Math Problem Statement

Diagonalize the matrix A = [[0, 4, -3], [-2, 6, -3], [-2, 4, -1]]

Solution

To diagonalize the given matrix AA, we need to find its eigenvalues and eigenvectors. The matrix AA is:

A=[043263241]A = \begin{bmatrix} 0 & 4 & -3 \\ -2 & 6 & -3 \\ -2 & 4 & -1 \end{bmatrix}

Steps to Diagonalize the Matrix:

  1. Find the Eigenvalues: Solve the characteristic polynomial det(AλI)=0\det(A - \lambda I) = 0.
  2. Find the Eigenvectors: For each eigenvalue λ\lambda, find a non-zero vector vv such that (AλI)v=0(A - \lambda I)v = 0.
  3. Construct the Diagonal Matrix DD: The eigenvalues will form the entries on the diagonal of DD.
  4. Construct the Matrix PP: Formed by the eigenvectors of AA as columns.
  5. Verify the Diagonalization: Check if A=PDP1A = PDP^{-1}.

I'll go ahead and calculate these details for you.

Step 1: Finding the Eigenvalues

Let’s solve for λ\lambda by calculating the characteristic polynomial:

det(AλI)=0\det\left( A - \lambda I \right) = 0

Step 2: Finding the Eigenvectors

For each eigenvalue, solve (AλI)v=0(A - \lambda I)v = 0 to find the corresponding eigenvectors.

Let me proceed with these computations.The eigenvalues of the matrix AA are:

λ1=1,λ2=2,λ3=2\lambda_1 = 1, \quad \lambda_2 = 2, \quad \lambda_3 = 2

The corresponding eigenvectors (normalized) are:

  1. For λ=1\lambda = 1: v1=[0.5770.5770.577]v_1 = \begin{bmatrix} -0.577 \\ -0.577 \\ -0.577 \end{bmatrix}
  2. For λ=2\lambda = 2: v2=[0.5090.6650.547]v_2 = \begin{bmatrix} 0.509 \\ 0.665 \\ 0.547 \end{bmatrix}
  3. For λ=2\lambda = 2: v3=[0.8210.0170.571]v_3 = \begin{bmatrix} -0.821 \\ 0.017 \\ 0.571 \end{bmatrix}

Step 3: Constructing the Diagonal Matrix DD

The diagonal matrix DD will have the eigenvalues on the diagonal:

D=[100020002]D = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}

Step 4: Constructing the Matrix PP

The matrix PP is formed by placing the eigenvectors as columns:

P=[0.5770.5090.8210.5770.6650.0170.5770.5470.571]P = \begin{bmatrix} -0.577 & 0.509 & -0.821 \\ -0.577 & 0.665 & 0.017 \\ -0.577 & 0.547 & 0.571 \end{bmatrix}

Step 5: Verification

To confirm the diagonalization, we can check if A=PDP1A = PDP^{-1}.

Would you like to proceed with this verification, or do you need further explanation on any step?


Related Questions:

  1. What is the significance of the eigenvalues and eigenvectors in diagonalization?
  2. Can every matrix be diagonalized? Why or why not?
  3. How does diagonalization simplify matrix computations?
  4. What is the role of matrix PP in the transformation process?
  5. How can diagonalization be applied in solving systems of differential equations?

Tip:

Diagonalization is especially powerful for computing powers of matrices, as An=PDnP1A^n = PD^nP^{-1} allows computations using only the diagonal elements of DD.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Linear Algebra
Matrix Diagonalization
Eigenvalues and Eigenvectors

Formulas

Characteristic Polynomial: det(A - λI) = 0
Eigenvector Equation: (A - λI)v = 0
Matrix Diagonalization: A = PDP^(-1)

Theorems

Diagonalization Theorem

Suitable Grade Level

Undergraduate Level