Math Problem Statement
-2 1 1 -11 4 5 -1 1 0 Find eigen values and vectors and check the diagonalisable
Solution
Let's find the eigenvalues, eigenvectors, and determine if the matrix is diagonalizable for the given matrix :
-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? --- 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.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Eigenvalues
Eigenvectors
Diagonalizability
Formulas
Characteristic equation
Determinant calculation
Theorems
Eigenvalue theorem
Diagonalizability theorem
Suitable Grade Level
Advanced Undergraduate
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