diagonalise this matrix : [[1,2,3],[3,4,5],[2,32,2]]

To diagonalize a matrix, we need to find its eigenvalues and eigenvectors. A matrix is diagonalizable if there are enough linearly independent eigenvectors to form a basis for the space, meaning we can write it as \( A = PDP^{-1} \), where:
- \( A \) is the original matrix,
- \( D \) is a diagonal matrix (with eigenvalues on the diagonal),
- \( P \) is the matrix of eigenvectors.

### Given matrix \( A \):
\[
A = \begin{bmatrix} 
1 & 2 & 3 \\
3 & 4 & 5 \\
2 & 32 & 2 
\end{bmatrix}
\]

#### Steps to diagonalize:
1. **Find the eigenvalues of \( A \)**: 
   The eigenvalues are found by solving the characteristic equation \( \det(A - \lambda I) = 0 \), where \( \lambda \) represents the eigenvalues and \( I \) is the identity matrix.

2. **Find the eigenvectors corresponding to each eigenvalue**: 
   For each eigenvalue \( \lambda \), solve the system \( (A - \lambda I)v = 0 \) to find the corresponding eigenvector \( v \).

3. **Form the diagonal matrix \( D \)**: 
   The diagonal elements of \( D \) are the eigenvalues of \( A \).

4. **Form the matrix \( P \)**: 
   The columns of \( P \) are the eigenvectors corresponding to each eigenvalue.

Let me calculate the eigenvalues and eigenvectors for you to complete the diagonalization process.### Diagonalization of the matrix:

1. **Eigenvalues** of the matrix \( A \):
   \[
   \lambda_1 = 16.8186, \quad \lambda_2 = -0.7903, \quad \lambda_3 = -9.0283
   \]

2. **Eigenvectors** corresponding to each eigenvalue:

   - Eigenvector corresponding to \( \lambda_1 = 16.8186 \):
     \[
     v_1 = \begin{bmatrix} -0.2193 \\ -0.3987 \\ -0.8905 \end{bmatrix}
     \]

   - Eigenvector corresponding to \( \lambda_2 = -0.7903 \):
     \[
     v_2 = \begin{bmatrix} -0.8615 \\ 0.0096 \\ 0.5077 \end{bmatrix}
     \]

   - Eigenvector corresponding to \( \lambda_3 = -9.0283 \):
     \[
     v_3 = \begin{bmatrix} -0.2163 \\ -0.3060 \\ 0.9271 \end{bmatrix}
     \]

3. **Diagonal matrix \( D \)**:
   \[
   D = \begin{bmatrix}
   16.8186 & 0 & 0 \\
   0 & -0.7903 & 0 \\
   0 & 0 & -9.0283
   \end{bmatrix}
   \]

4. **Matrix \( P \)** (with eigenvectors as columns):
   \[
   P = \begin{bmatrix}
   -0.2193 & -0.8615 & -0.2163 \\
   -0.3987 & 0.0096 & -0.3060 \\
   -0.8905 & 0.5077 & 0.9271
   \end{bmatrix}
   \]

Thus, the matrix \( A \) is diagonalizable as \( A = PDP^{-1} \).

Would you like further details or have any questions?

---

**Related questions:**
1. How is the characteristic equation derived for matrices?
2. What is the condition for a matrix to be diagonalizable?
3. Can all matrices be diagonalized? Why or why not?
4. How are eigenvalues and eigenvectors used in physical applications?
5. What are some practical applications of diagonalization?

**Tip:** Eigenvalues and eigenvectors provide insight into the behavior of linear transformations, especially in simplifying matrix operations.

Learn how to diagonalize a 3x3 matrix [[1,2,3],[3,4,5],[2,32,2]] by finding its eigenvalues and eigenvectors. This solution covers key linear algebra concepts like the characteristic equation, diagonalization, and the use of eigenvalues. Suitable for undergraduate-level linear algebra students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation