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{