Problems tagged with "linear transformations"

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:

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

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:

The eigenvectors are \((1, -1)^T\) with \(\lambda = 1\), and \((1, 1)^T\) with \(\lambda = -1\).

Geometrically, vectors along the line \(y = -x\)(i.e., multiples of \((1, -1)^T\)) are unchanged by reflection over that line, so they have eigenvalue \(1\). Vectors perpendicular to the line \(y = -x\)(i.e., multiples of \((1, 1)^T\)) are flipped to point in the opposite direction, so they have eigenvalue \(-1\).

The eigenvectors are \((1, 1)^T\) with \(\lambda = 2\), and \((1, -1)^T\) with \(\lambda = 3\).

Geometrically, vectors along the line \(y = x\)(i.e., multiples of \((1, 1)^T\)) are scaled by a factor of 2, so they have eigenvalue \(2\). Vectors along the line \(y = -x\)(i.e., multiples of \((1, -1)^T\)) are scaled by a factor of 3, so they have eigenvalue \(3\).

Any pair of orthogonal vectors works, such as \((1, 0)^T\) and \((0, 1)^T\). Both have eigenvalue \(\lambda = -1\).

Geometrically, rotating any vector by \(180°\) reverses its direction, so \(\vec f(\vec v) = -\vec v\) for all \(\vec v\). This means every nonzero vector is an eigenvector with eigenvalue \(-1\).

Since we need two orthogonal eigenvectors, any orthogonal pair will do.

False.

Recall from lecture that symmetric linear transformations have orthogonal axes of symmetry. In the visualization, this would appear as two perpendicular directions where the arrows point directly outward (or inward) from the circle.

In this figure, there are no such orthogonal axes of symmetry. The pattern of arrows does not exhibit the characteristic symmetry of a symmetric transformation.

True.

A matrix is diagonal if and only if the standard basis vectors are eigenvectors. In the visualization, eigenvectors correspond to directions where the arrows point radially (directly outward or inward).

Looking at the figure, the arrows at \((1, 0)\) and \((-1, 0)\) point horizontally, and the arrows at \((0, 1)\) and \((0, -1)\) point vertically. This means the standard basis vectors \(\hat e^{(1)} = (1, 0)^T\) and \(\hat e^{(2)} = (0, 1)^T\) are eigenvectors.

Since the standard basis vectors are eigenvectors, the matrix is diagonal in the standard basis.