Title: eigenvalues_deck

Q: Created by: Dr. Pang Chung Yin Amy, Dr. Chiang Shun Ling, and Naw Sai.
Department of Mathematics,
Hong Kong Baptist University (HKBU) © 2025.
A: These decks may be freely shared and distributed among students and educators. We hope they help in your linear algebra journey!

Q: Consider the matrix $A = \begin{bmatrix} 3 & 2 & 3 & 2 \\ 0 & 3 & 1 & 2 \\ 1 & 3 & 6 & 4 \\ 1 & 2 & 3 & 4 \end{bmatrix}$. Find a basis for the eigenspace corresponding to the eigenvalue $\lambda = 2$.
A: The eigenspace corresponding to $\lambda = 2$ is the solution set to $(A - 2I)\mathbf{x} = \mathbf{0}$.
In augmented matrix form this is $\left[\begin{array}{cccc|c} 1 & 2 & 3 & 2 & 0 \\ 0 & 1 & 1 & 2 & 0 \\ 1 & 3 & 4 & 4 & 0 \\ 1 & 2 & 3 & 2 & 0 \end{array}\right]$.
Row reducing gives $\left[\begin{array}{cccc|c} 1 & 0 & 1 & -2 & 0 \\ 0 & 1 & 1 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right]$, so the solution set is $\left\{ s\begin{bmatrix} -1 \\ -1 \\ 1 \\ 0 \end{bmatrix} + t\begin{bmatrix} 2 \\ -2 \\ 0 \\ 1 \end{bmatrix} \;\middle|\; s, t \in \mathbb{R} \right\}$.
So a basis for the eigenspace is $\left\{ \begin{bmatrix} -1 \\ -1 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 2 \\ -2 \\ 0 \\ 1 \end{bmatrix} \right\}$.

Q: Consider the matrix $A = \begin{bmatrix} 3 & 2 & 2 & 3 \\ 0 & 3 & 2 & 1 \\ 1 & 3 & 6 & 4 \\ 1 & 2 & 2 & 5 \end{bmatrix}$. Find the eigenspace corresponding to the eigenvalue $\lambda = 2$.
A: The eigenspace corresponding to $\lambda = 2$ is the solution set to $(A - 2I)\mathbf{x} = \mathbf{0}$.
In augmented matrix form this is $\left[\begin{array}{cccc|c} 1 & 2 & 2 & 3 & 0 \\ 0 & 1 & 2 & 1 & 0 \\ 1 & 3 & 4 & 4 & 0 \\ 1 & 2 & 2 & 3 & 0 \end{array}\right]$.
Row reducing gives $\left[\begin{array}{cccc|c} 1 & 0 & -2 & 1 & 0 \\ 0 & 1 & 2 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right]$, so the solution set is $\left\{ s\begin{bmatrix} 2 \\ -2 \\ 1 \\ 0 \end{bmatrix} + t\begin{bmatrix} -1 \\ -1 \\ 0 \\ 1 \end{bmatrix} \;\middle|\; s, t \in \mathbb{R} \right\}$.

Q: Consider the matrix $A = \begin{bmatrix} 4 & 1 & 2 & 3 \\ 0 & 4 & 2 & 2 \\ 2 & 3 & 9 & 8 \\ 1 & 1 & 2 & 6 \end{bmatrix}$. Find the set of eigenvectors corresponding to the eigenvalue $\lambda = 3$.
A: The eigenspace corresponding to $\lambda = 3$ is the solution set to $(A - 3I)\mathbf{x} = \mathbf{0}$.
In augmented matrix form this is $\left[\begin{array}{cccc|c} 1 & 1 & 2 & 3 & 0 \\ 0 & 1 & 2 & 2 & 0 \\ 2 & 3 & 6 & 8 & 0 \\ 1 & 1 & 2 & 3 & 0 \end{array}\right]$.
Row reducing gives $\left[\begin{array}{cccc|c} 1 & 0 & 0 & 1 & 0 \\ 0 & 1 & 2 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right]$, so the solution set is $\left\{ s\begin{bmatrix} 0 \\ -2 \\ 1 \\ 0 \end{bmatrix} + t\begin{bmatrix} -1 \\ -2 \\ 0 \\ 1 \end{bmatrix} \;\middle|\; s, t \in \mathbb{R} \right\}$.
The set of eigenvectors is this set excluding the zero vector, so $\left\{ s\begin{bmatrix} 0 \\ -2 \\ 1 \\ 0 \end{bmatrix} + t\begin{bmatrix} -1 \\ -2 \\ 0 \\ 1 \end{bmatrix} \;\middle|\; (s, t) \neq (0, 0) \right\}$.

