DSC 140B
Problems tagged with eigenvalues

Problems tagged with "eigenvalues"

Problem #28

Tags: linear algebra, quiz-03, eigenvalues, eigenvectors, lecture-04

Let \(A = \begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix}\) and let \(\vec v = (1, 1)^T\).

True or False: \(\vec v\) is an eigenvector of \(A\). If true, what is the corresponding eigenvalue?

Solution

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

We compute:

\[ A \vec v = \begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix}\begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 4 \\ 4 \end{pmatrix} = 4 \begin{pmatrix} 1 \\ 1 \end{pmatrix} = 4 \vec v \]

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

Problem #29

Tags: linear algebra, quiz-03, eigenvalues, eigenvectors, lecture-04

Let \(A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}\) and let \(\vec v = (1, 2)^T\).

True or False: \(\vec v\) is an eigenvector of \(A\). If true, what is the corresponding eigenvalue?

Solution

False.

We compute:

\[ A \vec v = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}\begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 4 \\ 5 \end{pmatrix}\]

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

\[\begin{pmatrix} 4 \\ 5 \end{pmatrix} = \lambda\begin{pmatrix} 1 \\ 2 \end{pmatrix}\]

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 \(A\).

Problem #30

Tags: linear algebra, quiz-03, eigenvalues, eigenvectors, lecture-04

Let \(A = \begin{pmatrix} 1 & 1 & 3 \\ 1 & 2 & 1 \\ 3 & 1 & 9 \end{pmatrix}\) and let \(\vec v = (1, 4, -1)^T\).

True or False: \(\vec v\) is an eigenvector of \(A\). If true, what is the corresponding eigenvalue?

Solution

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

We compute:

\[ A \vec v = \begin{pmatrix} 1 & 1 & 3 \\ 1 & 2 & 1 \\ 3 & 1 & 9 \end{pmatrix}\begin{pmatrix} 1 \\ 4 \\ -1 \end{pmatrix} = \begin{pmatrix} 1 + 4 - 3 \\ 1 + 8 - 1 \\ 3 + 4 - 9 \end{pmatrix} = \begin{pmatrix} 2 \\ 8 \\ -2 \end{pmatrix} = 2 \begin{pmatrix} 1 \\ 4 \\ -1 \end{pmatrix} = 2 \vec v \]

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

Problem #31

Tags: linear algebra, quiz-03, eigenvalues, eigenvectors, lecture-04

Let \(A = \begin{pmatrix} 5 & 1 & 1 & 1 \\ 1 & 5 & 1 & 1 \\ 1 & 1 & 5 & 1 \\ 1 & 1 & 1 & 5 \end{pmatrix}\) and let \(\vec v = (1, 2, -2, -1)^T\).

True or False: \(\vec v\) is an eigenvector of \(A\). If true, what is the corresponding eigenvalue?

Solution

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

We compute:

\[ A \vec v = \begin{pmatrix} 5 & 1 & 1 & 1 \\ 1 & 5 & 1 & 1 \\ 1 & 1 & 5 & 1 \\ 1 & 1 & 1 & 5 \end{pmatrix}\begin{pmatrix} 1 \\ 2 \\ -2 \\ -1 \end{pmatrix} = \begin{pmatrix} 5 + 2 - 2 - 1 \\ 1 + 10 - 2 - 1 \\ 1 + 2 - 10 - 1 \\ 1 + 2 - 2 - 5 \end{pmatrix} = \begin{pmatrix} 4 \\ 8 \\ -8 \\ -4 \end{pmatrix} = 4 \vec v \]

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

Problem #32

Tags: linear algebra, quiz-03, eigenvalues, eigenvectors, lecture-04

Let \(A = \begin{pmatrix} 1 & 6 \\ 6 & 6 \end{pmatrix}\) and let \(\vec v = (3, -2)^T\).

True or False: \(\vec v\) is an eigenvector of \(A\). If true, what is the corresponding eigenvalue?

