Problem 8 - Advanced Engineering Mathematics

Question

Use the Gram-Schmidt orthogonalization process (3) to transform the given basis \(B=\left\\{\mathbf{u}_{1}, \mathbf{u}_{2}\right\\}\) for \(R^{2}\) into an orthogonal basis \(B^{\prime}=\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}\right\\}\). Then form an orthonormal basis \(B^{\prime \prime}=\left\\{\mathbf{w}_{1}, \mathbf{w}_{2}\right\\}\) (a) First construct \(B^{\prime \prime}\) using \(\mathbf{v}_{1}, \mathbf{u}_{1}\). (b) Then construct \(B^{\prime \prime}\) using \(\mathbf{v}_{1}, \underline{u}_{2}\). (c) Sketch \(B\) and each basis \(B^{\prime \prime}\). $$ B=\\{\langle 5,7\rangle,\langle 1,-2\rangle\\} $$

Step-by-Step Solution

Answer

The orthogonal basis B' is \(\{\langle 5,7 \rangle, \langle \frac{119}{74}, \frac{-85}{74} \rangle \}\). The orthonormal basis B'' has vectors normalized from these.

Step 1: Set Up the Problem

We are given basis \(B=\{\mathbf{u}_{1}, \mathbf{u}_{2}\}\) where \(\mathbf{u}_1 = \langle 5, 7 \rangle\) and \(\mathbf{u}_2 = \langle 1, -2 \rangle\). We need to use the Gram-Schmidt process to find an orthogonal basis \(B' = \{\mathbf{v}_1, \mathbf{v}_2\}\).

Step 2: Start Gram-Schmidt Orthogonalization

Set \(\mathbf{v}_1 = \mathbf{u}_1 = \langle 5, 7 \rangle\). To find \(\mathbf{v}_2\), compute \(\mathbf{v}_2 = \mathbf{u}_2 - \text{proj}_{\mathbf{v}_1}(\mathbf{u}_2)\), where \(\text{proj}_{\mathbf{v}_1}(\mathbf{u}_2) = \frac{\mathbf{u}_2 \cdot \mathbf{v}_1}{\mathbf{v}_1 \cdot \mathbf{v}_1} \mathbf{v}_1\).

Step 3: Calculate Projection

Compute the dot products: \(\mathbf{u}_2 \cdot \mathbf{v}_1 = 5 \times 1 + 7 \times (-2) = 5 - 14 = -9\) and \(\mathbf{v}_1 \cdot \mathbf{v}_1 = 5^2 + 7^2 = 25 + 49 = 74\). Then, \(\text{proj}_{\mathbf{v}_1}(\mathbf{u}_2) = \frac{-9}{74}\mathbf{v}_1\).

Step 4: Find \(\mathbf{v}_2\)

Substitute the projection into the formula: \(\mathbf{v}_2 = \langle 1, -2 \rangle - \frac{-9}{74} \langle 5, 7 \rangle = \langle 1, -2 \rangle + \langle \frac{45}{74}, \frac{63}{74} \rangle = \langle \frac{74}{74} + \frac{45}{74}, \frac{-148}{74} + \frac{63}{74} \rangle = \langle \frac{119}{74}, \frac{-85}{74} \rangle\).

Step 5: Normalize to Find Orthonormal Basis

To find \(\mathbf{w}_1\) and \(\mathbf{w}_2\), normalize \(\mathbf{v}_1\) and \(\mathbf{v}_2\): \(\mathbf{w}_1 = \frac{\mathbf{v}_1}{\|\mathbf{v}_1\|}\) and \(\mathbf{w}_2 = \frac{\mathbf{v}_2}{\|\mathbf{v}_2\|}\). Compute \(\|\mathbf{v}_1\| = \sqrt{25 + 49} = \sqrt{74}\) and \(\|\mathbf{v}_2\| = \sqrt{\left(\frac{119}{74}\right)^2 + \left(\frac{-85}{74}\right)^2}\).

