Problem 5 - 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. $$\\{(1,2,-1,0,3),(1,1,0,2,-1),(4,2,-4,-5,-4)\\}$$

Step-by-Step Solution

Answer

The given set of vectors is orthogonal, and the corresponding orthonormal set is: $$\left\{ \begin{pmatrix}\frac{1}{\sqrt{15}}\\\frac{2}{\sqrt{15}}\\-\frac{1}{\sqrt{15}}\\0\\\frac{3}{\sqrt{15}}\end{pmatrix}, \begin{pmatrix}\frac{1}{\sqrt{7}}\\\frac{1}{\sqrt{7}}\\0\\\frac{2}{\sqrt{7}}\\-\frac{1}{\sqrt{7}}\end{pmatrix}, \begin{pmatrix}\frac{4}{\sqrt{77}}\\\frac{2}{\sqrt{77}}\\-\frac{4}{\sqrt{77}}\\-\frac{5}{\sqrt{77}}\\-\frac{4}{\sqrt{77}}\end{pmatrix} \right\}$$

Step 1: Find the dot products of each pair of vectors

Compute the dot product of the given vectors to check if they are orthogonal or not. Recall that two vectors are orthogonal if their dot product is 0. $$\begin{pmatrix}1\\2\\-1\\0\\3\end{pmatrix}\cdot\begin{pmatrix}1\\1\\0\\2\\-1\end{pmatrix} = 1(1) + 2(1) + (-1)(0) + 0(2)+ 3(-1) = 1 + 2 - 3 = 0$$ $$\begin{pmatrix}1\\2\\-1\\0\\3\end{pmatrix}\cdot\begin{pmatrix}4\\2\\-4\\-5\\-4\end{pmatrix} = 1(4) + 2(2) + (-1)(-4) + 0(-5)+ 3(-4) = 4 + 4 + 4 - 12 = 0$$ $$\begin{pmatrix}1\\1\\0\\2\\-1\end{pmatrix}\cdot\begin{pmatrix}4\\2\\-4\\-5\\-4\end{pmatrix} = 1(4) + 1(2) + 0(-4) + 2(-5)+ (-1)(-4) = 4 + 2 - 10 + 4 = 0$$ Since all three pairs of vectors have a dot product equal to 0, the given set of vectors is orthogonal.

Step 2: Normalize the given vectors to create an orthonormal set

To compute the orthonormal set, we need to normalize each vector by dividing each vector by its magnitude; i.e., we divide each vector by the square root of the sum of its elements. The magnitude of each vector can be found as follows: $$\|\begin{pmatrix}1\\2\\-1\\0\\3\end{pmatrix}\|=\sqrt{1^2+2^2+(-1)^2+0^2+3^2}=\sqrt{1+4+1+0+9}=\sqrt{15}$$ $$\|\begin{pmatrix}1\\1\\0\\2\\-1\end{pmatrix}\|=\sqrt{1^2+1^2+0^2+2^2+(-1)^2}=\sqrt{1+1+0+4+1}=\sqrt{7}$$ $$\|\begin{pmatrix}4\\2\\-4\\-5\\-4\end{pmatrix}\|=\sqrt{4^2+2^2+(-4)^2+(-5)^2+(-4)^2}=\sqrt{16+4+16+25+16}=\sqrt{77}$$ Now, normalize each vector by dividing by their magnitudes: $$\frac{1}{\sqrt{15}}\begin{pmatrix}1\\2\\-1\\0\\3\end{pmatrix}=\begin{pmatrix}\frac{1}{\sqrt{15}}\\\frac{2}{\sqrt{15}}\\-\frac{1}{\sqrt{15}}\\0\\\frac{3}{\sqrt{15}}\end{pmatrix}$$ $$\frac{1}{\sqrt{7}}\begin{pmatrix}1\\1\\0\\2\\-1\end{pmatrix}=\begin{pmatrix}\frac{1}{\sqrt{7}}\\\frac{1}{\sqrt{7}}\\0\\\frac{2}{\sqrt{7}}\\-\frac{1}{\sqrt{7}}\end{pmatrix}$$ $$\frac{1}{\sqrt{77}}\begin{pmatrix}4\\2\\-4\\-5\\-4\end{pmatrix}=\begin{pmatrix}\frac{4}{\sqrt{77}}\\\frac{2}{\sqrt{77}}\\-\frac{4}{\sqrt{77}}\\-\frac{5}{\sqrt{77}}\\-\frac{4}{\sqrt{77}}\end{pmatrix}$$ The corresponding orthonormal set of vectors is: $$\left\{ \begin{pmatrix}\frac{1}{\sqrt{15}}\\\frac{2}{\sqrt{15}}\\-\frac{1}{\sqrt{15}}\\0\\\frac{3}{\sqrt{15}}\end{pmatrix}, \begin{pmatrix}\frac{1}{\sqrt{7}}\\\frac{1}{\sqrt{7}}\\0\\\frac{2}{\sqrt{7}}\\-\frac{1}{\sqrt{7}}\end{pmatrix}, \begin{pmatrix}\frac{4}{\sqrt{77}}\\\frac{2}{\sqrt{77}}\\-\frac{4}{\sqrt{77}}\\-\frac{5}{\sqrt{77}}\\-\frac{4}{\sqrt{77}}\end{pmatrix} \right\}$$

