Problems tagged with "quiz-02"

5.

The "right" way to do this problem is not to calculate the entire dot product, but to recognize that The dot product \(\vec x \cdot\hat{e}^{(3)}\) extracts the third component of \(\vec x\)(which is 5).

\([\vec x]_{\mathcal{U}} = (3\sqrt{2}, -\sqrt{2})^T\).

To find the coordinates of \(\vec x\) in the basis \(\mathcal{U}\), we compute the dot product of \(\vec x\) with each basis vector:

Therefore, the coordinates of \(\vec x\) in the basis \(\mathcal{U}\) are:

\([\vec x]_{\mathcal{U}} = (2\sqrt{3}, \sqrt{2}, 0)^T\).

To find the coordinates of \(\vec x\) in the basis \(\mathcal{U}\), we compute the dot product of \(\vec x\) with each basis vector:

Therefore, the coordinates of \(\vec x\) in the basis \(\mathcal{U}\) are:

\([\vec x]_{\mathcal{U}} = (3\sqrt{2}, 3, -\sqrt{2})^T\).

To find the coordinates of \(\vec x\) in the basis \(\mathcal{U}\), we compute the dot product of \(\vec x\) with each basis vector:

Therefore, the coordinates of \(\vec x\) in the basis \(\mathcal{U}\) are:

\([\vec x]_{\mathcal{U}} = \left(-\frac{6}{5}, -\frac{17}{5}\right)^T\).

The coordinates of \(\vec x\) in basis \(\mathcal{U}\) are given by \([\vec x]_{\mathcal{U}} = (\vec x \cdot\hat{u}^{(1)}, \vec x \cdot\hat{u}^{(2)})^T\). We can compute each dot product using the given information:

Nope, they don't. The problem is that the second and third vectors are not orthogonal. To be an orthonormal basis, all pairs of vectors must be orthogonal (dot product zero) and each vector must have unit length.

Nope, they don't. While they are all orthogonal to each other, the second vector does not have unit length (its length is 2). To be an orthonormal basis, all vectors must have unit length.

Yes.

You can check that all vectors have unit length, and that each pair of vectors is orthogonal (dot product zero).

Yes, \(\vec f\) is a linear transformation.

To verify linearity, we must show that for all vectors \(\vec x, \vec y\) and scalars \(\alpha, \beta\):

Let \(\vec x = (x_1, x_2)^T\) and \(\vec y = (y_1, y_2)^T\). Then:

Therefore, \(\vec f\) is linear.

No, \(\vec f\) is not a linear transformation.

To show this, we find a counterexample. Consider \(\vec x = (1, 1)^T\) and \(\vec y = (1, 1)^T\), and let \(\alpha = \beta = 1\).

First, compute \(\vec f(\vec x + \vec y) = \vec f((2, 2)^T)\):

\(\vec f((2, 2)^T) = (2 \cdot 2, \, 2)^T = (4, 2)^T\) Now compute \(\vec f(\vec x) + \vec f(\vec y)\):

Since \((4, 2)^T \neq(2, 2)^T\), we have \(\vec f(\vec x + \vec y) \neq\vec f(\vec x) + \vec f(\vec y)\).

Therefore, \(\vec f\) is not linear. (The product \(x_1 x_2\) in the first component makes the function nonlinear.)

No, \(\vec f\) is not a linear transformation.

A quick way to check: for any linear transformation, we must have \(\vec f(c\vec x) = c \vec f(\vec x)\) for any scalar \(c\).

Let us check the scaling property with \(\vec x = (1, 0)^T\) and \(c = 2\):

Since \((4, 0)^T \neq(2, 0)^T\), we have \(\vec f(2\vec x) \neq 2\vec f(\vec x)\).

Therefore, \(\vec f\) is not linear. (The squared term \(x_1^2\) makes the function nonlinear.)

Yes, \(\vec f\) is a linear transformation.

To verify linearity, we must show that for all vectors \(\vec x, \vec y\) and scalars \(\alpha, \beta\):

Let \(\vec x = (x_1, x_2)^T\) and \(\vec y = (y_1, y_2)^T\). Then:

Therefore, \(\vec f\) is linear.

By the way, this transformation does something interesting geometrically: it rotates vectors by \(90°\) counterclockwise.

\(\vec f(\vec x) = (17, -1)^T\).

Since \(\vec f\) is linear, we can write:

Here, \(x_1 = 2\) and \(x_2 = 5\). Substituting the given values:

\(\vec f(\vec x) = (-8, 10)^T\).

Since \(\vec f\) is linear, we can write:

Substituting the given values:

True.

The sum of squared entries of a vector is equal to the length of the vector squared. If we express \(\vec x\) in a different basis, we get different coordinates, but it doesn't change the length of the vector itself. So whatever those new coordinates are, their sum of squares must still equal 1.

False.

The sum of absolute values (the \(\ell_1\) norm) is not preserved under a change of orthonormal basis. Only the sum of squares (the \(\ell_2\) norm squared) is preserved.

As a counterexample, consider \(\vec x = (1, 0)^T\) in \(\mathbb{R}^2\). The sum of absolute values is \(1\). Now consider the orthonormal basis \(\mathcal{U} = \{\frac{1}{\sqrt{2}}(1, 1)^T, \frac{1}{\sqrt{2}}(-1, 1)^T\}\). Then:

The sum of absolute values is now \(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}\neq 1\).

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

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 \(\vec f(\hat{u}^{(2)})\).

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:

We need to calculate all of these dot products:

Therefore, we have:

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

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

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 \(\vec f(\hat{u}^{(3)})\).

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:

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

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

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

Therefore, we have:

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

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

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

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:

This means we can immediately write:

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

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:

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:

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:

Placing these as columns:

That is the identity matrix we were expecting.

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

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

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

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

Remember that the columns of this matrix are:

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:

Therefore, the matrix is:

Remember that the columns of this matrix are:

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:

Therefore, the matrix is:

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

To convert from basis \(\mathcal{U}\) to the standard basis, we compute the linear combination of basis vectors: