DSC 140B
Problems tagged with lecture-03

Problems tagged with "lecture-03"

Problem #17

Tags: linear algebra, linear transformations, quiz-02, lecture-03

Let \(\vec f(\vec x) = (x_2, -x_1)^T\) be a linear transformation, and let \(\mathcal{U} = \{\hat{u}^{(1)}, \hat{u}^{(2)}\}\) be an orthonormal basis for \(\mathbb{R}^2\) with:

\[\hat{u}^{(1)} = \left(\frac{3}{5}, \frac{4}{5}\right)^T, \quad\hat{u}^{(2)} = \left(-\frac{4}{5}, \frac{3}{5}\right)^T \]

Express the transformation \(\vec f\) in the basis \(\mathcal{U}\). That is, if \([\vec x]_{\mathcal{U}} = (z_1, z_2)^T\), find \([\vec f(\vec x)]_{\mathcal{U}}\) in terms of \(z_1\) and \(z_2\).

Solution

\([\vec f(\vec x)]_{\mathcal{U}} = (z_2, -z_1)^T \).

To express \(\vec f\) in the basis \(\mathcal{U}\), we first need to find what \(\vec f\) does to the basis vectors \(\hat{u}^{(1)}\) and \(\hat{u}^{(2)}\), then express the results in the \(\mathcal{U}\) basis.

To start, we compute \(\vec f(\hat{u}^{(1)})\) and \(\vec f(\hat{u}^{(2)})\).

$$\begin{align*}\vec f(\hat{u}^{(1)}) &= \vec f\left(\frac{3}{5}, \frac{4}{5}\right)^T = \left(\frac{4}{5}, -\frac{3}{5}\right)^T \\\vec f(\hat{u}^{(2)}) &= \vec f\left(-\frac{4}{5}, \frac{3}{5}\right)^T = \left(\frac{3}{5}, \frac{4}{5}\right)^T \end{align*}$$

However, notice that this is \(\vec f(\hat{u}^{(1)})\) and \(\vec f(\hat{u}^{(2)})\) expressed in the standard basis. Therefore, we need to do a change of basis to express these vectors in the \(\mathcal{U}\) basis. We do the change of basis just like we would for any vector: by dotting with each basis vector. That is:

\[[\vec f(\hat{u}^{(1)})]_{\mathcal{U}} = \begin{pmatrix} \vec f(\hat{u}^{(1)}) \cdot \hat{u}^{(1)} \\ \vec f(\hat{u}^{(1)}) \cdot \hat{u}^{(2)} \end{pmatrix}, \quad[\vec f(\hat{u}^{(2)})]_{\mathcal{U}} = \begin{pmatrix} \vec f(\hat{u}^{(2)}) \cdot \hat{u}^{(1)} \\ \vec f(\hat{u}^{(2)}) \cdot \hat{u}^{(2)} \end{pmatrix}\]

We need to calculate all of these dot products:

$$\begin{align*}\vec f(\hat{u}^{(1)}) \cdot\hat{u}^{(1)}&= \left(\frac{4}{5}, -\frac{3}{5}\right)^T \cdot\left(\frac{3}{5}, \frac{4}{5}\right)^T = \frac{12}{25} - \frac{12}{25} = 0 \\\vec f(\hat{u}^{(1)}) \cdot\hat{u}^{(2)}&= \left(\frac{4}{5}, -\frac{3}{5}\right)^T \cdot\left(-\frac{4}{5}, \frac{3}{5}\right)^T = -\frac{16}{25} - \frac{9}{25} = -1 \\\vec f(\hat{u}^{(2)}) \cdot\hat{u}^{(1)}&= \left(\frac{3}{5}, \frac{4}{5}\right)^T \cdot\left(\frac{3}{5}, \frac{4}{5}\right)^T = \frac{9}{25} + \frac{16}{25} = 1 \\\vec f(\hat{u}^{(2)}) \cdot\hat{u}^{(2)}&= \left(\frac{3}{5}, \frac{4}{5}\right)^T \cdot\left(-\frac{4}{5}, \frac{3}{5}\right)^T = -\frac{12}{25} + \frac{12}{25} = 0 \end{align*}$$

