compute eigenvalue and eigenvectors for the following matrix:


2 0 0
0 1 0
0 1 2


Given the matrix:

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

### Step 1: Finding the Eigenvalues
The eigenvalues of a matrix \( A \) are found by solving the characteristic equation:

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

where \( \lambda \) is the eigenvalue and \( I \) is the identity matrix. Let's compute:

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

Next, we compute the determinant:

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

This simplifies to:

\[
(2 - \lambda) \times \begin{vmatrix} 1 - \lambda & 0 \\ 1 & 2 - \lambda \end{vmatrix} = (2 - \lambda) \times ((1 - \lambda)(2 - \lambda) - 0) = (2 - \lambda)(\lambda^2 - 3\lambda + 2)
\]

Setting the determinant to zero:

\[
(2 - \lambda)(\lambda^2 - 3\lambda + 2) = 0
\]

The roots of this equation give the eigenvalues. Solving each factor:

1. \( 2 - \lambda = 0 \Rightarrow \lambda = 2 \)
2. \( \lambda^2 - 3\lambda + 2 = 0 \Rightarrow (\lambda - 1)(\lambda - 2) = 0 \)

Thus, the eigenvalues are:

\[
\lambda = 2, \quad \lambda = 1, \quad \lambda = 2
\]

### Step 2: Finding the Eigenvectors
For each eigenvalue \( \lambda \), solve \( (A - \lambda I) \mathbf{v} = 0 \) to find the corresponding eigenvectors.

1. **Eigenvalue \( \lambda = 2 \)**:

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

Solve \( \begin{bmatrix} 0 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \).

This gives the equations:

\[
\begin{cases} -y = 0 \\ y + 0 = 0 \end{cases} \Rightarrow y = 0
\]

So, \( x \) and \( z \) are free variables. A general solution is:

\[
\mathbf{v} = \begin{bmatrix} x \\ 0 \\ z \end{bmatrix} = x \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} + z \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}
\]

Two linearly independent eigenvectors are:

\[
\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \quad \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}
\]

2. **Eigenvalue \( \lambda = 1 \)**:

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

Solve \( \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \).

This gives the equations:

\[
\begin{cases} x = 0 \\ y + z = 0 \end{cases} \Rightarrow x = 0, \, y = -z
\]

So, a general solution is:

\[
\mathbf{v} = \begin{bmatrix} 0 \\ -z \\ z \end{bmatrix} = z \begin{bmatrix} 0 \\ -1 \\ 1 \end{bmatrix}
\]

An eigenvector corresponding to \( \lambda = 1 \) is:

\[
\begin{bmatrix} 0 \\ -1 \\ 1 \end{bmatrix}
\]

### Final Answer
- **Eigenvalues**: \( \lambda = 2, \, 1, \, 2 \)
- **Eigenvectors**:

  - For \( \lambda = 2 \): \( \begin{bmatrix} 1 \\ 0 \\ 0 \end{

Compute eigenvalue and eigenvectors for the following matrix:

2 0 0
0 1 0
0 1 2

Learn how to compute eigenvalues and eigenvectors for the 3x3 matrix [2 0 0; 0 1 0; 0 1 2]. This solution covers key concepts in linear algebra, including the characteristic equation and eigenvector determination, suitable for advanced high school or undergraduate students.

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