Math Problem Statement

Suppose that a 2 x 2 matrix A has an eigenvalue 3 with corresponding eigenvector [-1 2] and an eigenvalue -2 with corresponding eigenvector [-1 1]. Find an invertible matrix P and a diagonal matrix D so that A = PDP^{-1}.

Solution

To solve this problem, we need to use the information about the matrix AA's eigenvalues and corresponding eigenvectors to construct matrices PP and DD such that A=PDP1A = PDP^{-1}, where:

  1. Matrix DD: This is a diagonal matrix containing the eigenvalues of AA.
  2. Matrix PP: This matrix has the eigenvectors of AA as its columns.

Step-by-Step Solution:

  1. Identify DD: The diagonal matrix DD will contain the eigenvalues of AA. We are given:

    • Eigenvalue 33 with its corresponding eigenvector [12]\begin{bmatrix} -1 \\ 2 \end{bmatrix}.
    • Eigenvalue 2-2 with its corresponding eigenvector [11]\begin{bmatrix} -1 \\ 1 \end{bmatrix}.

    So, D=[3002]D = \begin{bmatrix} 3 & 0 \\ 0 & -2 \end{bmatrix}.

  2. Construct PP: The matrix PP will have the eigenvectors as columns. Therefore, P=[1121].P = \begin{bmatrix} -1 & -1 \\ 2 & 1 \end{bmatrix}.

  3. Find P1P^{-1}: To complete the decomposition, we need the inverse of PP, P1P^{-1}.

    • P=[1121]P = \begin{bmatrix} -1 & -1 \\ 2 & 1 \end{bmatrix}.
    • To find P1P^{-1}, we use the formula for the inverse of a 2x2 matrix [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix}: P1=1adbc[dbca].P^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}.
    • For P=[1121]P = \begin{bmatrix} -1 & -1 \\ 2 & 1 \end{bmatrix}:
      • Determinant det(P)=(1)(1)(1)(2)=1+2=1\det(P) = (-1)(1) - (-1)(2) = -1 + 2 = 1.
      • Thus, P1=[1121].P^{-1} = \begin{bmatrix} 1 & 1 \\ -2 & -1 \end{bmatrix}.
  4. Express A=PDP1A = PDP^{-1}: Now, we have A=[1121][3002][1121].A = \begin{bmatrix} -1 & -1 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 3 & 0 \\ 0 & -2 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ -2 & -1 \end{bmatrix}.

Summary of the Answer:

  • P=[1121]P = \begin{bmatrix} -1 & -1 \\ 2 & 1 \end{bmatrix}
  • D=[3002]D = \begin{bmatrix} 3 & 0 \\ 0 & -2 \end{bmatrix}
  • P1=[1121]P^{-1} = \begin{bmatrix} 1 & 1 \\ -2 & -1 \end{bmatrix}

Would you like further details on any step or explanation on how to multiply these matrices if needed?


Related Questions:

  1. How do you find the inverse of a 3x3 matrix?
  2. What are the properties of diagonalizable matrices?
  3. How does the eigenvalue decomposition apply to larger matrices?
  4. What conditions must a matrix meet to be diagonalizable?
  5. How can you verify if a matrix is symmetric or orthogonal?

Tip:

Diagonalization is only possible if a matrix has enough linearly independent eigenvectors to form the matrix PP.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Matrix Diagonalization
Eigenvalues and Eigenvectors

Formulas

A = PDP^{-1}
P^{-1} for a 2x2 matrix: P^{-1} = (1/(ad - bc)) * [[d, -b], [-c, a]]

Theorems

Eigenvalue Decomposition
Diagonalization Theorem

Suitable Grade Level

Undergraduate Level