Q: Consider the matrix $A = \begin{bmatrix} 3 & 1 & 3 & 2 \\ 0 & 3 & 2 & 2 \\ 1 & 3 & 9 & 6 \\ 1 & 1 & 3 & 4 \end{bmatrix}$. Find a basis for the eigenspace corresponding to the eigenvalue $\lambda = 2$.
A: The eigenspace corresponding to $\lambda = 2$ is the solution set to $(A - 2I)\mathbf{x} = \mathbf{0}$.
In augmented matrix form this is $\left[\begin{array}{cccc|c} 1 & 1 & 3 & 2 & 0 \\ 0 & 1 & 2 & 2 & 0 \\ 1 & 3 & 7 & 6 & 0 \\ 1 & 1 & 3 & 2 & 0 \end{array}\right]$.
Row reducing gives $\left[\begin{array}{cccc|c} 1 & 0 & 1 & 0 & 0 \\ 0 & 1 & 2 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right]$, so the solution set is $\left\{ s\begin{bmatrix} -1 \\ -2 \\ 1 \\ 0 \end{bmatrix} + t\begin{bmatrix} 0 \\ -2 \\ 0 \\ 1 \end{bmatrix} \;\middle|\; s, t \in \mathbb{R} \right\}$.
So a basis for the eigenspace is $\left\{ \begin{bmatrix} -1 \\ -2 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ -2 \\ 0 \\ 1 \end{bmatrix} \right\}$.

Q: Consider the matrix $A = \begin{bmatrix} 4 & 2 & 3 & 1 \\ 0 & 4 & 2 & 2 \\ 1 & 3 & 8 & 3 \\ 1 & 2 & 3 & 4 \end{bmatrix}$. Find the eigenspace corresponding to the eigenvalue $\lambda = 3$.
A: The eigenspace corresponding to $\lambda = 3$ is the solution set to $(A - 3I)\mathbf{x} = \mathbf{0}$.
In augmented matrix form this is $\left[\begin{array}{cccc|c} 1 & 2 & 3 & 1 & 0 \\ 0 & 1 & 2 & 2 & 0 \\ 1 & 3 & 5 & 3 & 0 \\ 1 & 2 & 3 & 1 & 0 \end{array}\right]$.
Row reducing gives $\left[\begin{array}{cccc|c} 1 & 0 & -1 & -3 & 0 \\ 0 & 1 & 2 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right]$, so the solution set is $\left\{ s\begin{bmatrix} 1 \\ -2 \\ 1 \\ 0 \end{bmatrix} + t\begin{bmatrix} 3 \\ -2 \\ 0 \\ 1 \end{bmatrix} \;\middle|\; s, t \in \mathbb{R} \right\}$.

Q: Consider the matrix $A = \begin{bmatrix} 3 & 4 & -4 \\ -1 & -1 & 2 \\ 0 & 0 & 3 \end{bmatrix}$. Find the set of eigenvectors corresponding to the eigenvalue $\lambda = 1$.
A: The eigenspace corresponding to $\lambda = 1$ is the solution set to $(A - I)\mathbf{x} = \mathbf{0}$.
In augmented matrix form this is $\left[\begin{array}{ccc|c} 2 & 4 & -4 & 0 \\ -1 & -2 & 2 & 0 \\ 0 & 0 & 2 & 0 \end{array}\right]$.
Row reducing gives $\left[\begin{array}{ccc|c} 1 & 2 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]$, so the solution set is $\left\{ s\begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix} \;\middle|\; s \in \mathbb{R} \right\}$.
The set of eigenvectors is this set excluding the zero vector, so $\left\{ s\begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix} \;\middle|\; s \neq 0 \right\}$.

Q: Consider the matrix $A = \begin{bmatrix} 0 & -2 & -4 \\ 1 & 3 & 2 \\ 0 & 0 & 2 \end{bmatrix}$. Find a basis for the eigenspace corresponding to the eigenvalue $\lambda = 2$.
A: The eigenspace corresponding to $\lambda = 2$ is the solution set to $(A - 2I)\mathbf{x} = \mathbf{0}$.
In augmented matrix form this is $\left[\begin{array}{ccc|c} -2 & -2 & -4 & 0 \\ 1 & 1 & 2 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]$.
Row reducing gives $\left[\begin{array}{ccc|c} 1 & 1 & 2 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]$, so the solution set is $\left\{ s\begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix} + t\begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix} \;\middle|\; s, t \in \mathbb{R} \right\}$.
So a basis for the eigenspace is $\left\{ \begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix} \right\}$.

