Math Problem Statement

Question content area top Part 1 Diagonalize the following matrix. The real eigenvalues are given to the right of the matrix. Start 3 By 3 Table 1st Row 1st Column 13 2nd Column negative 12 3rd Column 6 2nd Row 1st Column 18 2nd Column negative 23 3rd Column 15 3rd Row 1st Column 18 2nd Column negative 28 3rd Column 20 EndTable ​; lambdaequals1​,4​,5 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 3 By 3 Table 1st Row 1st Column 1 2nd Column 0 3rd Column 0 2nd Row 1st Column 0 2nd Column 4 3rd Column 0 3rd Row 1st Column 0 2nd Column 0 3rd Column 5 EndTable . ​(Simplify your​ answer.) B. The matrix cannot be diagonalized.

Solution

To diagonalize the given matrix, we need to perform the following steps:

  1. Verify Eigenvalues: The given eigenvalues are λ=1,4,5\lambda = 1, 4, 5. These eigenvalues are used to compute the eigenvectors.

  2. Find Eigenvectors: For each eigenvalue λi\lambda_i, solve the equation (A−λiI)v=0(A - \lambda_i I)\mathbf{v} = \mathbf{0} to find the eigenvector v\mathbf{v}.

  3. Form Matrix PP: Construct matrix PP whose columns are the eigenvectors corresponding to λ1,λ2,λ3\lambda_1, \lambda_2, \lambda_3.

  4. Diagonal Matrix DD: The diagonal matrix DD has the eigenvalues λ1,λ2,λ3\lambda_1, \lambda_2, \lambda_3 on its diagonal.

  5. Verify A=PDP−1A = PDP^{-1}: Ensure A=PDP−1A = PDP^{-1} to confirm the diagonalization.

Matrix Details

The given matrix is:

\begin{bmatrix} 13 & -12 & 6 \\ 18 & -23 & 15 \\ 18 & -28 & 20 \end{bmatrix}$$ The eigenvalues are $$\lambda = 1, 4, 5$$, as given. #### Step 1: Eigenvectors for $$\lambda = 1$$ Solve $$(A - I)\mathbf{v} = \mathbf{0}$$: $$A - I = \begin{bmatrix} 12 & -12 & 6 \\ 18 & -24 & 15 \\ 18 & -28 & 19 \end{bmatrix}$$ Reducing $$A - I$$ to row echelon form, we find: $$\mathbf{v}_1 = \text{[1, 2, 1]}^\top$$ #### Step 2: Eigenvectors for $$\lambda = 4$$ Solve $$(A - 4I)\mathbf{v} = \mathbf{0}$$: $$A - 4I = \begin{bmatrix} 9 & -12 & 6 \\ 18 & -27 & 15 \\ 18 & -28 & 16 \end{bmatrix}$$ Reducing $$A - 4I$$ to row echelon form, we find: $$\mathbf{v}_2 = \text{[2, 3, 1]}^\top$$ #### Step 3: Eigenvectors for $$\lambda = 5$$ Solve $$(A - 5I)\mathbf{v} = \mathbf{0}$$: $$A - 5I = \begin{bmatrix} 8 & -12 & 6 \\ 18 & -28 & 15 \\ 18 & -28 & 15 \end{bmatrix}$$ Reducing $$A - 5I$$ to row echelon form, we find: $$\mathbf{v}_3 = \text{[1, 1, 1]}^\top$$ ### Construct $$P$$ and $$D$$ Matrix $$P$$: $$P = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 1 & 1 & 1 \end{bmatrix}$$ Matrix $$D$$: $$D = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 5 \end{bmatrix}$$ ### Final Verification Since we have found a full set of linearly independent eigenvectors, the matrix **can** be diagonalized. #### Choice: The correct choice is: $$A. \text{ For } P = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 1 & 1 & 1 \end{bmatrix}, \ D = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 5 \end{bmatrix}.$$ Would you like a detailed explanation for any step? --- ### Questions for Further Exploration: 1. What is the significance of eigenvectors in diagonalization? 2. How do you verify that $$PDP^{-1}$$ correctly reconstructs the original matrix? 3. What happens if the matrix has repeated eigenvalues? 4. Can a matrix without a full set of eigenvectors be diagonalized? 5. What are practical applications of diagonalizing matrices? **Tip**: Always check for linear independence of eigenvectors to confirm diagonalizability.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Matrix Diagonalization
Eigenvalues and Eigenvectors

Formulas

(A - λI)v = 0 (eigenvector calculation)
PDP⁻¹ = A (matrix diagonalization)

Theorems

Spectral Theorem (for diagonalizable matrices)

Suitable Grade Level

Undergraduate level (Linear Algebra course)

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