Solution

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

We compute:

\[ A \vec v = \begin{pmatrix} 1 & 6 \\ 6 & 6 \end{pmatrix}\begin{pmatrix} 3 \\ -2 \end{pmatrix} = \begin{pmatrix} 3 - 12 \\ 18 - 12 \end{pmatrix} = \begin{pmatrix} -9 \\ 6 \end{pmatrix} = -3 \begin{pmatrix} 3 \\ -2 \end{pmatrix} = -3 \vec v \]

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

Problem #33

Tags: linear algebra, quiz-03, eigenvalues, eigenvectors, lecture-04

Let \(\vec v\) be a unit vector in \(\mathbb R^d\), and let \(I\) be the \(d \times d\) identity matrix. Consider the matrix \(P\) defined as:

\[ P = I - 2 \vec v \vec v^T \]

True or False: \(\vec v\) is an eigenvector of \(P\).

Solution

True.

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

$$\begin{align*} P \vec v &= (I - 2 \vec v \vec v^T) \vec v \\&= I \vec v - 2 \vec v \vec v^T \vec v \\&= \vec v - 2 \vec v (\vec v^T \vec v) \\&= \vec v - 2 \vec v (1) \quad\text{($\vec v^T \vec v = \|\vec v\|^2$, and $\vec v$ is a unit vector)}\\&= \vec v - 2 \vec v \\&= -\vec v \end{align*}$$

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 \(\vec v\).

Problem #34

Tags: linear algebra, quiz-03, eigenvalues, eigenvectors, lecture-04

Let \(A\) be the matrix:

\[ A = \begin{pmatrix} 5 & 2 \\ 2 & 2 \end{pmatrix}\]

It can be verified that the vector \(\vec{u}^{(1)} = (2, 1)^T\) is an eigenvector of \(A\).

Part 1)

What is the eigenvalue associated with \(\vec{u}^{(1)}\)?

Solution

The eigenvalue is \(\lambda_1 = 6\).

To find the eigenvalue, we compute \(A \vec{u}^{(1)}\):

$$\begin{align*} A \vec{u}^{(1)}&= \begin{pmatrix} 5 & 2 \\ 2 & 2 \end{pmatrix}\begin{pmatrix} 2 \\ 1 \end{pmatrix}\\&= \begin{pmatrix} 5 \cdot 2 + 2 \cdot 1 \\ 2 \cdot 2 + 2 \cdot 1 \end{pmatrix}\\&= \begin{pmatrix} 12 \\ 6 \end{pmatrix}\\&= 6 \begin{pmatrix} 2 \\ 1 \end{pmatrix} = 6 \vec{u}^{(1)}\end{align*}$$

Therefore, \(\lambda_1 = 6\).

Part 2)

Find another eigenvector \(\vec{u}^{(2)}\) of \(A\). Your eigenvector should have an eigenvalue that is different from \(\vec{u}^{(1)}\)'s eigenvalue. It does not need to be normalized.

Solution

\(\vec{u}^{(2)} = (1, -2)^T\)(or any scalar multiple).

Since \(A\) is symmetric, we know that we can always find two orthogonal eigenvectors. This suggests that we should find a vector orthogonal to \(\vec{u}^{(1)} = (2, 1)^T\) and make sure that it is indeed an eigenvector.

We know from the math review in Week 01 that the vector \((1, -2)^T\) is orthogonal to \((2, 1)^T\)(in general, \((a, b)^T\) is orthogonal to \((-b, a)^T\)).

We can verify:

\[ A \vec{u}^{(2)} = \begin{pmatrix} 5 & 2 \\ 2 & 2 \end{pmatrix}\begin{pmatrix} 1 \\ -2 \end{pmatrix} = \begin{pmatrix} 5 - 4 \\ 2 - 4 \end{pmatrix} = \begin{pmatrix} 1 \\ -2 \end{pmatrix} = 1 \cdot\vec{u}^{(2)}\]

So the eigenvalue is \(\lambda_2 = 1\), which is indeed different from \(\lambda_1 = 6\).

Problem #35

Tags: linear algebra, quiz-03, eigenvalues, eigenvectors, lecture-04

Let \(D\) be the diagonal matrix:

\[ D = \begin{pmatrix} 2 & 0 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & 7 \end{pmatrix}\]

Part 1)

What is the top eigenvalue of \(D\)? What eigenvector corresponds to this eigenvalue?

Solution

The top eigenvalue is \(7\).

For a diagonal matrix, the eigenvalues are exactly the diagonal entries. The diagonal entries are \(2, -5, 7\), and the largest is \(7\).

