# TBIL Activities for Understanding Linear Algebra

## Section4.2Finding eigenvalues and eigenvectors

### Activity4.2.0.1.

Suppose that $$A$$ is a square matrix and that the nonzero vector $$\xvec$$ is a solution to the homogeneous equation $$A\xvec = \zerovec\text{.}$$

#### (a)

What can we conclude about the invertibility of $$A\text{?}$$
1. $$A$$ is invertible.
2. $$A$$ is not invertible.
3. $$A$$ could be either invertible or not invertible.

#### (b)

How does the determinant $$\det A$$ tell us if there is a nonzero solution to the homogeneous equation $$A\xvec = \zerovec\text{?}$$
1. The determinant of $$A$$ must be 0.
2. The determinant of $$A$$ must be nonzero.
3. The determinant of $$A$$ must be 1.

### Activity4.2.0.2.

Suppose that
\begin{equation*} A = \left[\begin{array}{rrr} 3 \amp -1 \amp 1 \\ 0 \amp 2 \amp 4 \\ 1 \amp 1 \amp 3 \\ \end{array}\right]\text{.} \end{equation*}

#### (a)

Find the determinant $$\det A\text{.}$$ What does this tell us about the solution space to the homogeneous equation $$A\xvec = \zerovec\text{?}$$

#### (b)

Find a basis for $$\nul(A)\text{.}$$

#### (c)

What is the relationship between the rank of a matrix and the dimension of its null space?

### Activity4.2.0.3.

The eigenvalues of a square matrix are defined by the condition that there be a nonzero solution to the homogeneous equation $$(A-\lambda I)\vvec=\zerovec\text{.}$$

#### (a)

If there is a nonzero solution to the homogeneous equation $$(A-\lambda I)\vvec = \zerovec\text{,}$$ what can we conclude about the invertibility of the matrix $$A-\lambda I\text{?}$$
1. $$A-\lambda I$$ is invertible.
2. $$A-\lambda I$$ is not invertible.
3. $$A-\lambda I$$ could be either invertible or not invertible, depending on the invertibility of $$A\text{.}$$

#### (b)

If there is a nonzero solution to the homogeneous equation $$(A-\lambda I)\vvec = \zerovec\text{,}$$ what can we conclude about the determinant $$\det(A-\lambda I)\text{?}$$
1. The determinant of $$A-\lambda I$$ must be 0.
2. The determinant of $$A-\lambda I$$ must be nonzero.
3. The determinant of $$A$$ must equal the determinant of $$\lambda I\text{.}$$

### Activity4.2.0.4.

Let's consider the matrix
\begin{equation*} A = \left[\begin{array}{rr} 1 \amp 2 \\ 2 \amp 1 \\ \end{array}\right] \end{equation*}
from which we construct
\begin{equation*} A-\lambda I = \left[\begin{array}{rr} 1 \amp 2 \\ 2 \amp 1 \\ \end{array}\right] - \lambda \left[\begin{array}{rr} 1 \amp 0 \\ 0 \amp 1 \\ \end{array}\right] = \left[\begin{array}{rr} 1-\lambda \amp 2 \\ 2 \amp 1-\lambda \\ \end{array}\right]\text{.} \end{equation*}

#### (a)

Find the determinant $$\det(A-\lambda I)\text{.}$$

#### (b)

What kind of equation do you obtain when we set this determinant to zero to obtain $$\det(A-\lambda I) = 0\text{?}$$

#### (c)

Use the determinant you found in the previous part to find the eigenvalues $$\lambda$$ by solving $$\det(A-\lambda I) = 0\text{.}$$ We considered this matrix in the previous section so we should find the same eigenvalues for $$A$$ that we found by reasoning geometrically there.

### Activity4.2.0.5.

Consider the matrix $$A = \left[\begin{array}{rr} 2 \amp 1 \\ 0 \amp 2 \\ \end{array}\right]$$ and find its eigenvalues by solving the equation $$\det(A-\lambda I) = 0\text{.}$$

### Activity4.2.0.6.

Consider the matrix $$A = \left[\begin{array}{rr} 0 \amp -1 \\ 1 \amp 0 \\ \end{array}\right]$$ and find its eigenvalues by solving the equation $$\det(A-\lambda I) = 0\text{.}$$

### Activity4.2.0.7.

This activity focuses on the eigenvalues of triangular matrices.

#### (a)

Find the eigenvalues of the triangular matrix $$\left[\begin{array}{rrr} 3 \amp -1 \amp 4 \\ 0 \amp -2 \amp 3 \\ 0 \amp 0 \amp 1 \\ \end{array}\right] \text{.}$$

#### (b)

What is generally true about the eigenvalues of a triangular matrix?
This activity demonstrates a technique that enables us to find the eigenvalues of a square matrix $$A\text{.}$$ Since an eigenvalue $$\lambda$$ is a scalar for which the equation $$(A-\lambda I)\vvec = \zerovec$$ has a nonzero solution, it must be the case that $$A-\lambda I$$ is not invertible. Therefore, its determinant is zero. This gives us the equation
\begin{equation*} \det(A-\lambda I) = 0 \end{equation*}
whose solutions are the eigenvalues of $$A\text{.}$$ This equation is called the characteristic equation of $$A\text{.}$$
In the next few activities, we will find the eigenvectors of a matrix as the null space of the matrix $$A-\lambda I\text{.}$$

### Activity4.2.0.8.

Let's begin with the matrix $$A = \left[\begin{array}{rr} 1 \amp 2 \\ 2 \amp 1 \\ \end{array}\right] \text{.}$$

#### (a)

We have seen that $$\lambda = 3$$ is an eigenvalue. Form the matrix $$A-3I$$ and find a basis for the eigenspace $$E_3 = \nul(A-3I)\text{.}$$

#### (b)

What is the dimension of this eigenspace?

