Problems tagged with "linear algebra"

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).

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:

True, with eigenvalue \(\lambda = 4\).

We compute:

Since \(A \vec v = 4 \vec v\), the vector \(\vec v\) is an eigenvector with eigenvalue \(4\).

False.

We compute:

For \(\vec v\) to be an eigenvector, we would need \(A \vec v = \lambda\vec v\) for some scalar \(\lambda\). This would require:

From the first component, we would need \(\lambda = 4\). From the second component, we would need \(\lambda = \frac{5}{2}\). Since these two values are not equal, there is no such \(\lambda\) that satisfies both equations. Therefore, \(\vec v\) is not an eigenvector of \(A\).

True, with eigenvalue \(\lambda = 2\).

We compute:

Since \(A \vec v = 2 \vec v\), the vector \(\vec v\) is an eigenvector with eigenvalue \(2\).

True, with eigenvalue \(\lambda = 4\).

We compute:

Since \(A \vec v = 4 \vec v\), the vector \(\vec v\) is an eigenvector with eigenvalue \(4\).

True, with eigenvalue \(\lambda = -3\).

We compute:

Since \(A \vec v = -3 \vec v\), the vector \(\vec v\) is an eigenvector with eigenvalue \(-3\).

True.

To verify this, we compute \(P \vec v\):

Since \(P \vec v = -\vec v = (-1) \vec v\), we see that \(\vec v\) is an eigenvector of \(P\) with eigenvalue \(-1\).

Note: The matrix \(P = I - 2 \vec v \vec v^T\) is called a Householder reflection matrix, which reflects vectors across the hyperplane orthogonal to \(\vec v\).

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.

\(\infty\) The upper-left \(2 \times 2\) block of \(A\) is the identity matrix. Recall from lecture that the identity matrix has infinitely many eigenvectors: every nonzero vector is an eigenvector of the identity with eigenvalue 1.

Similarly, any vector of the form \((a, b, 0)^T\) where \(a\) and \(b\) are not both zero satisfies:

So any such vector is an eigenvector with eigenvalue 1. There are infinitely many unit vectors of this form (they form a circle in the \(x\)-\(y\) plane), so \(A\) has infinitely many unit length eigenvectors.

Additionally, \((0, 0, 1)^T\) is an eigenvector with eigenvalue 5.

False.

The spectral theorem only applies to symmetric matrices. The matrix \(A\) is not symmetric because \(A^T \neq A\):

Since \(A\) is not symmetric, the spectral theorem does not apply, and we cannot use it to conclude anything about \(A\)'s eigenvectors.

False.

Consider the identity matrix \(I\). Every nonzero vector is an eigenvector of \(I\) with eigenvalue 1. For example, both \((1, 0)^T\) and \((1, 1)^T\) are eigenvectors of \(I\), but they are not orthogonal:

The spectral theorem says that you can find\(n\) orthogonal eigenvectors for an \(n \times n\) symmetric matrix, but it does not say that every pair of eigenvectors is orthogonal.

Aside: It can be shown that if \(\vec{u}^{(1)}\) and \(\vec{u}^{(2)}\) have different eigenvalues, then they must be orthogonal.

True.

By the spectral theorem, every \(d \times d\) symmetric matrix has \(d\) mutually orthogonal eigenvectors. If we normalize these eigenvectors, they form an orthonormal basis.

In this eigenbasis, the matrix \(A\) is diagonal: the diagonal entries are the eigenvalues of \(A\).

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.

\([\vec{f}(\vec{x})]_{\mathcal{U}} = (16, -12, -3)^T\) Using linearity and the eigenvector property:

Since \(\vec{u}^{(1)}\) is an eigenvector with eigenvalue \(8\), we have \(\vec{f}(\vec{u}^{(1)}) = 8\vec{u}^{(1)}\). Similarly, \(\vec{f}(\vec{u}^{(2)}) = 4\vec{u}^{(2)}\) and \(\vec{f}(\vec{u}^{(3)}) = -3\vec{u}^{(3)}\):

In the eigenbasis, this is simply \((16, -12, -3)^T\).

\([\vec{f}(\vec{x})]_{\mathcal{U}} = (20, -2, -6)^T\) Using linearity and the eigenvector property:

Since \(\vec{u}^{(1)}\) is an eigenvector with eigenvalue \(5\), we have \(\vec{f}(\vec{u}^{(1)}) = 5\vec{u}^{(1)}\). Similarly, \(\vec{f}(\vec{u}^{(2)}) = -2\vec{u}^{(2)}\) and \(\vec{f}(\vec{u}^{(3)}) = 3\vec{u}^{(3)}\):

In the eigenbasis, this is simply \((20, -2, -6)^T\).

False. The maximum is \(9\).

The maximum of \(\|A\vec{x}\|\) over unit vectors is achieved when \(\vec{x}\) is an eigenvector corresponding to the eigenvalue with the largest absolute value.

Here, \(|-9| = 9 > 6 = |6|\), so the maximum is achieved at the eigenvector with eigenvalue \(-9\).

If \(\vec{u}\) is a unit eigenvector with eigenvalue \(-9\), then:

True.

The main thing to realize is that \(A\) and \(B\) have the same eigenvectors. If \(\vec v\) is an eigenvector of \(A\) with eigenvalue \(\lambda\), then:

Thus, \(\vec v\) is also an eigenvector of \(B\) with eigenvalue \(\lambda + 5\).

This means that the eigenvalues of \(B\) are simply the eigenvalues of \(A\) shifted by 5. Therefore, if \(\lambda\) is the top eigenvalue of \(A\), then \(\lambda + 5\) is the top eigenvalue of \(B\).