An eigenvector corresponding to this eigenvalue is \(\vec v = (0, 0, 1)^T\).

Part 2)

What is the bottom (smallest) eigenvalue of \(D\)? What eigenvector corresponds to this eigenvalue?

Solution

The bottom eigenvalue is \(-5\).

An eigenvector corresponding to this eigenvalue is \(\vec w = (0, 1, 0)^T\).

Problem #36

Tags: linear algebra, quiz-03, eigenvalues, eigenvectors, lecture-04, linear transformations

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

Find two orthogonal eigenvectors of this transformation and their corresponding eigenvalues.

Solution

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 \(-1\).

Problem #37

Tags: linear algebra, quiz-03, eigenvalues, eigenvectors, lecture-04, linear transformations

Consider the linear transformation \(\vec f : \mathbb{R}^2 \to\mathbb{R}^2\) that scales vectors along the line \(y = x\) by a factor of 2, and scales vectors along the line \(y = -x\) by a factor of 3.

Find two orthogonal eigenvectors of this transformation and their corresponding eigenvalues.

Solution

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 \(3\).

Problem #38

Tags: linear algebra, quiz-03, eigenvalues, eigenvectors, lecture-04, linear transformations

Consider the linear transformation \(\vec f : \mathbb{R}^2 \to\mathbb{R}^2\) that rotates vectors by \(180°\).

Find two orthogonal eigenvectors of this transformation and their corresponding eigenvalues.

Solution

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.

Problem #39

Tags: linear algebra, quiz-03, eigenvalues, eigenvectors, lecture-04

Consider the diagonal matrix:

\[ A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 5 \end{pmatrix}\]

How many unit length eigenvectors does \(A\) have?

Solution

\(\infty\) 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:

\[ A \begin{pmatrix} a \\ b \\ 0 \end{pmatrix} = \begin{pmatrix} a \\ b \\ 0 \end{pmatrix} = 1 \cdot\begin{pmatrix} a \\ b \\ 0 \end{pmatrix}\]

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.

Problem #45

Tags: linear algebra, eigenbasis, quiz-03, eigenvalues, eigenvectors, lecture-04

Let \(\vec{u}^{(1)}, \vec{u}^{(2)}, \vec{u}^{(3)}\) be three unit length orthonormal eigenvectors of a linear transformation \(\vec{f}\), with eigenvalues \(8\), \(4\), and \(-3\) respectively.

Suppose a vector \(\vec{x}\) can be written as:

\[\vec{x} = 2\vec{u}^{(1)} - 3\vec{u}^{(2)} + \vec{u}^{(3)}\]

What is \(\vec{f}(\vec{x})\), expressed in coordinates with respect to the eigenbasis \(\{\vec{u}^{(1)}, \vec{u}^{(2)}, \vec{u}^{(3)}\}\)?

Solution

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

$$\begin{align*}\vec{f}(\vec{x}) &= \vec{f}(2\vec{u}^{(1)} - 3\vec{u}^{(2)} + \vec{u}^{(3)}) \\&= 2\vec{f}(\vec{u}^{(1)}) - 3\vec{f}(\vec{u}^{(2)}) + \vec{f}(\vec{u}^{(3)}) \\\end{align*}$$

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 \(\vec{f}(\vec{u}^{(3)}) = -3\vec{u}^{(3)}\):

$$\begin{align*}&= 2(8\vec{u}^{(1)}) - 3(4\vec{u}^{(2)}) + (-3)\vec{u}^{(3)}\\&= 16\vec{u}^{(1)} - 12\vec{u}^{(2)} - 3\vec{u}^{(3)}\end{align*}$$

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

Problem #46

Tags: linear algebra, eigenbasis, quiz-03, eigenvalues, eigenvectors, lecture-04

Let \(\vec{u}^{(1)}, \vec{u}^{(2)}, \vec{u}^{(3)}\) be three unit length orthonormal eigenvectors of a linear transformation \(\vec{f}\), with eigenvalues \(5\), \(-2\), and \(3\) respectively.

Suppose a vector \(\vec{x}\) can be written as:

\[\vec{x} = 4\vec{u}^{(1)} + \vec{u}^{(2)} - 2\vec{u}^{(3)}\]

What is \(\vec{f}(\vec{x})\), expressed in coordinates with respect to the eigenbasis \(\{\vec{u}^{(1)}, \vec{u}^{(2)}, \vec{u}^{(3)}\}\)?

