-2 1 1
-11 4 5
-1 1 0
Find eigen values and vectors and check the diagonalisable

Let's find the eigenvalues, eigenvectors, and determine if the matrix is diagonalizable for the given \(3 \times 3\) matrix \(A\):

\[
A = \begin{pmatrix}
-2 & 1 & 1 \\
-11 & 4 & 5 \\
-1 & 1 & 0
\end{pmatrix}
\]

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

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

Where \(I\) is the identity matrix of the same dimension as \(A\). Let's calculate the determinant of \(A - \lambda I\):

\[
A - \lambda I = \begin{pmatrix}
-2 - \lambda & 1 & 1 \\
-11 & 4 - \lambda & 5 \\
-1 & 1 & -\lambda
\end{pmatrix}
\]

The determinant is calculated as:

\[
\text{det}(A - \lambda I) = (-2 - \lambda) \begin{vmatrix} 4 - \lambda & 5 \\ 1 & -\lambda \end{vmatrix} - 1 \begin{vmatrix} -11 & 5 \\ -1 & -\lambda \end{vmatrix} + 1 \begin{vmatrix} -11 & 4 - \lambda \\ -1 & 1 \end{vmatrix}
\]

Now let's calculate this determinant to find the characteristic polynomial.The characteristic polynomial of the matrix \( A \) is:

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

To find the eigenvalues, we need to solve the characteristic equation:

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

Let's solve this cubic equation for \(\lambda\).The eigenvalues of the matrix \( A \) are:

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

### Step 2: Finding the Eigenvectors
For each eigenvalue \(\lambda\), we find the corresponding eigenvector \(v\) by solving the equation:

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

Let's find the eigenvectors for each eigenvalue.The eigenvectors corresponding to the eigenvalues are:

- For \(\lambda_1 = -1\), the eigenvector is \( \begin{pmatrix} 0 \\ -1 \\ 1 \end{pmatrix} \).
- For \(\lambda_2 = 1\), the eigenvector is \( \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} \).
- For \(\lambda_3 = 2\), the eigenvector is \( \begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix} \).

### Step 3: Checking Diagonalizability
A matrix is diagonalizable if there are \(n\) linearly independent eigenvectors for an \(n \times n\) matrix. Here, the matrix \(A\) is \(3 \times 3\), and we have three distinct eigenvectors, which indicates that the matrix is diagonalizable.

### Summary
- **Eigenvalues**: \(-1\), \(1\), \(2\)
- **Eigenvectors**:
  - \( \lambda_1 = -1 \), eigenvector: \( \begin{pmatrix} 0 \\ -1 \\ 1 \end{pmatrix} \)
  - \( \lambda_2 = 1 \), eigenvector: \( \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} \)
  - \( \lambda_3 = 2 \), eigenvector: \( \begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix} \)

- **Diagonalizability**: The matrix \(A\) is diagonalizable.

Would you like further details or explanations on any step?

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Here are some related questions:
1. What does it mean for a matrix to be diagonalizable?
2. How do you find the null space of a matrix?
3. Can a matrix with repeated eigenvalues be diagonalizable?
4. What are the practical applications of diagonalizing a matrix?
5. How can you determine the algebraic and geometric multiplicities of eigenvalues?
6. Why is the determinant important in finding eigenvalues?
7. What is the significance of distinct eigenvalues?
8. How do you diagonalize a matrix after finding eigenvalues and eigenvectors?

**Tip:** Always verify the linear independence of eigenvectors when determining if a matrix is diagonalizable.

Learn how to find eigenvalues, eigenvectors, and check diagonalizability of a 3x3 matrix with a detailed step-by-step guide. This problem covers key concepts in linear algebra, including characteristic equations and determinant calculations. Suitable for advanced undergraduate students and enthusiasts of linear algebra.

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