DSC 140B
Problems tagged with lecture-05

Problems tagged with "lecture-05"

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

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

Problem #053

Tags: linear algebra, quiz-03, lecture-05, eigenvalues, eigenvectors, diagonalization

Let \(\vec f\) be a linear transformation with eigenvectors and eigenvalues:

$$\begin{align*}\hat{u}^{(1)}&= \frac{1}{\sqrt 2}(1, 1)^T & \lambda_1 &= 2\\\hat{u}^{(2)}&= \frac{1}{\sqrt 2}(-1, 1)^T & \lambda_2 &= -1 \end{align*}$$

What is \(\vec f(\vec x)\) for \(\vec x = (3, 1)^T\)?

Solution

\(\vec f(\vec x) = (3, 5)^T\).

We'll take the three step approach of 1) finding the coordinates of \(\vec x\) in the eigenbasis, 2) applying the transformation in that basis (where the matrix of the transformation \(A_\mathcal{U}\) is diagonal and consists of the eigenvalues), and 3) converting back to the standard basis.

We saw in lecture that we can do this all in one go using the change of basis matrix \(U\):

\[\vec f(\vec x) = U^T A_\mathcal{U} U \vec x \]

where \(U\) is the change of basis matrix with the eigenvectors as rows and \(A_\mathcal{U}\) is the diagonal matrix with the eigenvalues on the diagonal. In this case, they are:

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

Then:

$$\begin{align*}\vec f(\vec x) &= U^T A_\mathcal{U} U \vec x \\&= \begin{pmatrix} \frac{1}{\sqrt 2} & \frac{-1}{\sqrt 2}\\[0.5em] \frac{1}{\sqrt 2} & \frac{1}{\sqrt 2} \end{pmatrix}\begin{pmatrix} 2 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} \frac{1}{\sqrt 2} & \frac{1}{\sqrt 2}\\[0.5em] \frac{-1}{\sqrt 2} & \frac{1}{\sqrt 2} \end{pmatrix}\begin{pmatrix} 3 \\ 1 \end{pmatrix}\\&= \begin{pmatrix} 3 \\ 5 \end{pmatrix}\end{align*}$$

Problem #054

Tags: linear algebra, quiz-03, lecture-05, eigenvalues, eigenvectors, diagonalization

Let \(\vec f\) be a linear transformation with eigenvectors and eigenvalues:

$$\begin{align*}\hat{u}^{(1)}&= \frac{1}{\sqrt{2}}(1, 1, 0)^T & \lambda_1 &= 3\\\hat{u}^{(2)}&= \frac{1}{\sqrt{2}}(1, -1, 0)^T & \lambda_2 &= 2\\\hat{u}^{(3)}&= (0, 0, 1)^T & \lambda_3 &= 1 \end{align*}$$

What is \(\vec f(\vec x)\) for \(\vec x = (1, 1, 2)^T\)?

Solution

\(\vec f(\vec x) = (3, 3, 2)^T\).

We'll take the three step approach of 1) finding the coordinates of \(\vec x\) in the eigenbasis, 2) applying the transformation in that basis (where the matrix of the transformation \(A_\mathcal{U}\) is diagonal and consists of the eigenvalues), and 3) converting back to the standard basis.

We saw in lecture that we can do this all in one go using the change of basis matrix \(U\):

\[\vec f(\vec x) = U^T A_\mathcal{U} U \vec x \]

where \(U\) is the change of basis matrix with the eigenvectors as rows and \(A_\mathcal{U}\) is the diagonal matrix with the eigenvalues on the diagonal. In this case, they are:

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

Then:

$$\begin{align*}\vec f(\vec x) &= U^T A_\mathcal{U} U \vec x \\&= \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0\\[0.5em] \frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{2}} & 0\\[0.5em] 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} 3 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0\\[0.5em] \frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{2}} & 0\\[0.5em] 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}\\&= \begin{pmatrix} 3 \\ 3 \\ 2 \end{pmatrix}\end{align*}$$

Problem #055

Tags: linear algebra, quiz-03, lecture-05, eigenvalues, eigenvectors, diagonalization