Solution

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

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

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 \(\vec{f}(\vec{u}^{(3)}) = 3\vec{u}^{(3)}\):

$$\begin{align*}&= 4(5\vec{u}^{(1)}) + (-2)\vec{u}^{(2)} - 2(3\vec{u}^{(3)}) \\&= 20\vec{u}^{(1)} - 2\vec{u}^{(2)} - 6\vec{u}^{(3)}\end{align*}$$

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

Problem #47

Tags: optimization, linear algebra, quiz-03, eigenvalues, eigenvectors, lecture-04

Let \(A\) be a symmetric matrix with eigenvalues \(6\) and \(-9\).

True or False: The maximum value of \(\|A\vec{x}\|\) over all unit vectors \(\vec{x}\) is \(6\).

Solution

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:

\[\|A\vec{u}\| = \|(-9)\vec{u}\| = |-9| \cdot\|\vec{u}\| = 9 \cdot 1 = 9 \]

Problem #48

Tags: optimization, linear algebra, quiz-03, eigenvalues, eigenvectors, quadratic forms, lecture-04

Let \(A\) be a \(3 \times 3\) symmetric matrix with the following eigenvectors and corresponding eigenvalues: \(\vec{u}^{(1)} = \frac{1}{3}(1, 2, 2)^T\) has eigenvalue \(4\), \(\vec{u}^{(2)} = \frac{1}{3}(2, 1, -2)^T\) has eigenvalue \(1\), and \(\vec{u}^{(3)} = \frac{1}{3}(2, -2, 1)^T\) has eigenvalue \(-10\).

Consider the quadratic form \(\vec{x}\cdot A\vec{x}\).

Part 1)

What unit vector \(\vec{x}\) maximizes \(\vec{x}\cdot A\vec{x}\)?

Solution

\(\vec{u}^{(1)} = \frac{1}{3}(1, 2, 2)^T\) The quadratic form \(\vec{x}\cdot A\vec{x}\) is maximized by the eigenvector with the largest eigenvalue. Among \(4\), \(1\), and \(-10\), the largest is \(4\), so the maximizer is \(\vec{u}^{(1)}\).

Part 2)

What is the maximum value of \(\vec{x}\cdot A\vec{x}\) over all unit vectors?

Solution

\(4\) The maximum value equals the largest eigenvalue. We can verify:

\[\vec{u}^{(1)}\cdot A\vec{u}^{(1)} = \vec{u}^{(1)}\cdot(4\vec{u}^{(1)}) = 4(\vec{u}^{(1)}\cdot\vec{u}^{(1)}) = 4 \cdot 1 = 4 \]

Part 3)

What unit vector \(\vec{x}\) minimizes \(\vec{x}\cdot A\vec{x}\)?

Solution

\(\vec{u}^{(3)} = \frac{1}{3}(2, -2, 1)^T\) The quadratic form is minimized by the eigenvector with the smallest eigenvalue. Among \(4\), \(1\), and \(-10\), the smallest is \(-10\), so the minimizer is \(\vec{u}^{(3)}\).

Part 4)

What is the minimum value of \(\vec{x}\cdot A\vec{x}\) over all unit vectors?

Solution

\(-10\) The minimum value equals the smallest eigenvalue. We can verify:

\[\vec{u}^{(3)}\cdot A\vec{u}^{(3)} = \vec{u}^{(3)}\cdot(-10\vec{u}^{(3)}) = -10(\vec{u}^{(3)}\cdot\vec{u}^{(3)}) = -10 \cdot 1 = -10 \]

Problem #49

Tags: linear algebra, quiz-03, eigenvalues, eigenvectors, lecture-04

Let \(A\) be a symmetric matrix with top eigenvalue \(\lambda\). Let \(B = A + 5I\), where \(I\) is the identity matrix.

True or False: The top eigenvalue of \(B\) must be \(\lambda + 5\).

Solution

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:

$$\begin{align*} B \vec v &= (A + 5I) \vec v \\&= A \vec v + 5I \vec v \\&= \lambda\vec v + 5 \vec v \\&= (\lambda + 5) \vec v \end{align*}$$

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 \(B\).