Q: Consider the matrix $A = \begin{bmatrix} -1 & -4 & 4 \\ 2 & 5 & -2 \\ 0 & 0 & -1 \end{bmatrix}$. Find the eigenspace corresponding to the eigenvalue $\lambda = 3$.
A: The eigenspace corresponding to $\lambda = 3$ is the solution set to $(A - 3I)\mathbf{x} = \mathbf{0}$.
In augmented matrix form this is $\left[\begin{array}{ccc|c} -4 & -4 & 4 & 0 \\ 2 & 2 & -2 & 0 \\ 0 & 0 & -4 & 0 \end{array}\right]$.
Row reducing gives $\left[\begin{array}{ccc|c} 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]$, so the solution set is $\left\{ s\begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix} \;\middle|\; s \in \mathbb{R} \right\}$.

Q: Consider the matrix $A = \begin{bmatrix} 2 & -1 & 1 \\ 0 & 2 & 0 \\ -1 & -1 & 0 \end{bmatrix}$. Find the set of eigenvectors corresponding to the eigenvalue $\lambda = 1$.
A: The eigenspace corresponding to $\lambda = 1$ is the solution set to $(A - I)\mathbf{x} = \mathbf{0}$.
In augmented matrix form this is $\left[\begin{array}{ccc|c} 1 & -1 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ -1 & -1 & -1 & 0 \end{array}\right]$.
Row reducing gives $\left[\begin{array}{ccc|c} 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]$, so the solution set is $\left\{ s\begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} \;\middle|\; s \in \mathbb{R} \right\}$.
The set of eigenvectors is this set excluding the zero vector, so $\left\{ s\begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} \;\middle|\; s \neq 0 \right\}$.

Q: Consider the matrix $A = \begin{bmatrix} 5 & -3 & 6 \\ 0 & 2 & 0 \\ -3 & 3 & -4 \end{bmatrix}$. Find a basis for the eigenspace corresponding to the eigenvalue $\lambda = 2$.
A: The eigenspace corresponding to $\lambda = 2$ is the solution set to $(A - 2I)\mathbf{x} = \mathbf{0}$.
In augmented matrix form this is $\left[\begin{array}{ccc|c} 3 & -3 & 6 & 0 \\ 0 & 0 & 0 & 0 \\ -3 & 3 & -6 & 0 \end{array}\right]$.
Row reducing gives $\left[\begin{array}{ccc|c} 1 & -1 & 2 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]$, so the solution set is $\left\{ s\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} + t\begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix} \;\middle|\; s, t \in \mathbb{R} \right\}$.
So a basis for the eigenspace is $\left\{ \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix} \right\}$.

Q: Consider the matrix $A = \begin{bmatrix} 3 & -2 & 1 \\ 0 & -1 & 0 \\ -2 & 4 & 0 \end{bmatrix}$. Find the eigenspace corresponding to the eigenvalue $\lambda = 2$.
A: The eigenspace corresponding to $\lambda = 2$ is the solution set to $(A - 2I)\mathbf{x} = \mathbf{0}$.
In augmented matrix form this is $\left[\begin{array}{ccc|c} 1 & -2 & 1 & 0 \\ 0 & -3 & 0 & 0 \\ -2 & 4 & -2 & 0 \end{array}\right]$.
Row reducing gives $\left[\begin{array}{ccc|c} 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]$, so the solution set is $\left\{ s\begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} \;\middle|\; s \in \mathbb{R} \right\}$.

Q: Consider the matrix $A = \begin{bmatrix} 4 & -2 & 4 \\ 0 & 1 & 0 \\ -1 & 1 & 0 \end{bmatrix}$. Find the set of eigenvectors corresponding to the eigenvalue $\lambda = 2$.
A: The eigenspace corresponding to $\lambda = 2$ is the solution set to $(A - 2I)\mathbf{x} = \mathbf{0}$.
In augmented matrix form this is $\left[\begin{array}{ccc|c} 2 & -2 & 4 & 0 \\ 0 & -1 & 0 & 0 \\ -1 & 1 & -2 & 0 \end{array}\right]$.
Row reducing gives $\left[\begin{array}{ccc|c} 1 & 0 & 2 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]$, so the solution set is $\left\{ s\begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix} \;\middle|\; s \in \mathbb{R} \right\}$.
The set of eigenvectors is this set excluding the zero vector, so $\left\{ s\begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix} \;\middle|\; s \neq 0 \right\}$.