Step 6: Construct Orthonormal Basis B'' Using \(\mathbf{v}_1, \mathbf{u}_1\)

Since the Gram-Schmidt process involves only \(\mathbf{v}_1\) and \(\mathbf{u}_1\), this is equivalent to \(\mathbf{w}_1\), derived earlier.

Step 7: Construct Orthonormal Basis B'' Using \(\mathbf{v}_1, \underline{u}_2\)

For this step, use \(\mathbf{w}_1\) already found and normalize \(\underline{u}_2\) to find \(\mathbf{w}_2\) for this basis.

Step 8: Sketch Basis Vectors

Graphical representation: draw the original basis vectors \(\mathbf{u}_1\) and \(\mathbf{u}_2\), then draw the orthogonal basis vectors \(\mathbf{v}_1\) and \(\mathbf{v}_2\), and finally draw the orthonormal basis vectors \(\mathbf{w}_1\) and \(\mathbf{w}_2\).

Key Concepts

Orthogonal Basis

In a vector space, a set of vectors is considered an orthogonal basis when each pair of distinct vectors in the set is orthogonal, meaning their dot product is zero. This ensures that the vectors are independent and will not interfere with each other during computations. The Gram-Schmidt process is a popular method used to convert any given set of linearly independent vectors into an orthogonal basis.

• Starting with a given vector basis, select the first vector as it is, and assign it to the first vector of the orthogonal basis.
• For each subsequent vector, subtract from it the projection onto every previously determined orthogonal vector.
• This ensures that the resulting vectors are orthogonal to each other.

In our example exercise, we transformed the basis \( B = \{\mathbf{u}_1, \mathbf{u}_2\} \) into an orthogonal basis \( B' = \{\mathbf{v}_1, \mathbf{v}_2\} \). The vector \( \mathbf{v}_1 \) remained as \( \mathbf{u}_1 \), while \( \mathbf{v}_2 \) was computed by subtracting the projection of \( \mathbf{u}_2 \) onto \( \mathbf{v}_1 \). This process ensured that \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \) are orthogonal to each other.

Orthonormal Basis

Once we have an orthogonal basis, we can easily convert it into an orthonormal basis. An orthonormal basis means that not only are the vectors orthogonal, but each vector also has a unit length (norm of 1). Normalizing a vector involves dividing it by its magnitude, ensuring its length is 1.

• To normalize a vector, calculate its magnitude using the square root of the sum of the squares of its components.
• Divide each component of the vector by this magnitude.

In the exercise, we took the orthogonal basis \( B' = \{\mathbf{v}_1, \mathbf{v}_2\} \) and normalized each vector to form the orthonormal basis \( B'' = \{\mathbf{w}_1, \mathbf{w}_2\} \). This made \( \mathbf{w}_1 \) and \( \mathbf{w}_2 \) not only orthogonal but also ensured they had unit lengths. This transformation is crucial in many applications such as simplifying calculations in quantum mechanics and performing dimensionality reduction in machine learning.

Vector Spaces

Vector spaces, also known as linear spaces, are fundamental in linear algebra and mathematics. They consist of a set of vectors along with vector addition and scalar multiplication operations satisfying certain conditions. A vector space provides a context within which vectors exist and interact.

• Vector addition must be commutative and associative.
• There is a zero vector, which acts as an additive identity.
• Scalar multiplication must distribute over both vector addition and field addition.
• There must be an element-wise scalar multiplication identity.

In our exercise, we worked within the vector space \( R^2 \), which means we dealt with vectors containing two real number components. The basis \( B = \{\mathbf{u}_1, \mathbf{u}_2\} \) was linearly independent, forming a complete basis for this vector space, allowing the representation of any other vector in \( R^2 \) as a linear combination of \( \mathbf{u}_1 \) and \( \mathbf{u}_2 \). Understanding these vector spaces is essential for anyone delving into applied or theoretical fields that use vector and matrix operations extensively.