Math Problem Statement
Find the eigenvalues and corresponding vectors of the following matrix, and diagonalise if possible. Find the matrix which diagonolise it. A= 1 -1 2 2 -2 4 3 -3 6
Solution
We are given a matrix and we need to find its eigenvalues, eigenvectors, and diagonalize it if possible.
The matrix is:
1 & -1 & 2 \\ 2 & -2 & 4 \\ 3 & -3 & 6 \end{pmatrix}$$ ### Step 1: Find the eigenvalues To find the eigenvalues of matrix $$A$$, we solve the characteristic equation: $$\text{det}(A - \lambda I) = 0$$ where $$\lambda$$ is the eigenvalue and $$I$$ is the identity matrix. The characteristic matrix is: $$A - \lambda I = \begin{pmatrix} 1 - \lambda & -1 & 2 \\ 2 & -2 - \lambda & 4 \\ 3 & -3 & 6 - \lambda \end{pmatrix}$$ We now compute the determinant of $$A - \lambda I$$. ### Step 2: Solve for eigenvectors Once we have the eigenvalues, we solve for the eigenvectors by solving: $$(A - \lambda I) v = 0$$ for each eigenvalue $$\lambda$$, where $$v$$ is the corresponding eigenvector. ### Step 3: Diagonalization If the matrix has enough independent eigenvectors, we can diagonalize it. The matrix that diagonalizes $$A$$ is: $$P = \begin{pmatrix} v_1 & v_2 & v_3 \end{pmatrix}$$ where $$v_1, v_2, v_3$$ are the eigenvectors corresponding to the eigenvalues, and the diagonal matrix $$D$$ is: $$D = \begin{pmatrix} \lambda_1 & 0 & 0 \\ 0 & \lambda_2 & 0 \\ 0 & 0 & \lambda_3 \end{pmatrix}$$ Let me calculate the eigenvalues and eigenvectors for you.### Eigenvalues: The eigenvalues of matrix $$A$$ are approximately: $$\lambda_1 = 5, \quad \lambda_2 \approx 0, \quad \lambda_3 \approx 0$$ ### Eigenvectors: The corresponding eigenvectors are: $$v_1 = \begin{pmatrix} 0.2673 \\ 0.5345 \\ 0.8018 \end{pmatrix}, \quad v_2 = \begin{pmatrix} 0.3537 \\ -0.7538 \\ -0.5538 \end{pmatrix}, \quad v_3 = \begin{pmatrix} -0.4987 \\ -0.8489 \\ -0.1751 \end{pmatrix}$$ ### Diagonalization: Since we have enough independent eigenvectors, the matrix $$A$$ can be diagonalized. The matrix $$P$$, which diagonalizes $$A$$, is formed by the eigenvectors: $$P = \begin{pmatrix} 0.2673 & 0.3537 & -0.4987 \\ 0.5345 & -0.7538 & -0.8489 \\ 0.8018 & -0.5538 & -0.1751 \end{pmatrix}$$ The diagonal matrix $$D$$ is: $$D = \begin{pmatrix} 5 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$ ### Final Result: The matrix $$A$$ can be diagonalized as: $$A = P D P^{-1}$$ Let me know if you'd like further clarification! ### Follow-up Questions: 1. What does it mean for a matrix to be diagonalizable? 2. How do we interpret eigenvalues and eigenvectors geometrically? 3. Can all matrices be diagonalized? 4. How do we compute the inverse of a matrix in this context? 5. What happens when eigenvalues are repeated? ### Tip: Diagonalization is particularly useful for simplifying matrix powers, making complex calculations much easier.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Eigenvalues
Eigenvectors
Matrix Diagonalization
Formulas
det(A - λI) = 0
(A - λI)v = 0
Theorems
Eigenvalue theorem
Diagonalization theorem
Suitable Grade Level
Undergraduate level