Problem 1 - Advanced Engineering Mathematics

Question

Verify that the basis \(B\) for the given vector space is orthonormal. Use Theorem \(7.7 .1\) to find the coordinates of the vector \(\mathbf{u}\) relative to the basis \(B .\) Then write \(\mathbf{u}\) as a linear combination of the basis vectors. $$ B=\left\\{\left\langle\frac{12}{13}, \frac{5}{13}\right\rangle,\left\langle\frac{5}{13},-\frac{12}{13}\right\rangle\right\\}, \quad R^{2} ; \quad \mathbf{u}=\langle 4,2\rangle $$

Step-by-Step Solution

Answer

The basis \( B \) is orthonormal. The coordinates are \( c_1 = \frac{58}{13} \), \( c_2 = -\frac{4}{13} \), and \( \mathbf{u} = c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 \).

Step 1: Verify Orthonormality

To verify that the basis \( B \) is orthonormal, we need to check two things for the vectors \( \mathbf{v}_1 = \left\langle \frac{12}{13}, \frac{5}{13} \right\rangle \) and \( \mathbf{v}_2 = \left\langle \frac{5}{13}, -\frac{12}{13} \right\rangle \):1. Each vector must be a unit vector (norm equals 1).2. The vectors must be orthogonal (dot product equals 0).Calculate the norm of \( \mathbf{v}_1 \):\[ \| \mathbf{v}_1 \| = \sqrt{ \left( \frac{12}{13} \right)^2 + \left( \frac{5}{13} \right)^2 } = \sqrt{\frac{144}{169} + \frac{25}{169}} = \sqrt{\frac{169}{169}} = 1 \]Calculate the norm of \( \mathbf{v}_2 \):\[ \| \mathbf{v}_2 \| = \sqrt{ \left( \frac{5}{13} \right)^2 + \left( -\frac{12}{13} \right)^2 } = \sqrt{\frac{25}{169} + \frac{144}{169}} = \sqrt{\frac{169}{169}} = 1 \]Calculate the dot product of \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \):\[ \mathbf{v}_1 \cdot \mathbf{v}_2 = \left( \frac{12}{13} \right)\left( \frac{5}{13} \right) + \left( \frac{5}{13} \right)\left( -\frac{12}{13} \right) = \frac{60}{169} - \frac{60}{169} = 0 \]Since both conditions are satisfied, the vectors in \( B \) form an orthonormal basis.

Step 2: Find Coordinates Relative to Basis

Use Theorem 7.7.1, which states that if \( B = \{ \mathbf{v}_1, \mathbf{v}_2 \} \) is an orthonormal basis for \( \mathbb{R}^2 \) and \( \mathbf{u} \) is a vector in \( \mathbb{R}^2 \), then the coordinates of \( \mathbf{u} \) relative to \( B \) are given by:\[ c_1 = \mathbf{u} \cdot \mathbf{v}_1, \quad c_2 = \mathbf{u} \cdot \mathbf{v}_2 \]First, compute \( c_1 \):\[ c_1 = \langle 4, 2 \rangle \cdot \left\langle \frac{12}{13}, \frac{5}{13} \right\rangle = 4\left( \frac{12}{13} \right) + 2\left( \frac{5}{13} \right) = \frac{48}{13} + \frac{10}{13} = \frac{58}{13} \]Next, compute \( c_2 \):\[ c_2 = \langle 4, 2 \rangle \cdot \left\langle \frac{5}{13}, -\frac{12}{13} \right\rangle = 4\left( \frac{5}{13} \right) + 2\left( -\frac{12}{13} \right) = \frac{20}{13} - \frac{24}{13} = -\frac{4}{13} \]

Step 3: Write \( \mathbf{u} \) as a Linear Combination

Using the coordinates \( c_1 \) and \( c_2 \) found in the previous step, write \( \mathbf{u} \) as a linear combination of the basis vectors:\[ \mathbf{u} = c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 = \frac{58}{13} \left\langle \frac{12}{13}, \frac{5}{13} \right\rangle + \left(-\frac{4}{13}\right) \left\langle \frac{5}{13}, -\frac{12}{13} \right\rangle \]Substitute and simplify to confirm:First calculation:\[ \frac{58}{13} \left( \langle \frac{12}{13}, \frac{5}{13} \rangle \right) = \langle \frac{696}{169}, \frac{290}{169} \rangle \]Second calculation:\[ -\frac{4}{13} \left( \langle \frac{5}{13}, -\frac{12}{13} \rangle \right) = \langle -\frac{20}{169}, \frac{48}{169} \rangle \]Combine both results:\[ \langle \frac{696}{169} - \frac{20}{169}, \frac{290}{169} + \frac{48}{169} \rangle = \langle \frac{676}{169}, \frac{338}{169} \rangle = \langle 4, 2 \rangle \]This confirms that \( \mathbf{u} = \langle 4, 2 \rangle \) is correctly expressed as a linear combination of the orthonormal basis vectors.

Key Concepts

Vector Spaces

Vector spaces are foundational elements of linear algebra. They provide the setting where vectors, which are arrays of numbers, exist and interact. In a vector space, you have the following properties:

• Vectors can be added together.
• Vectors can be multiplied by scalars, which are numbers or constants.
• There are zero vectors that function as an additive identity.

These vector space operations abide by specific rules, such as distributive and associative laws. One practical characteristic of vector spaces is the presence of bases. A basis is a set of vectors from which any vector in the space can be expressed, often through linear combinations. This leads us to the idea of orthonormal bases, where each basis vector is orthogonal and normalized. Using these bases, we can efficiently analyze and manipulate vector coordinates.

Linear Algebra

Linear algebra involves the study of vectors, vector spaces, and linear transformations, among other things. It's a crucial part of mathematics that finds applications in various fields, from data science to engineering. A core concept in linear algebra is the understanding of linear combinations. When solving problems involving vectors, like expressing a vector using a basis, we construct linear combinations of basis vectors to represent the original vector.

In an orthonormal basis within linear algebra, each vector in the basis is orthogonal to every other vector, and each has a magnitude of one (is normalized). These properties allow for easy computation, such as finding coordinates relative to the basis using dot products, as demonstrated in the problem.

Linear transformations also play a substantial role in these calculations and are often represented as matrices. The orthonormal basis simplifies these transformations by making matrix operations straightforward.

Coordinate Systems

Coordinate systems are used to uniquely identify the position of a point in space. In the realm of vector spaces, they allow us to pinpoint vectors as a combination of basis vectors, facilitating their manipulation. When you have an orthonormal basis, determining the coordinates of any vector becomes much simpler.

To find these coordinates, you need to perform a dot product of the vector with each of the basis vectors—these results are the coordinates relative to that basis. This step-by-step approach was illustrated in the solution where we transformed vector \( \mathbf{u} = \langle 4, 2 \rangle \) into a coordinate representation in terms of an orthonormal basis \( B \).

Thus, coordinate systems provide a structured way to handle vectors mathematically, ensuring precision and efficiency in expressing complex calculations and transformations.