Find the eigen values and eigen vectors of the following matrices: [
1 −2 2
−2 1 2
−2 0 3]

Let's find the eigenvalues and eigenvectors of the given matrix:

\[
A = \begin{bmatrix} 1 & -2 & 2 \\ -2 & 1 & 2 \\ -2 & 0 & 3 \end{bmatrix}
\]

### Step 1: Find the Eigenvalues
The eigenvalues of matrix \( A \) are the roots of the characteristic equation:

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

Where \( \lambda \) is the eigenvalue and \( I \) is the identity matrix. The matrix \( A - \lambda I \) is:

\[
A - \lambda I = \begin{bmatrix} 
1 - \lambda & -2 & 2 \\ 
-2 & 1 - \lambda & 2 \\ 
-2 & 0 & 3 - \lambda 
\end{bmatrix}
\]

The determinant of this matrix is:

\[
\text{det}(A - \lambda I) = \begin{vmatrix} 
1 - \lambda & -2 & 2 \\ 
-2 & 1 - \lambda & 2 \\ 
-2 & 0 & 3 - \lambda 
\end{vmatrix}
\]

Now, let's compute this determinant to find the characteristic polynomial:

\[
\text{det}(A - \lambda I) = (1 - \lambda) \begin{vmatrix} 
1 - \lambda & 2 \\ 
0 & 3 - \lambda 
\end{vmatrix} 
- (-2) \begin{vmatrix} 
-2 & 2 \\ 
-2 & 3 - \lambda 
\end{vmatrix}
+ 2 \begin{vmatrix} 
-2 & 1 - \lambda \\ 
-2 & 2 
\end{vmatrix}
\]

Let's evaluate each of these 2x2 determinants and simplify:

\[
\text{det}(A - \lambda I) = (1 - \lambda)[(1 - \lambda)(3 - \lambda) - (2)(0)] - (-2)[(-2)(3 - \lambda) - 2(-2)] + 2[(-2)(2) - (-2)(1 - \lambda)]
\]

Let's simplify further to get the characteristic polynomial, and then solve for \( \lambda \).

### Step 2: Solve for Eigenvalues
We solve the characteristic polynomial equation to find the eigenvalues \( \lambda_1, \lambda_2, \lambda_3 \).

### Step 3: Find the Eigenvectors
For each eigenvalue \( \lambda_i \), we solve the system:

\[
(A - \lambda_i I)\mathbf{v}_i = \mathbf{0}
\]

Where \( \mathbf{v}_i \) is the eigenvector corresponding to eigenvalue \( \lambda_i \).

Let's calculate these steps.

#### 1. Calculation of Determinant:

First, I'll compute the determinant and solve for eigenvalues.The characteristic polynomial is:

\[
-\lambda^3 + 5\lambda^2 - 7\lambda + 3 = 0
\]

The eigenvalues of the matrix are:

\[
\lambda_1 = 1, \quad \lambda_2 = 3 \quad \text{(with multiplicity 2)}
\]

### Step 4: Find the Eigenvectors
For each eigenvalue, we find the corresponding eigenvectors by solving the system \((A - \lambda_i I) \mathbf{v}_i = 0\).

Let's compute the eigenvectors for \( \lambda_1 = 1 \) and \( \lambda_2 = 3 \).The eigenvectors corresponding to each eigenvalue are:

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

- **For \( \lambda_2 = 3 \)** (with multiplicity 2):
  The eigenvector is:
  \[
  \mathbf{v}_2 = \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}
  \]

### Summary
- **Eigenvalues**: \( \lambda_1 = 1 \), \( \lambda_2 = 3 \) (with multiplicity 2)
- **Eigenvectors**:
  - For \( \lambda_1 = 1 \): \( \mathbf{v}_1 = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \)
  - For \( \lambda_2 = 3 \): \( \mathbf{v}_2 = \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} \)

Would you like more details, or do you have any questions?

### Related Questions
1. How do you find the characteristic polynomial of a matrix?
2. What are the properties of eigenvalues and eigenvectors in symmetric matrices?
3. How do multiplicities of eigenvalues affect the eigenvectors?
4. Can a matrix have complex eigenvalues? Under what conditions?
5. What is the geometric interpretation of eigenvalues and eigenvectors?
6. How can eigenvectors be used to diagonalize a matrix?
7. What are the applications of eigenvalues and eigenvectors in real-world problems?
8. How does the determinant relate to the eigenvalues of a matrix?

### Tip
When dealing with repeated eigenvalues, the number of independent eigenvectors may be less than the multiplicity of the eigenvalue, which requires further analysis for diagonalization.

Learn how to find eigenvalues and eigenvectors of the matrix [1 -2 2; -2 1 2; -2 0 3]. Detailed step-by-step solution provided including characteristic polynomial calculation. Suitable for advanced undergraduate students studying linear algebra.

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