Therefore, we have:

\[[\vec f(\hat{u}^{(1)})]_{\mathcal{U}} = \begin{pmatrix} 0 \\ -1 \end{pmatrix}, \quad[\vec f(\hat{u}^{(2)})]_{\mathcal{U}} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}\]

Finally, since \(\vec f\) is linear, we can express \(\vec f(\vec x)\) in the \(\mathcal{U}\) basis as:

$$\begin{align*}[\vec f(\vec x)]_{\mathcal{U}}&= z_1 [\vec f(\hat{u}^{(1)})]_{\mathcal{U}} + z_2 [\vec f(\hat{u}^{(2)})]_{\mathcal{U}}\\&= z_1 \begin{pmatrix} 0 \\ -1 \end{pmatrix} + z_2 \begin{pmatrix} 1 \\ 0 \end{pmatrix}\\&= \begin{pmatrix} z_2 \\ -z_1 \end{pmatrix}\end{align*}$$

Problem #18

Tags: linear algebra, linear transformations, quiz-02, lecture-03

Let \(\vec f(\vec x) = (x_1, -x_2, x_3)^T\) be a linear transformation, and let \(\mathcal{U} = \{\hat{u}^{(1)}, \hat{u}^{(2)}, \hat{u}^{(3)}\}\) be an orthonormal basis for \(\mathbb{R}^3\) with:

\[\hat{u}^{(1)} = \frac{1}{\sqrt{2}}(1, 1, 0)^T, \quad\hat{u}^{(2)} = \frac{1}{\sqrt{2}}(-1, 1, 0)^T, \quad\hat{u}^{(3)} = (0, 0, 1)^T \]

Express the transformation \(\vec f\) in the basis \(\mathcal{U}\). That is, if \([\vec x]_{\mathcal{U}} = (z_1, z_2, z_3)^T\), find \([\vec f(\vec x)]_{\mathcal{U}}\) in terms of \(z_1\), \(z_2\), and \(z_3\).

Solution

\([\vec f(\vec x)]_{\mathcal{U}} = (-z_2, -z_1, z_3)^T \).

To express \(\vec f\) in the basis \(\mathcal{U}\), we first need to find what \(\vec f\) does to the basis vectors \(\hat{u}^{(1)}\), \(\hat{u}^{(2)}\), and \(\hat{u}^{(3)}\), then express the results in the \(\mathcal{U}\) basis.

To start, we compute \(\vec f(\hat{u}^{(1)})\), \(\vec f(\hat{u}^{(2)})\), and \(\vec f(\hat{u}^{(3)})\).

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

However, notice that these are expressed in the standard basis. Therefore, we need to do a change of basis to express these vectors in the \(\mathcal{U}\) basis. We do the change of basis just like we would for any vector: by dotting with each basis vector. That is:

\[[\vec f(\hat{u}^{(i)})]_{\mathcal{U}} = \begin{pmatrix} \vec f(\hat{u}^{(i)}) \cdot \hat{u}^{(1)} \\ \vec f(\hat{u}^{(i)}) \cdot \hat{u}^{(2)} \\ \vec f(\hat{u}^{(i)}) \cdot \hat{u}^{(3)} \end{pmatrix}\]

We calculate the dot products for \(\vec f(\hat{u}^{(1)})\):

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

We calculate the dot products for \(\vec f(\hat{u}^{(2)})\):

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

For \(\vec f(\hat{u}^{(3)})\), we note that \(\vec f(\hat{u}^{(3)}) = (0, 0, 1)^T = \hat{u}^{(3)}\), so:

\[[\vec f(\hat{u}^{(3)})]_{\mathcal{U}} = (0, 0, 1)^T \]

