Math Problem Statement

Question content area top Part 1 Diagonalize the following matrix. Start 4 By 4 Table 1st Row 1st Column 6 2nd Column negative 3 3rd Column 0 4st Column 3 2nd Row 1st Column 0 2nd Column 4 3rd Column 1 4st Column negative 3 3rd Row 1st Column 0 2nd Column 0 3rd Column 3 4st Column 0 4st Row 1st Column 0 2nd Column 0 3rd Column 0 4st Column 3 EndTable Question content area bottom Part 1 Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice. A. For Pequals enter your response here​, DequalsStart 4 By 4 Table 1st Row 1st Column 3 2nd Column 0 3rd Column 0 4st Column 0 2nd Row 1st Column 0 2nd Column 3 3rd Column 0 4st Column 0 3rd Row 1st Column 0 2nd Column 0 3rd Column 4 4st Column 0 4st Row 1st Column 0 2nd Column 0 3rd Column 0 4st Column 6 EndTable B. The matrix cannot be diagonalized.

Solution

To determine if the matrix can be diagonalized, and if so, to find the matrices PP and DD, we follow these steps:

Step 1: Write the given matrix AA

6 & -3 & 0 & 3 \\ 0 & 4 & 1 & -3 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 3 \end{bmatrix}$$ ### Step 2: Find the eigenvalues of $$A$$ The eigenvalues are found by solving the characteristic equation: $$\det(A - \lambda I) = 0$$ This involves subtracting $$\lambda$$ along the diagonal of $$A$$ and finding the determinant of the resulting matrix. Subtract $$\lambda$$ from the diagonal: $$A - \lambda I = \begin{bmatrix} 6 - \lambda & -3 & 0 & 3 \\ 0 & 4 - \lambda & 1 & -3 \\ 0 & 0 & 3 - \lambda & 0 \\ 0 & 0 & 0 & 3 - \lambda \end{bmatrix}$$ The determinant is: $$\det(A - \lambda I) = (6 - \lambda) \det \begin{bmatrix} 4 - \lambda & 1 & -3 \\ 0 & 3 - \lambda & 0 \\ 0 & 0 & 3 - \lambda \end{bmatrix}$$ The smaller determinant is: $$\det \begin{bmatrix} 4 - \lambda & 1 & -3 \\ 0 & 3 - \lambda & 0 \\ 0 & 0 & 3 - \lambda \end{bmatrix} = (4 - \lambda)(3 - \lambda)^2$$ Thus: $$\det(A - \lambda I) = (6 - \lambda)(4 - \lambda)(3 - \lambda)^2$$ The eigenvalues are: $$\lambda_1 = 6, \quad \lambda_2 = 4, \quad \lambda_3 = 3 \, (\text{with multiplicity } 2)$$ ### Step 3: Verify diagonalizability A matrix is diagonalizable if it has a complete set of linearly independent eigenvectors. This is equivalent to the total number of independent eigenvectors equaling the size of the matrix (4 in this case). - $$\lambda_1 = 6$$: One eigenvector - $$\lambda_2 = 4$$: One eigenvector - $$\lambda_3 = 3$$: Two eigenvectors (due to multiplicity) Since the total number of eigenvectors equals 4, the matrix is diagonalizable. ### Step 4: Construct $$D$$ and $$P$$ The diagonal matrix $$D$$ contains the eigenvalues along its diagonal: $$D = \begin{bmatrix} 6 & 0 & 0 & 0 \\ 0 & 4 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 3 \end{bmatrix}$$ The matrix $$P$$ is formed by placing the eigenvectors of $$A$$ as its columns. The eigenvectors are computed by solving $$(A - \lambda I)v = 0$$ for each $$\lambda$$. The exact eigenvectors depend on solving linear systems. ### Final Answer: **Choice A**: $$P = \text{(Matrix of eigenvectors)}, \quad D = \begin{bmatrix} 6 & 0 & 0 & 0 \\ 0 & 4 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 3 \end{bmatrix}$$ Would you like a detailed computation of the eigenvectors to construct $$P$$? --- ### Related Questions: 1. How do you compute eigenvectors for $$\lambda = 6$$? 2. What happens if the eigenvalues do not yield enough linearly independent eigenvectors? 3. Can diagonalization always be performed for square matrices? 4. How can you verify that $$P^{-1}AP = D$$? 5. What is the significance of the multiplicity of eigenvalues in diagonalization? **Tip:** Ensure that all eigenvalues and eigenvectors are computed accurately, as errors here can lead to incorrect diagonalization.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Matrix Diagonalization
Eigenvalues and Eigenvectors

Formulas

Characteristic equation: det(A - λI) = 0
Eigenvector equation: (A - λI)v = 0

Theorems

Diagonalizability theorem
Eigenvalue multiplicity and eigenvector basis

Suitable Grade Level

Undergraduate Level (Linear Algebra)

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