Problem 2 - Differential Equations and Linear Algebra

Question

Determine whether the given set of vectors is an orthogonal set in \(\mathbb{R}^{n} .\) For those that are, determine a corresponding orthonormal set of vectors. $$\\{(2,-1,1),(1,1,-1),(0,1,1)\\}$$

Step-by-Step Solution

Answer

The given set of vectors \(\{ (2,-1,1), (1,1,-1), (0,1,1) \}\) is orthogonal. The corresponding orthonormal set of vectors is: \(\left\{{\left(\frac{2}{\sqrt{6}}, \frac{-1}{\sqrt{6}}, \frac{1}{\sqrt{6}}\right), \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}\right), \left(0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)}\right\}\).

Step 1: Write down the given vectors

The three given vectors are: \( v_1 = (2,-1,1) \) ,\( v_2 = (1,1,-1) \),\( v_3 = (0,1,1) \)

Step 2: Check if vectors are orthogonal by computing their dot products

To determine if these vectors are orthogonal, we need to verify if their dot products equal to zero. So, calculate the dot products between the following pairs: \( \langle\, v_1,v_2 \rangle ,\ \langle\, v_1, v_3 \rangle ,\ \text{and }\, \langle\,v_2, v_3 \rangle \) \(\, \langle\, v_1, v_2 \rangle = 2(1) + (-1)(1) + 1(-1) = 2 - 1 - 1 = 0 \) \\ \(\, \langle\, v_1, v_3 \rangle = 2(0) + (-1)(1) + 1(1) = 0 - 1 + 1 = 0 \) \\ \(\, \langle\, v_2, v_3 \rangle = 1(0) + 1(1) + (-1)(1) = 0 + 1 - 1 = 0 \) Since all these dot products equal zero, the given set of vectors is indeed an orthogonal set.

Step 3: Determine orthonormal set by normalizing each vector

Now that we know the set is orthogonal, we can find the corresponding orthonormal set by dividing each vector by its magnitude. \( ||v_1|| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{6} \) \\ \( ||v_2|| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{3} \) \\ \( ||v_3|| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2} \) Then, normalize each vector: \(u_1 = \frac{v_1}{||v_1||} = \frac{1}{\sqrt{6}} (2, -1, 1) = \left(\frac{2}{\sqrt{6}}, \frac{-1}{\sqrt{6}}, \frac{1}{\sqrt{6}}\right)\) \\ \(u_2 = \frac{v_2}{||v_2||} = \frac{1}{\sqrt{3}} (1, 1, -1) = \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}\right)\) \\ \(u_3 = \frac{v_3}{||v_3||} = \frac{1}{\sqrt{2}} (0, 1, 1) = \left(0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\) The corresponding orthonormal set of vectors is: \(\left\{{\left(\frac{2}{\sqrt{6}}, \frac{-1}{\sqrt{6}}, \frac{1}{\sqrt{6}}\right), \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}\right), \left(0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)}\right\}\)

Key Concepts

Orthonormal Set

An orthonormal set is made up of vectors that have two special properties. First, each pair of different vectors in the set is orthogonal. This means their dot product equals zero. Second, every vector in the set is a unit vector, meaning its length is 1.

When dealing with a set of vectors in \(\mathbb{R}^n\), these properties help in simplifying complex operations. For instance, orthonormal vectors are crucial in formulas for rotation and transformation in space.

In this exercise, the initial task was to check if the vectors were orthogonal. Then, by normalizing them, they were transformed into unital (unit-length) vectors, forming an orthonormal set.

• Orthogonal means the dot product is zero.
• Orthogonality ensures independence among vectors.
• Normalization turns orthogonal vectors into orthonormal.

These features make orthonormal sets highly desirable in fields like computer graphics and data analysis, where efficient calculations are needed.

Dot Product

The dot product is a fundamental tool in vector mathematics. It's a way to multiply two vectors and get a scalar (a single number).

For vectors \( \mathbf{a} = (a_1, a_2, ..., a_n) \) and \( \mathbf{b} = (b_1, b_2, ..., b_n) \), the dot product is computed as:

\[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + ... + a_nb_n \]

In this exercise, the dot product was used to test if vectors are orthogonal. If their dot product is zero, the vectors are orthogonal. This indicates they are at right angles to each other in space.

• Dot product results in a scalar.
• Useful to measure angles and to determine orthogonality.
• Zero result indicates orthogonality.

In practical terms, the dot product helps intersect vectors and form relationships between them, which is essential in physical applications like mechanics and computer animations.

Vector Normalization

Vector normalization is the process of converting a vector into a unit vector. A unit vector has a length of 1. This is done by dividing each component of the vector by the vector's magnitude.

The magnitude of a vector \( \mathbf{v} = (v_1, v_2, ..., v_n) \) is calculated as:
\[ ||\mathbf{v}|| = \sqrt{v_1^2 + v_2^2 + ... + v_n^2} \]

After finding the magnitude, the normalized vector \( \mathbf{u} \) is:
\[ \mathbf{u} = \frac{\mathbf{v}}{||\mathbf{v}||} \]

This transformation retains the direction of the vector but changes its length to 1.

• Normalization keeps direction but changes length.
• Essential for creating orthonormal sets.
• Makes calculations simpler and more efficient.

Normalization is vital in many algorithms, especially in machine learning, where weighted vectors need to be compared fairly.