Q: Consider the matrix $A = \begin{bmatrix} 5 & -4 & 8 \\ 0 & 1 & 0 \\ -4 & 4 & -7 \end{bmatrix}$. Find a basis for the eigenspace corresponding to the eigenvalue $\lambda = 1$.
A: The eigenspace corresponding to $\lambda = 1$ is the solution set to $(A - I)\mathbf{x} = \mathbf{0}$.
In augmented matrix form this is $\left[\begin{array}{ccc|c} 4 & -4 & 8 & 0 \\ 0 & 0 & 0 & 0 \\ -4 & 4 & -8 & 0 \end{array}\right]$.
Row reducing gives $\left[\begin{array}{ccc|c} 1 & -1 & 2 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]$, so the solution set is $\left\{ s\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} + t\begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix} \;\middle|\; s, t \in \mathbb{R} \right\}$.
So a basis for the eigenspace is $\left\{ \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix} \right\}$.

Q: Consider the matrix $A = \begin{bmatrix} 0 & -2 & -1 \\ 0 & -1 & 0 \\ 2 & 2 & 3 \end{bmatrix}$. Find the eigenspace corresponding to the eigenvalue $\lambda = 1$.
A: The eigenspace corresponding to $\lambda = 1$ is the solution set to $(A - I)\mathbf{x} = \mathbf{0}$.
In augmented matrix form this is $\left[\begin{array}{ccc|c} -1 & -2 & -1 & 0 \\ 0 & -2 & 0 & 0 \\ 2 & 2 & 2 & 0 \end{array}\right]$.
Row reducing gives $\left[\begin{array}{ccc|c} 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]$, so the solution set is $\left\{ s\begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} \;\middle|\; s \in \mathbb{R} \right\}$.

Q: Find the eigenvalues of the matrix $A = \begin{bmatrix} 3 & 4 & -4 \\ -1 & -1 & 2 \\ 0 & 0 & 3 \end{bmatrix}$.
A: The eigenvalues are the solutions to $\det(A - \lambda I) = 0$.
We have $A - \lambda I = \begin{bmatrix} 3 - \lambda & 4 & -4 \\ -1 & -1 - \lambda & 2 \\ 0 & 0 & 3 - \lambda \end{bmatrix}$, so $\det(A - \lambda I) = - \left(\lambda - 3\right) \left(\lambda - 1\right)^{2}$.
Setting this equal to zero and solving, the eigenvalues are $\lambda_{1} = 1, \lambda_{2} = 1, \lambda_{3} = 3$.

Q: Find the eigenvalues of the matrix $A = \begin{bmatrix} 0 & -2 & -4 \\ 1 & 3 & 2 \\ 0 & 0 & 2 \end{bmatrix}$.
A: The eigenvalues are the solutions to $\det(A - \lambda I) = 0$.
We have $A - \lambda I = \begin{bmatrix} -\lambda & -2 & -4 \\ 1 & 3 - \lambda & 2 \\ 0 & 0 & 2 - \lambda \end{bmatrix}$, so $\det(A - \lambda I) = - \left(\lambda - 2\right)^{2} \left(\lambda - 1\right)$.
Setting this equal to zero and solving, the eigenvalues are $\lambda_{1} = 1, \lambda_{2} = 2, \lambda_{3} = 2$.

Q: Find the eigenvalues of the matrix $A = \begin{bmatrix} -1 & -4 & 4 \\ 2 & 5 & -2 \\ 0 & 0 & -1 \end{bmatrix}$.
A: The eigenvalues are the solutions to $\det(A - \lambda I) = 0$.
We have $A - \lambda I = \begin{bmatrix} -1 - \lambda & -4 & 4 \\ 2 & 5 - \lambda & -2 \\ 0 & 0 & -1 - \lambda \end{bmatrix}$, so $\det(A - \lambda I) = - \left(\lambda - 3\right) \left(\lambda - 1\right) \left(\lambda + 1\right)$.
Setting this equal to zero and solving, the eigenvalues are $\lambda_{1} = -1, \lambda_{2} = 1, \lambda_{3} = 3$.

