DSC 140B
Problems tagged with orthogonal matrices

Problems tagged with "orthogonal matrices"

Problem #050

Tags: orthogonal matrices, linear algebra, quiz-03, change of basis, lecture-05

Let \(\hat{u}^{(1)}\) and \(\hat{u}^{(2)}\) be an orthonormal basis for \(\mathbb R^2\):

$$\begin{align*}\hat{u}^{(1)}&= \frac{1}{\sqrt 2}(1, 1)^T\\\hat{u}^{(2)}&= \frac{1}{\sqrt 2}(-1, 1)^T \end{align*}$$

Part 1)

What is the change of basis matrix \(U\)?

Solution

The change of basis matrix \(U\) has the new basis vectors as its rows:

\[ U = \begin{pmatrix} \frac{1}{\sqrt 2} & \frac{1}{\sqrt 2}\\[0.5em] \frac{-1}{\sqrt 2} & \frac{1}{\sqrt 2} \end{pmatrix}\]

This matrix represents a 45-degree rotation.

Part 2)

Let \(\vec x = (2, 0)^T\) be a vector in the standard basis. What are the coordinates of \(\vec x\) in the basis \(\mathcal{U}\)?

Solution

We compute \([\vec x]_{\mathcal{U}} = U \vec x\):

\[[\vec x]_{\mathcal{U}} = \begin{pmatrix} \frac{1}{\sqrt 2} & \frac{1}{\sqrt 2}\\[0.5em] \frac{-1}{\sqrt 2} & \frac{1}{\sqrt 2} \end{pmatrix}\begin{pmatrix} 2 \\ 0 \end{pmatrix} = \begin{pmatrix} \frac{2}{\sqrt 2} \\[0.5em] \frac{-2}{\sqrt 2} \end{pmatrix} = \begin{pmatrix} \sqrt 2 \\ -\sqrt 2 \end{pmatrix}\]

Problem #051

Tags: orthogonal matrices, linear algebra, quiz-03, change of basis, lecture-05

Let \(\hat{u}^{(1)}\), \(\hat{u}^{(2)}\), and \(\hat{u}^{(3)}\) be an orthonormal basis for \(\mathbb R^3\):

$$\begin{align*}\hat{u}^{(1)}&= \left(\frac{\sqrt 3}{2}, \frac{1}{2}, 0\right)^T\\\hat{u}^{(2)}&= \left(\frac{-1}{2}, \frac{\sqrt 3}{2}, 0\right)^T\\\hat{u}^{(3)}&= (0, 0, 1)^T \end{align*}$$

Part 1)

What is the change of basis matrix \(U\)?

Solution

The change of basis matrix \(U\) has the new basis vectors as its rows:

\[ U = \begin{pmatrix} \frac{\sqrt 3}{2} & \frac{1}{2} & 0\\[0.5em] \frac{-1}{2} & \frac{\sqrt 3}{2} & 0\\[0.5em] 0 & 0 & 1 \end{pmatrix}\]

This matrix represents a 30-degree rotation in the \(xy\)-plane while leaving the \(z\)-axis unchanged.

Part 2)

Let \(\vec x = (\sqrt 3, 1, 2)^T\) be a vector in the standard basis. What are the coordinates of \(\vec x\) in the basis \(\mathcal{U}\)?

Solution

We compute \([\vec x]_{\mathcal{U}} = U \vec x\):

\[[\vec x]_{\mathcal{U}} = \begin{pmatrix} \frac{\sqrt 3}{2} & \frac{1}{2} & 0\\[0.5em] \frac{-1}{2} & \frac{\sqrt 3}{2} & 0\\[0.5em] 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} \sqrt 3 \\ 1 \\ 2 \end{pmatrix} = \begin{pmatrix} \frac{3}{2} + \frac{1}{2} \\[0.5em] \frac{-\sqrt 3}{2} + \frac{\sqrt 3}{2} \\[0.5em] 2 \end{pmatrix} = \begin{pmatrix} 2 \\ 0 \\ 2 \end{pmatrix}\]

Problem #052

Tags: orthogonal matrices, linear algebra, quiz-03, change of basis, lecture-05

Let \(\hat{u}^{(1)}\), \(\hat{u}^{(2)}\), \(\hat{u}^{(3)}\), and \(\hat{u}^{(4)}\) be an orthonormal basis for \(\mathbb R^4\):

$$\begin{align*}\hat{u}^{(1)}&= (0, 1, 0, 0)^T\\\hat{u}^{(2)}&= (-1, 0, 0, 0)^T\\\hat{u}^{(3)}&= (0, 0, 0, 1)^T\\\hat{u}^{(4)}&= (0, 0, -1, 0)^T \end{align*}$$

Part 1)

What is the change of basis matrix \(U\)?

Solution

The change of basis matrix \(U\) has the new basis vectors as its rows:

\[ U = \begin{pmatrix} 0 & 1 & 0 & 0\\ -1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & -1 & 0 \end{pmatrix}\]

Part 2)

Let \(\vec x = (2, 3, 1, 4)^T\) be a vector in the standard basis. What are the coordinates of \(\vec x\) in the basis \(\mathcal{U}\)?

Solution

We compute \([\vec x]_{\mathcal{U}} = U \vec x\):

\[[\vec x]_{\mathcal{U}} = \begin{pmatrix} 0 & 1 & 0 & 0\\ -1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & -1 & 0 \end{pmatrix}\begin{pmatrix} 2 \\ 3 \\ 1 \\ 4 \end{pmatrix} = \begin{pmatrix} 3 \\ -2 \\ 4 \\ -1 \end{pmatrix}\]