Therefore, we have:

\[[\vec f(\hat{u}^{(1)})]_{\mathcal{U}} = \begin{pmatrix} 0 \\ -1 \\ 0 \end{pmatrix}, \quad[\vec f(\hat{u}^{(2)})]_{\mathcal{U}} = \begin{pmatrix} -1 \\ 0 \\ 0 \end{pmatrix}, \quad[\vec f(\hat{u}^{(3)})]_{\mathcal{U}} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\]

Finally, since \(\vec f\) is linear, we can express \(\vec f(\vec x)\) in the \(\mathcal{U}\) basis as:

$$\begin{align*}[\vec f(\vec x)]_{\mathcal{U}}&= z_1 [\vec f(\hat{u}^{(1)})]_{\mathcal{U}} + z_2 [\vec f(\hat{u}^{(2)})]_{\mathcal{U}} + z_3 [\vec f(\hat{u}^{(3)})]_{\mathcal{U}}\\&= z_1 \begin{pmatrix} 0 \\ -1 \\ 0 \end{pmatrix} + z_2 \begin{pmatrix} -1 \\ 0 \\ 0 \end{pmatrix} + z_3 \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\\&= \begin{pmatrix} -z_2 \\ -z_1 \\ z_3 \end{pmatrix}\end{align*}$$

Problem #19

Tags: linear algebra, linear transformations, quiz-02, lecture-03

Let \(\vec f(\vec x) = (x_2, x_1)^T\) be a linear transformation, and let \(\mathcal{U} = \{\hat{u}^{(1)}, \hat{u}^{(2)}\}\) be an orthonormal basis for \(\mathbb{R}^2\) with:

\[\hat{u}^{(1)} = \frac{1}{\sqrt{2}}(1, 1)^T, \quad\hat{u}^{(2)} = \frac{1}{\sqrt{2}}(1, -1)^T \]

Express the transformation \(\vec f\) in the basis \(\mathcal{U}\). That is, if \([\vec x]_{\mathcal{U}} = (z_1, z_2)^T\), find \([\vec f(\vec x)]_{\mathcal{U}}\) in terms of \(z_1\) and \(z_2\).

Solution

\([\vec f(\vec x)]_{\mathcal{U}} = (z_1, -z_2)^T \).

To express \(\vec f\) in the basis \(\mathcal{U}\), we first compute what \(\vec f\) does to the basis vectors \(\hat{u}^{(1)}\) and \(\hat{u}^{(2)}\).

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

Now, these results are expressed in the standard basis, and we need to convert them to the \(\mathcal{U}\) basis. In principle, we could do this by dotting with each basis vector, but we can also notice that:

\[\vec f(\hat{u}^{(1)}) = \hat{u}^{(1)}, \quad\vec f(\hat{u}^{(2)}) = -\hat{u}^{(2)}\]

This means we can immediately write:

\[[\vec f(\hat{u}^{(1)})]_{\mathcal{U}} = (1, 0)^T, \quad[\vec f(\hat{u}^{(2)})]_{\mathcal{U}} = (0, -1)^T \]

Finally, since \(\vec f\) is linear:

$$\begin{align*}[\vec f(\vec x)]_{\mathcal{U}}&= z_1 [\vec f(\hat{u}^{(1)})]_{\mathcal{U}} + z_2 [\vec f(\hat{u}^{(2)})]_{\mathcal{U}}\\&= z_1 (1, 0)^T + z_2 (0, -1)^T \\&= (z_1, -z_2)^T \end{align*}$$

Problem #20

Tags: linear algebra, linear transformations, quiz-02, lecture-03

Suppose \(\vec f\) is a linear transformation with:

\[\vec f(\hat{e}^{(1)}) = \begin{pmatrix} 3 \\ -1 \end{pmatrix}, \quad\vec f(\hat{e}^{(2)}) = \begin{pmatrix} 2 \\ 4 \end{pmatrix}\]

Write the matrix \(A\) that represents \(\vec f\) with respect to the standard basis.

