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To solve this problem, we need to find both the eigenvalues and the eigenvectors of the given matrix \( A \). The matrix is:

\[
A = \begin{pmatrix}
0 & 2 & 2 \\
2 & 2 & -4 \\
0 & 1 & 1
\end{pmatrix}
\]

### Steps to find eigenvalues and eigenvectors:

1. **Find the characteristic equation**: The eigenvalues \( \lambda \) are the roots of the characteristic equation, which is derived from the matrix determinant:
   \[
   \text{det}(A - \lambda I) = 0
   \]
   where \( I \) is the identity matrix of the same dimension as \( A \).

2. **Solve for \( \lambda \)**: The characteristic equation is a cubic equation in \( \lambda \), and solving it will give us the eigenvalues.

3. **Find eigenvectors**: For each eigenvalue \( \lambda \), substitute it into the equation \( (A - \lambda I) \mathbf{v} = 0 \) to find the corresponding eigenvector \( \mathbf{v} \).

I'll start by calculating the determinant of \( A - \lambda I \).

\[
A - \lambda I = \begin{pmatrix}
0 - \lambda & 2 & 2 \\
2 & 2 - \lambda & -4 \\
0 & 1 & 1 - \lambda
\end{pmatrix}
= \begin{pmatrix}
-\lambda & 2 & 2 \\
2 & 2 - \lambda & -4 \\
0 & 1 & 1 - \lambda
\end{pmatrix}
\]

Now, calculating the determinant of this 3x3 matrix:

\[
\text{det}(A - \lambda I) = \text{det} \begin{pmatrix}
-\lambda & 2 & 2 \\
2 & 2 - \lambda & -4 \\
0 & 1 & 1 - \lambda
\end{pmatrix}
\]

\[
= -\lambda \left( (2 - \lambda)(1 - \lambda) - (-4) \right) - 2 \left( 2(1 - \lambda) - 0 \right)
\]

This gives a cubic polynomial. I will solve this and then proceed with eigenvectors based on the eigenvalues found.

Let me complete this calculation.The characteristic polynomial of the matrix is:

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

This can be factored as:

\[
\lambda (\lambda - 1)(\lambda - 2) = 0
\]

Thus, the eigenvalues are:

\[
\lambda_1 = 0, \quad \lambda_2 = 1, \quad \lambda_3 = 2
\]

### Now, let's find the eigenvectors corresponding to each eigenvalue.

1. **For \( \lambda_1 = 0 \):**
   Solve \( A \mathbf{v} = 0 \mathbf{v} \).

2. **For \( \lambda_2 = 1 \):**
   Solve \( (A - I) \mathbf{v} = 0 \).

3. **For \( \lambda_3 = 2 \):**
   Solve \( (A - 2I) \mathbf{v} = 0 \).

I'll proceed with solving for the eigenvectors now.The eigenvectors corresponding to the eigenvalues are as follows:

1. **For \( \lambda_1 = 0 \):** The eigenvector is:
   \[
   \mathbf{v}_1 = \begin{pmatrix} 3 \\ -1 \\ 1 \end{pmatrix}
   \]

2. **For \( \lambda_2 = 1 \):** The eigenvector is:
   \[
   \mathbf{v}_2 = \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix}
   \]

3. **For \( \lambda_3 = 2 \):** The eigenvector is:
   \[
   \mathbf{v}_3 = \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}
   \]

In summary, the eigenvalues are \( \lambda_1 = 0 \), \( \lambda_2 = 1 \), \( \lambda_3 = 2 \), and the corresponding eigenvectors are as shown above.

Would you like further details or any clarifications on the solution? 

Here are five related questions to explore:
1. How would the eigenvalues change if one element in the matrix is altered?
2. How do eigenvalues help in determining matrix stability?
3. Can eigenvectors be scaled, and how does it affect their properties?
4. What does it mean geometrically when an eigenvalue is zero?
5. How are eigenvalues related to the determinant of a matrix?

**Tip:** Eigenvalues and eigenvectors play a crucial role in simplifying matrix transformations like diagonalization and in understanding systems of linear equations.

Find all eigenvalues and corresponding eigenvectors of the matrix A = [[0, 2, 2], [2, 2, -4], [0, 1, 1]].

Find the eigenvalues and eigenvectors of the matrix A = [[0, 2, 2], [2, 2, -4], [0, 1, 1]]. This solution includes the steps to calculate the characteristic equation and solve for eigenvalues and eigenvectors. It covers linear algebra concepts such as matrix operations and determinants. Suitable for university-level linear algebra students.

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