Let \(\vec f\) be a linear transformation with eigenvectors and eigenvalues:

$$\begin{align*}\hat{u}^{(1)}&= \frac{1}{2}(1, 1, 1, 1)^T & \lambda_1 &= 4\\\hat{u}^{(2)}&= \frac{1}{2}(1, 1, -1, -1)^T & \lambda_2 &= 2\\\hat{u}^{(3)}&= \frac{1}{2}(1, -1, 1, -1)^T & \lambda_3 &= 1\\\hat{u}^{(4)}&= \frac{1}{2}(1, -1, -1, 1)^T & \lambda_4 &= 0 \end{align*}$$

What is \(\vec f(\vec x)\) for \(\vec x = (4, 0, 0, 0)^T\)?

Solution

\(\vec f(\vec x) = (7, 5, 3, 1)^T\).

We'll take the three step approach of 1) finding the coordinates of \(\vec x\) in the eigenbasis, 2) applying the transformation in that basis (where the matrix of the transformation \(A_\mathcal{U}\) is diagonal and consists of the eigenvalues), and 3) converting back to the standard basis.

We saw in lecture that we can do this all in one go using the change of basis matrix \(U\):

\[\vec f(\vec x) = U^T A_\mathcal{U} U \vec x \]

where \(U\) is the change of basis matrix with the eigenvectors as rows and \(A_\mathcal{U}\) is the diagonal matrix with the eigenvalues on the diagonal. In this case, they are:

\[ U = \frac{1}{2}\begin{pmatrix} 1 & 1 & 1 & 1\\ 1 & 1 & -1 & -1\\ 1 & -1 & 1 & -1\\ 1 & -1 & -1 & 1 \end{pmatrix}\qquad A_\mathcal{U} = \begin{pmatrix} 4 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}\]

Then:

$$\begin{align*}\vec f(\vec x) &= U^T A_\mathcal{U} U \vec x \\&= \frac{1}{2}\begin{pmatrix} 1 & 1 & 1 & 1\\ 1 & 1 & -1 & -1\\ 1 & -1 & 1 & -1\\ 1 & -1 & -1 & 1 \end{pmatrix}\begin{pmatrix} 4 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}\frac{1}{2}\begin{pmatrix} 1 & 1 & 1 & 1\\ 1 & 1 & -1 & -1\\ 1 & -1 & 1 & -1\\ 1 & -1 & -1 & 1 \end{pmatrix}\begin{pmatrix} 4 \\ 0 \\ 0 \\ 0 \end{pmatrix}\\&= \begin{pmatrix} 7 \\ 5 \\ 3 \\ 1 \end{pmatrix}\end{align*}$$

Problem #056

Tags: linear algebra, quiz-03, symmetric matrices, spectral theorem, eigenvalues, eigenvectors, lecture-05

Find a symmetric \(2 \times 2\) matrix \(A\) whose eigenvectors are \(\hat{u}^{(1)} = \frac{1}{\sqrt{5}}(1, 2)^T\) with eigenvalue \(6\) and \(\hat{u}^{(2)} = \frac{1}{\sqrt{5}}(2, -1)^T\) with eigenvalue \(1\).

Solution

\(A = \begin{pmatrix} 2 & 2 \\ 2 & 5 \end{pmatrix}\).

We saw in lecture that a symmetric matrix \(A\) can be written as \(A = U^T \Lambda U\), where \(U\) is the matrix whose rows are the eigenvectors of \(A\) and \(\Lambda\) is the diagonal matrix of eigenvalues.

This means that if we know the eigenvectors and eigenvalues of a symmetric matrix, we can "work backwards" to find the matrix itself.