Solution
\[ A = \begin{pmatrix} 3 & 2 \\ -1 & 4 \end{pmatrix}\]

To make the matrix representing the linear transformation \(\vec f\) with respect to the standard basis, we simply place \(f(\hat{e}^{(1)})\) and \(\vec f(\hat{e}^{(2)})\) as the first and second columns of the matrix, respectively. That is:

\[ A = \begin{pmatrix} \uparrow & \uparrow \\ \vec f(\hat{e}^{(1)}) & \vec f(\hat{e}^{(2)}) \\ \downarrow & \downarrow \end{pmatrix} = \begin{pmatrix} 3 & 2 \\ -1 & 4 \end{pmatrix}\]

Problem #21

Tags: linear algebra, linear transformations, quiz-02, lecture-03

Suppose \(\vec f\) is a linear transformation with:

\[\vec f(\hat{e}^{(1)}) = \begin{pmatrix} 2 \\ 0 \\ -1 \end{pmatrix}, \quad\vec f(\hat{e}^{(2)}) = \begin{pmatrix} 1 \\ 3 \\ 0 \end{pmatrix}, \quad\vec f(\hat{e}^{(3)}) = \begin{pmatrix} -1 \\ 2 \\ 4 \end{pmatrix}\]

Write the matrix \(A\) that represents \(\vec f\) with respect to the standard basis.

Solution
\[ A = \begin{pmatrix} 2 & 1 & -1 \\ 0 & 3 & 2 \\ -1 & 0 & 4 \end{pmatrix}\]

To make the matrix representing the linear transformation \(\vec f\) with respect to the standard basis, we simply place \(\vec f(\hat{e}^{(1)})\), \(\vec f(\hat{e}^{(2)})\), and \(\vec f(\hat{e}^{(3)})\) as the columns of the matrix. That is:

\[ A = \begin{pmatrix} \uparrow & \uparrow & \uparrow \\ \vec f(\hat{e}^{(1)}) & \vec f(\hat{e}^{(2)}) & \vec f(\hat{e}^{(3)}) \\ \downarrow & \downarrow & \downarrow \end{pmatrix} = \begin{pmatrix} 2 & 1 & -1 \\ 0 & 3 & 2 \\ -1 & 0 & 4 \end{pmatrix}\]

Problem #22

Tags: linear algebra, linear transformations, quiz-02, lecture-03

Consider the linear transformation \(\vec f : \mathbb{R}^3 \to\mathbb{R}^3\) that takes in a vector \(\vec x\) and simply outputs that same vector. That is, \(\vec f(\vec x) = \vec x\) for all \(\vec x\).

What is the matrix \(A\) that represents \(\vec f\) with respect to the standard basis?

Solution
\[ A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\]

You probably could have written down the answer immediately, since this is a very well-known transformation called the identity transformation and the matrix is the identity matrix. But here's why the identity matrix is what it is:

To find the matrix, we need to determine what \(\vec f\) does to each basis vector:

$$\begin{align*}\vec f(\hat{e}^{(1)}) &= \hat{e}^{(1)} = (1, 0, 0)^T \\\vec f(\hat{e}^{(2)}) &= \hat{e}^{(2)} = (0, 1, 0)^T \\\vec f(\hat{e}^{(3)}) &= \hat{e}^{(3)} = (0, 0, 1)^T \end{align*}$$

Placing these as columns:

\[ A = \begin{pmatrix} \uparrow & \uparrow & \uparrow \\ \vec f(\hat{e}^{(1)}) & \vec f(\hat{e}^{(2)}) & \vec f(\hat{e}^{(3)}) \\ \downarrow & \downarrow & \downarrow \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\]

That is the identity matrix we were expecting.

Problem #23

Tags: linear algebra, linear transformations, quiz-02, lecture-03

Suppose \(\vec f\) is represented by the following matrix:

\[ A = \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix}\]

Let \(\vec x = (5, 2)^T\). What is \(\vec f(\vec x)\)?