Q: Find the eigenvalues of the matrix $A = \begin{bmatrix} 2 & -1 & 1 \\ 0 & 2 & 0 \\ -1 & -1 & 0 \end{bmatrix}$.
A: The eigenvalues are the solutions to $\det(A - \lambda I) = 0$.
We have $A - \lambda I = \begin{bmatrix} 2 - \lambda & -1 & 1 \\ 0 & 2 - \lambda & 0 \\ -1 & -1 & -\lambda \end{bmatrix}$, so $\det(A - \lambda I) = - \left(\lambda - 2\right) \left(\lambda - 1\right)^{2}$.
Setting this equal to zero and solving, the eigenvalues are $\lambda_{1} = 1, \lambda_{2} = 1, \lambda_{3} = 2$.

Q: Find the eigenvalues of the matrix $A = \begin{bmatrix} 5 & -3 & 6 \\ 0 & 2 & 0 \\ -3 & 3 & -4 \end{bmatrix}$.
A: The eigenvalues are the solutions to $\det(A - \lambda I) = 0$.
We have $A - \lambda I = \begin{bmatrix} 5 - \lambda & -3 & 6 \\ 0 & 2 - \lambda & 0 \\ -3 & 3 & -4 - \lambda \end{bmatrix}$, so $\det(A - \lambda I) = - \left(\lambda - 2\right)^{2} \left(\lambda + 1\right)$.
Setting this equal to zero and solving, the eigenvalues are $\lambda_{1} = -1, \lambda_{2} = 2, \lambda_{3} = 2$.

Q: Find the eigenvalues of the matrix $A = \begin{bmatrix} 3 & -2 & 1 \\ 0 & -1 & 0 \\ -2 & 4 & 0 \end{bmatrix}$.
A: The eigenvalues are the solutions to $\det(A - \lambda I) = 0$.
We have $A - \lambda I = \begin{bmatrix} 3 - \lambda & -2 & 1 \\ 0 & -1 - \lambda & 0 \\ -2 & 4 & -\lambda \end{bmatrix}$, so $\det(A - \lambda I) = - \left(\lambda - 2\right) \left(\lambda - 1\right) \left(\lambda + 1\right)$.
Setting this equal to zero and solving, the eigenvalues are $\lambda_{1} = -1, \lambda_{2} = 1, \lambda_{3} = 2$.

Q: Find the eigenvalues of the matrix $A = \begin{bmatrix} 4 & -2 & 4 \\ 0 & 1 & 0 \\ -1 & 1 & 0 \end{bmatrix}$.
A: The eigenvalues are the solutions to $\det(A - \lambda I) = 0$.
We have $A - \lambda I = \begin{bmatrix} 4 - \lambda & -2 & 4 \\ 0 & 1 - \lambda & 0 \\ -1 & 1 & -\lambda \end{bmatrix}$, so $\det(A - \lambda I) = - \left(\lambda - 2\right)^{2} \left(\lambda - 1\right)$.
Setting this equal to zero and solving, the eigenvalues are $\lambda_{1} = 1, \lambda_{2} = 2, \lambda_{3} = 2$.

Q: Find the eigenvalues of the matrix $A = \begin{bmatrix} 5 & -4 & 8 \\ 0 & 1 & 0 \\ -4 & 4 & -7 \end{bmatrix}$.
A: The eigenvalues are the solutions to $\det(A - \lambda I) = 0$.
We have $A - \lambda I = \begin{bmatrix} 5 - \lambda & -4 & 8 \\ 0 & 1 - \lambda & 0 \\ -4 & 4 & -7 - \lambda \end{bmatrix}$, so $\det(A - \lambda I) = - \left(\lambda - 1\right)^{2} \left(\lambda + 3\right)$.
Setting this equal to zero and solving, the eigenvalues are $\lambda_{1} = -3, \lambda_{2} = 1, \lambda_{3} = 1$.

Q: Find the eigenvalues of the matrix $A = \begin{bmatrix} 0 & -2 & -1 \\ 0 & -1 & 0 \\ 2 & 2 & 3 \end{bmatrix}$.
A: The eigenvalues are the solutions to $\det(A - \lambda I) = 0$.
We have $A - \lambda I = \begin{bmatrix} -\lambda & -2 & -1 \\ 0 & -1 - \lambda & 0 \\ 2 & 2 & 3 - \lambda \end{bmatrix}$, so $\det(A - \lambda I) = - \left(\lambda - 2\right) \left(\lambda - 1\right) \left(\lambda + 1\right)$.
Setting this equal to zero and solving, the eigenvalues are $\lambda_{1} = -1, \lambda_{2} = 1, \lambda_{3} = 2$.