Show that if \(\mathrm{C}\) is a matrix whose columns are the components \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\) of two perpendicular vectors each of unit length, then \(\mathrm{C}\) is an orthogonal matrix. Hint: Find \(\mathrm{C}^{\mathrm{T}} \mathrm{C}\)

The transpose of a matrix is created by switching the rows and columns of the original matrix. For example, if you have a matrix \( A \) defined as:
\[ A = \begin{pmatrix} a & b \ c & d \end{pmatrix} \]
The transpose of this matrix, denoted as \( A^T \), is:
\[ A^T = \begin{pmatrix} a & c \ b & d \end{pmatrix} \]
Transposing a matrix is a crucial step in various mathematical operations and proofs. In the exercise you're working on, finding the transpose of matrix \( C \) is necessary to show that it is an orthogonal matrix. Essentially, you rearrange the components of the original matrix to switch its rows and columns.

The dot product of two vectors is a way to multiply two vectors that results in a scalar. For two vectors \( \mathbf{a} = (a_1, a_2) \) and \( \mathbf{b} = (b_1, b_2) \), the dot product is calculated as follows:
\[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \]
The dot product can tell us about the angle between two vectors. If the dot product is zero, it means the vectors are perpendicular. In your exercise, this property is crucial as it is given that the vectors are perpendicular, leading to one of the simplifying steps where the dot product equals zero. This property becomes part of the reasoning showing that the matrix \( C \) is orthogonal.

A unit vector is a vector that has a magnitude (length) of 1. For example, a vector \( \mathbf{u} \) is a unit vector if \( \|\mathbf{u}\| = 1 \), where \( \| \cdot \| \) represents the magnitude defined as:
\[ \|\mathbf{u}\| = \sqrt{u_1^2 + u_2^2} \]
In the exercise, the given vectors are unit vectors, which means their magnitudes are 1. This simplifies calculations significantly because it implies that the sum of the squares of their components is 1. Specifically:

• For \( (x_1, y_1) \), \( x_1^2 + y_1^2 = 1 \)
• For \( (x_2, y_2) \), \( x_2^2 + y_2^2 = 1 \)

This property is one of the reasons why the resulting matrix \( C^T C \) becomes the identity matrix \( I \).

Matrix multiplication involves taking the rows of the first matrix and the columns of the second matrix to produce a new matrix. For matrices \( A \) and \( B \), where \( A \) is of size \( m \times n \) and \( B \) is of size \( n \times p \), the resulting matrix \( AB \) will be of size \( m \times p \). Each element \( (i, j) \) in \( AB \) is calculated by:
\[ (AB)_{ij} = \sum_{k=1}^{n} A_{ik} B_{kj} \]
In your exercise, you multiply the transpose of matrix \( C \) by matrix \( C \) itself:
\[ C^T C = \begin{pmatrix} x_1 & y_1 \ x_2 & y_2 \end{pmatrix} \begin{pmatrix} x_1 & x_2 \ y_1 & y_2 \end{pmatrix} \]
The result is a matrix where each element is derived using the dot product of the corresponding vectors. Because of the unit length and perpendicular properties, this particular multiplication simplifies to the identity matrix, \( I \). This shows that matrix \( C \) is orthogonal.