Solution

\(\vec f(\vec x) = (8, 23)^T\).

Since the matrix \(A\) represents \(\vec f\), we have \(\vec f(\vec x) = A \vec x\). We compute:

$$\begin{align*}\vec f(\vec x) = A \vec x &= \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix}\begin{pmatrix} 5 \\ 2 \end{pmatrix}\\&= 5 \begin{pmatrix} 2 \\ 3 \end{pmatrix} + 2 \begin{pmatrix} -1 \\ 4 \end{pmatrix}\\&= \begin{pmatrix} 10 \\ 15 \end{pmatrix} + \begin{pmatrix} -2 \\ 8 \end{pmatrix}\\&= \begin{pmatrix} 8 \\ 23 \end{pmatrix}\end{align*}$$

Problem #24

Tags: linear algebra, linear transformations, quiz-02, lecture-03

Suppose \(\vec f\) is represented by the following matrix:

\[ A = \begin{pmatrix} 1 & 2 & 0 \\ 0 & 1 & 3 \\ 2 & 0 & 1 \end{pmatrix}\]

Let \(\vec x = (2, 1, -1)^T\). What is \(\vec f(\vec x)\)?

Solution

\(\vec f(\vec x) = (4, -2, 3)^T\).

Since the matrix \(A\) represents \(\vec f\), we have \(\vec f(\vec x) = A \vec x\). We compute:

$$\begin{align*}\vec f(\vec x) = A \vec x &= \begin{pmatrix} 1 & 2 & 0 \\ 0 & 1 & 3 \\ 2 & 0 & 1 \end{pmatrix}\begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}\\&= 2 \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + 1 \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix} + (-1) \begin{pmatrix} 0 \\ 3 \\ 1 \end{pmatrix}\\&= \begin{pmatrix} 2 \\ 0 \\ 4 \end{pmatrix} + \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix} + \begin{pmatrix} 0 \\ -3 \\ -1 \end{pmatrix}\\&= \begin{pmatrix} 4 \\ -2 \\ 3 \end{pmatrix}\end{align*}$$

Problem #25

Tags: linear algebra, linear transformations, quiz-02, lecture-03

Consider the linear transformation \(\vec f : \mathbb{R}^2 \to\mathbb{R}^2\) that rotates vectors by \(90^\circ\) clockwise.

Let \(\mathcal{U} = \{\hat{u}^{(1)}, \hat{u}^{(2)}\}\) be an orthonormal basis with:

\[\hat{u}^{(1)} = \frac{1}{\sqrt{2}}(1, 1)^T, \quad\hat{u}^{(2)} = \frac{1}{\sqrt{2}}(1, -1)^T \]

What is the matrix \(A\) that represents \(\vec f\) with respect to the basis \(\mathcal{U}\)?

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

Remember that the columns of this matrix are:

\[ A = \begin{pmatrix} \uparrow & \uparrow \\ \vec[f(\hat{u}^{(1)})]_{\mathcal{U}} & \vec[f(\hat{u}^{(2)})]_{\mathcal{U}} \\ \downarrow & \downarrow \end{pmatrix}\]

So we need to compute \(\vec f(\hat{u}^{(1)})\) and \(\vec f(\hat{u}^{(2)})\), then express those results in the \(\mathcal{U}\) basis.

Let's start with \(\vec f(\hat{u}^{(1)})\). \(\vec f\) takes the vector \(\hat{u}^{(1)} = \frac{1}{\sqrt{2}}(1, 1)^T\), which points up and to the right at a \(45^\circ\) angle, and rotates it \(90^\circ\) clockwise, resulting in the vector \(\frac{1}{\sqrt{2}}(1, -1)^T = \hat{u}^{(2)}\)(that is, the unit vector pointing down and to the right at a \(45^\circ\) angle). As it so happens, this is exactly the second basis vector. That is, \(\vec f(\hat{u}^{(1)}) = \hat{u}^{(2)}\).

