The matrix provided is:

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

To find the eigenvalues and eigenvectors, we follow these steps:

1. Find Eigenvalues:

The eigenvalues $\lambda$ are found by solving the characteristic equation:

$\det(A - \lambda I) = 0$

Where $I$ is the identity matrix and $\lambda$ is the eigenvalue.

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

Now, find the determinant:

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

Since it's an upper triangular matrix, the determinant is simply the product of the diagonal elements:

$\det(A - \lambda I) = (2-\lambda)(2-\lambda)(2-\lambda) = (2-\lambda)^3$

Therefore, the eigenvalues are:

$\lambda_1 = \lambda_2 = \lambda_3 = 2$

2. Find Eigenvectors:

Now that we know the eigenvalue $\lambda = 2$, we can find the eigenvectors by solving:

$(A - 2I)v = 0$

This simplifies to solving the system:

$\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$

From the first row: $x_2 = 0$.

From the second row: $x_3 = 0$.

Thus, the eigenvectors take the form:

$v = \begin{pmatrix} x_1 \\ 0 \\ 0 \end{pmatrix}$

where $x_1$ can be any non-zero value. Hence, the eigenvectors corresponding to the eigenvalue $\lambda = 2$ are of the form:

$v = c \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$

where $c$ is any constant.

Would you like more detailed steps on the process? Here are 5 related questions to consider:

1. What is the geometric multiplicity of the eigenvalue $\lambda = 2$?
2. How do the properties of upper triangular matrices simplify finding eigenvalues?
3. Can a matrix with repeated eigenvalues still be diagonalizable?
4. How do you find a basis for the eigenspace corresponding to a specific eigenvalue?
5. What is the relationship between eigenvectors and the diagonalization of a matrix?

Tip: For upper triangular matrices, the eigenvalues are always the entries on the diagonal.