis are not only orthogonal but also normalized, signifying that they span the same subspace effectively with optimized calculations. Orthonormal bases are often preferred in computational algorithms due to their numerical stability and simplicity in handling matrix operations.