In this case:

\[ U = \frac{1}{\sqrt{5}}\begin{pmatrix} 1 & 2 \\ 2 & -1 \end{pmatrix}, \quad\Lambda = \begin{pmatrix} 6 & 0 \\ 0 & 1 \end{pmatrix}\]

Therefore:

$$\begin{align*} A &= U^T \Lambda U \\[0.5em]&= \frac{1}{\sqrt{5}}\begin{pmatrix} 1 & 2 \\ 2 & -1 \end{pmatrix}\begin{pmatrix} 6 & 0 \\ 0 & 1 \end{pmatrix}\frac{1}{\sqrt{5}}\begin{pmatrix} 1 & 2 \\ 2 & -1 \end{pmatrix}\\[0.5em]&= \frac{1}{5}\begin{pmatrix} 1 & 2 \\ 2 & -1 \end{pmatrix}\begin{pmatrix} 6 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 2 \\ 2 & -1 \end{pmatrix}\\[0.5em]&= \frac{1}{5}\begin{pmatrix} 6 & 2 \\ 12 & -1 \end{pmatrix}\begin{pmatrix} 1 & 2 \\ 2 & -1 \end{pmatrix}\\[0.5em]&= \frac{1}{5}\begin{pmatrix} 6 + 4 & 12 - 2 \\ 12 - 2 & 24 + 1 \end{pmatrix}\\[0.5em]&= \frac{1}{5}\begin{pmatrix} 10 & 10 \\ 10 & 25 \end{pmatrix}\\[0.5em]&= \begin{pmatrix} 2 & 2 \\ 2 & 5 \end{pmatrix}\end{align*}$$

Problem #057

Tags: linear algebra, quiz-03, symmetric matrices, spectral theorem, eigenvalues, eigenvectors, lecture-05

Find a symmetric \(3 \times 3\) matrix \(A\) whose eigenvectors are \(\hat{u}^{(1)} = \frac{1}{3}(1, 2, 2)^T\) with eigenvalue \(9\), \(\hat{u}^{(2)} = \frac{1}{3}(2, 1, -2)^T\) with eigenvalue \(18\), and \(\hat{u}^{(3)} = \frac{1}{3}(2, -2, 1)^T\) with eigenvalue \(-9\).

Solution

\(A = \begin{pmatrix} 5 & 10 & -8 \\ 10 & 2 & 2 \\ -8 & 2 & 11 \end{pmatrix}\).

We saw in lecture that a symmetric matrix \(A\) can be written as \(A = U^T \Lambda U\), where \(U\) is the matrix whose rows are the eigenvectors of \(A\) and \(\Lambda\) is the diagonal matrix of eigenvalues.

In this case:

\[ U = \frac{1}{3}\begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{pmatrix}, \quad\Lambda = \begin{pmatrix} 9 & 0 & 0 \\ 0 & 18 & 0 \\ 0 & 0 & -9 \end{pmatrix}\]

Therefore:

$$\begin{align*} A &= U^T \Lambda U \\[0.5em]&= \frac{1}{3}\begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{pmatrix}\begin{pmatrix} 9 & 0 & 0 \\ 0 & 18 & 0 \\ 0 & 0 & -9 \end{pmatrix}\frac{1}{3}\begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{pmatrix}\\[0.5em]&= \frac{1}{9}\begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{pmatrix}\begin{pmatrix} 9 & 0 & 0 \\ 0 & 18 & 0 \\ 0 & 0 & -9 \end{pmatrix}\begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{pmatrix}\\[0.5em]&= \frac{1}{9}\begin{pmatrix} 9 & 36 & -18 \\ 18 & 18 & 18 \\ 18 & -36 & -9 \end{pmatrix}\begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{pmatrix}\\[0.5em]&= \frac{1}{9}\begin{pmatrix} 9 + 72 - 36 & 18 + 36 + 36 & 18 - 72 - 18 \\ 18 + 36 + 36 & 36 + 18 - 36 & 36 - 36 + 18 \\ 18 - 72 - 18 & 36 - 36 + 18 & 36 + 72 - 9 \end{pmatrix}\\[0.5em]&= \frac{1}{9}\begin{pmatrix} 45 & 90 & -72 \\ 90 & 18 & 18 \\ -72 & 18 & 99 \end{pmatrix}\\[0.5em]&= \begin{pmatrix} 5 & 10 & -8 \\ 10 & 2 & 2 \\ -8 & 2 & 11 \end{pmatrix}\end{align*}$$