Math Problem Statement

diagonalize the matrix orthogonally a= [ 3 -1 1, -1 5 -1, 1 -1 3 ]

Solution

To orthogonally diagonalize a matrix AA, we need to follow these steps:

  1. Verify if AA is symmetric: A symmetric matrix can be orthogonally diagonalized, meaning that there exists an orthogonal matrix PP such that PTAP=DP^T A P = D, where DD is a diagonal matrix.

    For AA: A=[311151113]A = \begin{bmatrix} 3 & -1 & 1 \\ -1 & 5 & -1 \\ 1 & -1 & 3 \end{bmatrix} The matrix is symmetric because AT=AA^T = A.

  2. Find the eigenvalues of AA: To find the eigenvalues, solve the characteristic equation det(AλI)=0\text{det}(A - \lambda I) = 0, where λ\lambda is the eigenvalue, and II is the identity matrix.

  3. Find the eigenvectors of AA: For each eigenvalue λi\lambda_i, solve (AλiI)v=0(A - \lambda_i I) \mathbf{v} = 0 to find the corresponding eigenvectors vi\mathbf{v}_i.

  4. Construct the orthogonal matrix PP: Normalize the eigenvectors and arrange them as columns of PP. The matrix PP will be orthogonal, meaning PTP=IP^T P = I.

  5. Diagonal matrix DD: The diagonal matrix DD will contain the eigenvalues of AA along its diagonal.

Let me compute the eigenvalues and eigenvectors of matrix AA.The eigenvalues of matrix AA are:

λ1=6,λ2=2,λ3=3\lambda_1 = 6, \quad \lambda_2 = 2, \quad \lambda_3 = 3

The corresponding eigenvectors (normalized) are:

v1=[0.40820.81650.4082],v2=[0.70719.7×10160.7071],v3=[0.57740.57740.5774]\mathbf{v}_1 = \begin{bmatrix} -0.4082 \\ 0.8165 \\ -0.4082 \end{bmatrix}, \quad \mathbf{v}_2 = \begin{bmatrix} 0.7071 \\ 9.7 \times 10^{-16} \\ -0.7071 \end{bmatrix}, \quad \mathbf{v}_3 = \begin{bmatrix} 0.5774 \\ 0.5774 \\ 0.5774 \end{bmatrix}

Orthogonal Matrix PP:

The matrix PP, whose columns are the normalized eigenvectors, is:

-0.4082 & 0.7071 & 0.5774 \\ 0.8165 & 9.7 \times 10^{-16} & 0.5774 \\ -0.4082 & -0.7071 & 0.5774 \end{bmatrix}$$ ### Diagonal Matrix $$D$$: The diagonal matrix $$D$$, with the eigenvalues along the diagonal, is: $$D = \begin{bmatrix} 6 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix}$$ Thus, the matrix $$A$$ can be orthogonally diagonalized as $$A = P D P^T$$. Would you like more details on any part of this process? Here are a few related questions for deeper understanding: 1. How is the characteristic equation used to find eigenvalues? 2. What does it mean for a matrix to be orthogonally diagonalizable? 3. Why is orthogonality important in diagonalization? 4. How can you verify the orthogonality of the matrix $$P$$? 5. What are some applications of diagonalization in mathematics and physics? **Tip:** When working with eigenvectors, always check if they are normalized. If not, normalize them to ensure the orthogonal matrix has unit-length columns.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Matrix Diagonalization
Eigenvalues and Eigenvectors
Orthogonality

Formulas

Characteristic equation det(A - λI) = 0
Eigenvector equation (A - λI)v = 0
Orthogonal matrix condition P^T P = I
Diagonalization formula A = P D P^T

Theorems

Spectral Theorem: A real symmetric matrix can be orthogonally diagonalized
Eigenvalue Theorem: The eigenvalues of a matrix are found by solving the characteristic equation

Suitable Grade Level

Undergraduate (Linear Algebra)