#### (c)

For each of the basis vectors $$\vvec\text{,}$$ verify that $$A\vvec = 3\vvec\text{.}$$

#### (d)

We also saw that $$\lambda = -1$$ is an eigenvalue. Form the matrix $$A-(-1)I$$ and find a basis for the eigenspace $$E_{-1}\text{.}$$

#### (e)

What is the dimension of this eigenspace?

#### (f)

For each of the basis vectors $$\vvec\text{,}$$ verify that $$A\vvec = -\vvec\text{.}$$

#### (g)

Is it possible to form a basis of $$\real^2$$ consisting of eigenvectors of $$A\text{?}$$

### Activity4.2.0.9.

Now consider the matrix $$A = \left[\begin{array}{rr} 3 \amp 0 \\ 0 \amp 3 \\ \end{array}\right] \text{.}$$

#### (a)

Write the characteristic equation for $$A$$ and use it to find the eigenvalues of $$A\text{.}$$

#### (b)

For each eigenvalue, find a basis for its eigenspace $$E_\lambda\text{.}$$

#### (c)

Is it possible to form a basis of $$\real^2$$ consisting of eigenvectors of $$A\text{?}$$

### Activity4.2.0.10.

Next, consider the matrix $$A = \left[\begin{array}{rr} 2 \amp 1 \\ 0 \amp 2 \\ \end{array}\right] \text{.}$$

#### (a)

Write the characteristic equation for $$A$$ and use it to find the eigenvalues of $$A\text{.}$$

#### (b)

For each eigenvalue, find a basis for its eigenspace $$E_\lambda\text{.}$$

#### (c)

Is it possible to form a basis of $$\real^2$$ consisting of eigenvectors of $$A\text{?}$$

### Activity4.2.0.11.

Let $$A = \left[\begin{array}{rr} 4 \amp 0 \\ 0 \amp -1 \\ \end{array}\right] \text{.}$$

#### (a)

Find the eigenvalues and eigenvectors of the diagonal matrix $$A\text{.}$$

#### (b)

Explain your result by considering the geometric effect of the matrix transformation defined by $$A\text{.}$$
Once we find the eigenvalues of a matrix $$A\text{,}$$ describing the eigenspace $$E_\lambda$$ amounts to the familiar task of describing the null space $$\nul(A-\lambda I)\text{.}$$

### Activity4.2.0.12.

Suppose you have an $$n\times n$$ matrix whose characteristic polynomial is $$(2-\lambda)^3(-3-\lambda)^{10}(5-\lambda)\text{.}$$

#### (a)

Identify the eigenvalues, and their multiplicities, of the matrix based on the characteristic polynomial.

#### (b)

What can you conclude about the dimensions of the eigenspaces?

#### (c)

What is the dimension of the matrix?

#### (d)

Do you have enough information to guarantee that there is a basis of $$\real^n$$ consisting of eigenvectors?

### Activity4.2.0.13.

Let $$A= \left[\begin{array}{rr} 0 \amp -1 \\ 4 \amp -4 \\ \end{array}\right] \text{.}$$

#### (a)

Find the eigenvalues of $$A$$ and state their multiplicities.

#### (b)

Can you find a basis of $$\real^2$$ consisting of eigenvectors of this matrix?

### Activity4.2.0.14.

Consider the matrix $$A = \left[\begin{array}{rrr} -1 \amp 0 \amp 2 \\ -2 \amp -2 \amp -4 \\ 0 \amp 0 \amp -2 \\ \end{array}\right]$$ whose characteristic equation is
\begin{equation*} (-2-\lambda)^2(-1-\lambda) = 0\text{.} \end{equation*}

#### (a)

Identify the eigenvalues and their multiplicities.

#### (b)

For each eigenvalue $$\lambda\text{,}$$ find a basis of the eigenspace $$E_\lambda$$ and state its dimension.

#### (c)

Is there a basis of $$\real^3$$ consisting of eigenvectors of $$A\text{?}$$

### Activity4.2.0.15.

Now consider the matrix $$A = \left[\begin{array}{rrr} -5 \amp -2 \amp -6 \\ -2 \amp -2 \amp -4 \\ 2 \amp 1 \amp 2 \\ \end{array}\right]$$ whose characteristic equation is also
\begin{equation*} (-2-\lambda)^2(-1-\lambda) = 0\text{.} \end{equation*}

#### (a)

Identify the eigenvalues and their multiplicities.

#### (b)

For each eigenvalue $$\lambda\text{,}$$ find a basis of the eigenspace $$E_\lambda$$ and state its dimension.

#### (c)

Is there a basis of $$\real^3$$ consisting of eigenvectors of $$A\text{?}$$

### Activity4.2.0.16.

Consider the matrix $$A = \left[\begin{array}{rrr} -5 \amp -2 \amp -6 \\ 4 \amp 1 \amp 8 \\ 2 \amp 1 \amp 2 \\ \end{array}\right]$$ whose characteristic equation is
\begin{equation*} (-2-\lambda)(1-\lambda)(-1-\lambda) = 0\text{.} \end{equation*}

#### (a)

Identify the eigenvalues and their multiplicities.

#### (b)

For each eigenvalue $$\lambda\text{,}$$ find a basis of the eigenspace $$E_\lambda$$ and state its dimension.

#### (c)

Is there a basis of $$\real^3$$ consisting of eigenvectors of $$A\text{?}$$