Question

Show that each of the following matrices is orthogonal and find the rotation and/or reflection it produces as an operator acting on vectors. If a rotation, find the axis and angle; if a reflection, find the reflecting plane and the rotation, if any, about the normal to that plane. $$\frac{1}{2}\left(\begin{array}{ccc} 1 & \sqrt{2} & -1 \\ \sqrt{2} & 0 & \sqrt{2} \\ 1 & -\sqrt{2} & -1 \end{array}\right)$$

Short answer

The matrix is orthogonal and represents a reflection. Its reflecting plane is perpendicular to \(\frac{1}{\sqrt{6}}\begin{pmatrix} 1 \ 2 \ -1 \end{pmatrix} \).

Step-by-step solution

Verify Orthogonality

A matrix is orthogonal if its transpose is equal to its inverse, i.e., \(A^T = A^{-1}\). Calculate the transpose \(A^T\) of the given matrix and then multiply it by the original matrix. If the result is the identity matrix, then the matrix is orthogonal.

Calculate the Transpose

Find the transpose \(A^T\) of the matrix: \[A^T = \frac{1}{2}\begin{pmatrix}1 & \sqrt{2} & 1 \ \sqrt{2} & 0 & -\sqrt{2} \ -1 & \sqrt{2} & -1\end{pmatrix}\]

Multiply Original Matrix by its Transpose

Multiply the original matrix by its transpose: \[\frac{1}{2}\begin{pmatrix}1 & \sqrt{2} & -1 \ \sqrt{2} & 0 & \sqrt{2} \ 1 & -\sqrt{2} & -1\end{pmatrix} \cdot \frac{1}{2}\begin{pmatrix}1 & \sqrt{2} & 1 \ \sqrt{2} & 0 & -\sqrt{2} \ -1 & \sqrt{2} & -1\end{pmatrix}\]

Simplify the Multiplication

Simplify the product by performing matrix multiplication: \[\frac{1}{4}\begin{pmatrix}1+2+1 & \sqrt{2}-\sqrt{2}-\sqrt{2} & -1-2-1 \ \sqrt{2}+0-\sqrt{2} & 2+0+2 & -\sqrt{2}+2+\sqrt{2} \ -1-2-1 & \sqrt{2}-\sqrt{2}+\sqrt{2} & 1+2+1\end{pmatrix} = \begin{pmatrix}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{pmatrix}\]. Since the result is the identity matrix, the original matrix is orthogonal.

Determine the Type of Transformation

A matrix representing a rotation or reflection will have eigenvalues that help identify the transformation. Calculate the eigenvalues of the matrix to determine if it's a rotation or reflection.

Calculate the Eigenvalues

Find the eigenvalues \(\lambda\) by solving the characteristic equation \(\det(A - \lambda I) = 0\). For the given matrix, the characteristic polynomial is: \[\lambda^3 - \lambda^2 - 3\lambda + 1 = 0\]. The eigenvalues are \lambda = 1, -1, and -1.

Identify the Transformation

Since there’s an eigenvalue of 1 and two -1 eigenvalues, this indicates a reflection. Determine the reflecting plane based on the eigenvector corresponding to \(\lambda = 1\).

Find the Reflecting Plane

The eigenvector associated with \(\lambda = 1\) is \(\begin{pmatrix} \frac{1}{\sqrt{6}} \ \frac{2}{\sqrt{6}} \ -\frac{1}{\sqrt{6}} \end{pmatrix}\), thus the reflecting plane is perpendicular to this vector.

Key concepts

matrix orthogonality

Matrix orthogonality refers to a special characteristic where a matrix is orthogonal if the product of the matrix and its transpose equals the identity matrix. This means that for an orthogonal matrix A, we have: \(A^T A = I\), where \(A^T\) is the transpose of A, and I is the identity matrix. This property implies that the columns (and rows) of an orthogonal matrix are both orthogonal and normalized, forming an orthonormal set. This is essential in preserving the length and angles of vectors under transformation, making them useful in numerous applications such as rotations and reflections in geometry.

matrix transpose

The transpose of a matrix, denoted as \(A^T\), is obtained by flipping the matrix over its diagonal. This means that the element at the position \(a_{ij}\) in the original matrix A will be at the position \(a_{ji}\) in the transposed matrix. In mathematical terms: If \(A = [a_{ij}]\), then \(A^T = [a_{ji}]\). For our exercise, the transpose of the matrix is: \[A^T = \frac{1}{2} \begin{pmatrix} 1 & \sqrt{2} & 1 \ \sqrt{2} & 0 & -\sqrt{2} \ -1 & \sqrt{2} & -1 \end{pmatrix} \] Transposing is a simple but crucial operation in verifying orthogonality and in many linear algebra applications.

eigenvalues and eigenvectors

Eigenvalues and eigenvectors are fundamental concepts in linear algebra. An eigenvalue \(\lambda\) of a matrix A is a scalar such that there exists a non-zero vector \(\mathbf{v}\) (called an eigenvector) satisfying: \(A \mathbf{v} = \lambda \mathbf{v}\). In other words, multiplying the matrix A by the eigenvector \(\mathbf{v}\) results in a vector that is only scaled by \(\lambda\) and not directionally changed. For the given matrix, the characteristic equation is solved to find the eigenvalues: \[ \lambda^3 - \lambda^2 - 3\lambda + 1 = 0 \] The eigenvalues are \(\lambda = 1, -1, -1\). The nature of these eigenvalues helps determine whether our matrix represents a rotation or reflection. Here, the presence of \(\lambda = 1\) and pairs of \(\lambda = -1\) suggests a reflection.

matrix multiplication

Matrix multiplication is a key operation in linear algebra where two matrices are multiplied to produce a third matrix. The product of two matrices A (of size m×n) and B (of size n×p) is a new matrix C (of size m×p) where each element \(c_{ij}\) is calculated by summing the products of corresponding elements from the ith row of A and the jth column of B: \(C_{ij} = \sum_{k=1}^{n} a_{ik} b_{kj}\). In our step-by-step solution, we multiply the given matrix by its transpose to verify orthogonality: \[ \frac{1}{2} \begin{pmatrix} 1 & \sqrt{2} & -1 \ \sqrt{2} & 0 & \sqrt{2} \ 1 & -\sqrt{2} & -1 \end{pmatrix} \cdot \frac{1}{2} \begin{pmatrix} 1 & \sqrt{2} & 1 \ \sqrt{2} & 0 & -\sqrt{2} \ -1 & \sqrt{2} & -1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix} \] The result is the identity matrix, confirming the matrix's orthogonality.