Key Concepts

Orthonormal Vectors

When dealing with vectors, orthonormality combines two concepts: orthogonality and normalization. Orthonormal vectors are both orthogonal to each other and are of unit length. To achieve this, not only must the dot product between each pair of vectors be zero, indicating they're orthogonal, but each vector must also have a magnitude, or length, of one. This means every vector in an orthonormal set is perpendicular to each other and scales to length one.

Why are orthonormal vectors important? They simplify many mathematical operations. For instance, in computer graphics, orthonormal sets reduce computational complexity. They also have applications in signal processing and statistical analysis, where ease and efficiency are essential.

To verify if a set of vectors is orthonormal, first ensure they're orthogonal by checking that their dot products are zero. Next, confirm they have unit magnitudes through the process of vector normalization.

Dot Product

The dot product is a fundamental operation in vector analysis. It involves multiplying corresponding components of two vectors and adding those products. For vectors \(\mathbf{a} = (a_1, a_2, ..., a_n)\) and \(\mathbf{b} = (b_1, b_2, ..., b_n)\), their dot product is calculated as \(\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + ... + a_nb_n\).

The dot product has several properties:

• It measures the cosine of the angle between two vectors, providing insight into their alignment.
• A dot product of zero implies the vectors are orthogonal, meaning they intersect at a right angle.
• If the dot product is positive, the vectors point in similar directions, while a negative result indicates opposite directions.

Computing the dot product efficiently checks for orthogonality. For instance, in the provided solution, confirming that each vector pair had a zero dot product showed the vectors were orthogonal. This concept is crucial in higher mathematics and physics because it integrates geometry and algebra.

Vector Normalization

Vector normalization is a process where each vector is adjusted to have a magnitude of one, forming a unit vector. This is achieved by dividing each component of the vector by its own length or magnitude. For a vector \(\mathbf{v} = (v_1, v_2, ..., v_n)\), its magnitude is calculated as \(\|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2 + ... + v_n^2}\). The normalized vector is then expressed as \(\mathbf{u} = \frac{1}{\|\mathbf{v}\|}\mathbf{v}\).

Normalizing vectors is key to creating orthonormal sets. It scales each vector in the set so that every one is of unit length, which is vital for mathematical operations where scale consistency is required.

Why normalize? Beyond forming orthonormal sets, normalization helps in applications needing consistent scales, such as simplifying complex mathematical models and algorithms. In our original solution, dividing each vector by its magnitude helped create the desired orthonormal set. This process ensures data integrity and stability in various computational applications.