Math Problem Statement
L ̊at A = 1 1 1 1 1 1 1 1 1 . Best ̈am alla egenv ̈arden och motsvarande egenrum till A. Finns det en bas f ̈or R3 som best ̊ar av egenvektorer till A? Best ̈am, om m ̈ojligt en matris P och en diagonal matris D s ̊a att D = P −1AP
Solution
We are given the matrix :
1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}$$ We need to: 1. Determine the eigenvalues and corresponding eigenspaces of $$A$$. 2. Check if there is a basis for $$\mathbb{R}^3$$ that consists of eigenvectors of $$A$$. 3. If possible, find a matrix $$P$$ and a diagonal matrix $$D$$ such that $$D = P^{-1} A P$$. ### Step 1: Finding the Eigenvalues To find the eigenvalues, we solve the characteristic equation: $$\det(A - \lambda I) = 0$$ Where $$I$$ is the identity matrix and $$\lambda$$ is the eigenvalue. The matrix $$A - \lambda I$$ is: $$A - \lambda I = \begin{bmatrix} 1 - \lambda & 1 & 1 \\ 1 & 1 - \lambda & 1 \\ 1 & 1 & 1 - \lambda \end{bmatrix}$$ The determinant of this matrix is: $$\det(A - \lambda I) = \det\begin{bmatrix} 1 - \lambda & 1 & 1 \\ 1 & 1 - \lambda & 1 \\ 1 & 1 & 1 - \lambda \end{bmatrix}$$ Expanding this determinant, we get: $$(1 - \lambda) \left( (1 - \lambda)^2 - 1 \right) - 1 \left( (1 - \lambda) - 1 \right) + 1 \left( (1 - \lambda) - 1 \right)$$ Simplifying this expression gives: $$(1 - \lambda)\left((1 - \lambda)^2 - 1 \right) = (1 - \lambda) \left( \lambda^2 - 2 \lambda \right) = \lambda (\lambda^2 - 2 \lambda)$$ This simplifies to: $$\lambda(\lambda - 3) = 0$$ Thus, the eigenvalues are $$\lambda = 0$$ and $$\lambda = 3$$. ### Step 2: Finding the Eigenspaces #### Eigenvalue $$\lambda = 3$$: We solve $$(A - 3I) \mathbf{v} = 0$$ for the eigenvectors: $$A - 3I = \begin{bmatrix} -2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2 \end{bmatrix}$$ Solving this system, we find that the eigenspace corresponding to $$\lambda = 3$$ is spanned by the vector: $$\mathbf{v}_1 = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$$ #### Eigenvalue $$\lambda = 0$$: We solve $$A \mathbf{v} = 0$$ for the eigenvectors. The matrix $$A$$ becomes: $$A - 0I = A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}$$ The null space of this matrix corresponds to the eigenspace for $$\lambda = 0$$. Solving this system, we find that the eigenspace corresponding to $$\lambda = 0$$ is spanned by two vectors: $$\mathbf{v}_2 = \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix}, \quad \mathbf{v}_3 = \begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix}$$ ### Step 3: Basis of Eigenvectors for $$\mathbb{R}^3$$ Since we have three linearly independent eigenvectors $$\mathbf{v}_1$$, $$\mathbf{v}_2$$, and $$\mathbf{v}_3$$, these form a basis for $$\mathbb{R}^3$$. ### Step 4: Diagonalization We can form the matrix $$P$$ from the eigenvectors: $$P = \begin{bmatrix} 1 & 1 & 0 \\ 1 & -1 & 1 \\ 1 & 0 & -1 \end{bmatrix}$$ The diagonal matrix $$D$$ will have the eigenvalues on its diagonal: $$D = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$ Finally, we can verify that: $$D = P^{-1} A P$$ This completes the diagonalization of the matrix $$A$$. Would you like further details or have any questions? ### 5 Related Questions: 1. How do you calculate the inverse of a matrix? 2. Can all matrices be diagonalized? 3. What does it mean for a matrix to be defective? 4. How does the Jordan canonical form differ from diagonalization? 5. What are some real-world applications of eigenvalue decomposition? ### Tip: Diagonalization is useful in simplifying matrix functions, like computing powers or exponentials of matrices.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Eigenvalues
Eigenvectors
Matrix Diagonalization
Formulas
Characteristic equation: det(A - λI) = 0
Matrix diagonalization: D = P^{-1} A P
Theorems
Eigenvalue decomposition
Diagonalization theorem
Suitable Grade Level
University level (Linear Algebra course)