Now for \(\vec f(\hat{u}^{(2)})\). \(\vec f\) takes the vector \(\hat{u}^{(2)} = \frac{1}{\sqrt{2}}(1, -1)^T\), which points down and to the right at a \(45^\circ\) angle, and rotates it \(90^\circ\) clockwise, resulting in the vector \(\frac{1}{\sqrt{2}}(-1, -1)^T\). This isn't \(\vec f(\hat{u}^{(1)})\), exactly, but it is \(-1\) times \(\hat{u}^{(1)}\). That is, \(\vec f(\hat{u}^{(2)}) = -\hat{u}^{(1)}\).

What we have found is:

$$\begin{align*}[\vec f(\hat{u}^{(1)})]_{\mathcal{U}}&= [\hat{u}^{(2)}]_{\mathcal{U}} = (0, 1)^T \\{}[\vec f(\hat{u}^{(2)})]_{\mathcal{U}}&= [-\hat{u}^{(1)}]_{\mathcal{U}} = (-1, 0)^T \end{align*}$$

Therefore, the matrix is:

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

Problem #26

Tags: linear algebra, linear transformations, quiz-02, lecture-03

Consider the linear transformation \(\vec f : \mathbb{R}^2 \to\mathbb{R}^2\) that reflects vectors over the \(x\)-axis.

Let \(\mathcal{U} = \{\hat{u}^{(1)}, \hat{u}^{(2)}\}\) be an orthonormal basis with:

\[\hat{u}^{(1)} = \frac{1}{\sqrt{2}}(1, 1)^T, \quad\hat{u}^{(2)} = \frac{1}{\sqrt{2}}(1, -1)^T \]

What is the matrix \(A\) that represents \(\vec f\) with respect to the basis \(\mathcal{U}\)?

Solution
\[ A = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\]

Remember that the columns of this matrix are:

\[ A = \begin{pmatrix} \uparrow & \uparrow \\ [\vec f(\hat{u}^{(1)})]_{\mathcal{U}} & [\vec f(\hat{u}^{(2)})]_{\mathcal{U}} \\ \downarrow & \downarrow \end{pmatrix}\]

So we need to compute \(\vec f(\hat{u}^{(1)})\) and \(\vec f(\hat{u}^{(2)})\), then express those results in the \(\mathcal{U}\) basis.

Let's start with \(\vec f(\hat{u}^{(1)})\). \(\vec f\) takes the vector \(\hat{u}^{(1)} = \frac{1}{\sqrt{2}}(1, 1)^T\), which points up and to the right at a \(45^\circ\) angle, and reflects it over the \(x\)-axis, resulting in the vector \(\frac{1}{\sqrt{2}}(1, -1)^T = \hat{u}^{(2)}\)(the unit vector pointing down and to the right at a \(45^\circ\) angle). That is, \(\vec f(\hat{u}^{(1)}) = \hat{u}^{(2)}\).

Now for \(\vec f(\hat{u}^{(2)})\). \(\vec f\) takes the vector \(\hat{u}^{(2)} = \frac{1}{\sqrt{2}}(1, -1)^T\), which points down and to the right at a \(45^\circ\) angle, and reflects it over the \(x\)-axis, resulting in the vector \(\frac{1}{\sqrt{2}}(1, 1)^T = \hat{u}^{(1)}\). That is, \(\vec f(\hat{u}^{(2)}) = \hat{u}^{(1)}\).

What we have found is:

$$\begin{align*}[\vec f(\hat{u}^{(1)})]_{\mathcal{U}}&= [\hat{u}^{(2)}]_{\mathcal{U}} = (0, 1)^T \\{}[\vec f(\hat{u}^{(2)})]_{\mathcal{U}}&= [\hat{u}^{(1)}]_{\mathcal{U}} = (1, 0)^T \end{align*}$$

Therefore, the matrix is:

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