Find the eigenvalues of the matrix 0 1 0, 0 2 1, 0 2 0 3.

We are tasked with finding the eigenvalues of the given \(3 \times 3\) matrix. The matrix is:

\[
A = \begin{pmatrix}
0 & 1 & 0 \\
0 & 2 & 1 \\
0 & 2 & 0 \\
\end{pmatrix}
\]

### Step 1: Eigenvalue equation
To find the eigenvalues, we need to solve the **characteristic equation**, which is given by:

\[
\det(A - \lambda I) = 0
\]

Where \(A\) is the matrix, \(\lambda\) is the eigenvalue, and \(I\) is the identity matrix.

Let's compute \(A - \lambda I\), where \(\lambda\) is a scalar:

\[
A - \lambda I = \begin{pmatrix}
0 - \lambda & 1 & 0 \\
0 & 2 - \lambda & 1 \\
0 & 2 & 0 - \lambda \\
\end{pmatrix} = \begin{pmatrix}
-\lambda & 1 & 0 \\
0 & 2 - \lambda & 1 \\
0 & 2 & -\lambda \\
\end{pmatrix}
\]

### Step 2: Determinant of \(A - \lambda I\)

Now, we need to compute the determinant of this matrix:

\[
\det(A - \lambda I) = \det \begin{pmatrix}
-\lambda & 1 & 0 \\
0 & 2 - \lambda & 1 \\
0 & 2 & -\lambda
\end{pmatrix}
\]

We expand the determinant along the first row (since it has zeros that simplify the calculation):

\[
\det(A - \lambda I) = (-\lambda) \cdot \det \begin{pmatrix} 2 - \lambda & 1 \\ 2 & -\lambda \end{pmatrix}
\]

Now, we compute the determinant of the \(2 \times 2\) matrix:

\[
\det \begin{pmatrix} 2 - \lambda & 1 \\ 2 & -\lambda \end{pmatrix} = (2 - \lambda)(-\lambda) - (1)(2) = -\lambda(2 - \lambda) - 2
\]

Simplify:

\[
-\lambda(2 - \lambda) - 2 = -2\lambda + \lambda^2 - 2
\]

Thus, the characteristic equation becomes:

\[
\det(A - \lambda I) = -\lambda \left( \lambda^2 - 2\lambda - 2 \right)
\]

### Step 3: Solve the characteristic equation

The characteristic equation is:

\[
\lambda \left( \lambda^2 - 2\lambda - 2 \right) = 0
\]

This gives us two equations to solve:
1. \( \lambda = 0 \)
2. \( \lambda^2 - 2\lambda - 2 = 0 \)

The second equation is a quadratic equation. We solve it using the quadratic formula:

\[
\lambda = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-2)}}{2(1)} = \frac{2 \pm \sqrt{4 + 8}}{2} = \frac{2 \pm \sqrt{12}}{2} = \frac{2 \pm 2\sqrt{3}}{2}
\]

Simplify:

\[
\lambda = 1 \pm \sqrt{3}
\]

### Step 4: Eigenvalues

Thus, the eigenvalues of the matrix are:

\[
\lambda_1 = 0, \quad \lambda_2 = 1 + \sqrt{3}, \quad \lambda_3 = 1 - \sqrt{3}
\]

Would you like to go through the steps in more detail or have any questions about the process?

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Here are five related questions for further exploration:

1. How do you find the eigenvectors corresponding to these eigenvalues?
2. What is the geometric interpretation of eigenvalues and eigenvectors?
3. How do the eigenvalues relate to the trace and determinant of the matrix?
4. Can a matrix with all real entries have complex eigenvalues?
5. What are the applications of eigenvalues in physics and engineering?

**Tip**: Always check the characteristic polynomial for errors by re-expanding to ensure accuracy in solving eigenvalue problems.

Learn how to find the eigenvalues of a 3x3 matrix using the characteristic equation. This step-by-step solution involves calculating the determinant and solving a quadratic equation, suitable for advanced high school or undergraduate students studying linear algebra.

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