A=（3,2,4 2,0,2 4,2,3） You are given that the characteristic equation of A is (λ + 1)2(λ − 8) = 0, i.e. the eigenvalues of A are λ = −1, 8. Find an orthogonal matrix P that orthogonally diagonalizes A
Question
A=（3,2,4 2,0,2 4,2,3） You are given that the characteristic equation of A is (λ + 1)2(λ − 8) = 0, i.e. the eigenvalues of A are λ = −1, 8. Find an orthogonal matrix P that orthogonally diagonalizes A
Solution
To find an orthogonal matrix P that orthogonally diagonalizes A, we first need to find the eigenvectors corresponding to each eigenvalue.
Step 1: Find the eigenvectors for each eigenvalue
For λ = -1, we solve the equation (A - λI)v = 0, where I is the identity matrix and v is the eigenvector we're trying to find.
Substituting A and λ into the equation, we get:
(3+1, 2, 4 2, 0+1, 2 4, 2, 3+1)v = 0
Solving this system of equations, we get the eigenvector v1 = (1, -2, 1) for λ = -1.
For λ = 8, we do the same:
(3-8, 2, 4 2, 0-8, 2 4, 2, 3-8)v = 0
Solving this system of equations, we get the eigenvector v2 = (1, 1, 1) for λ = 8.
Step 2: Normalize the eigenvectors
To form an orthogonal matrix, the eigenvectors must be normalized (i.e., each must have a length of 1).
The length of a vector v = (x, y, z) is given by √(x² + y² + z²).
So, the length of v1 = √(1² + (-2)² + 1²) = √6. Therefore, the normalized v1 = (1/√6, -2/√6, 1/√6).
The length of v2 = √(1² + 1² + 1²) = √3. Therefore, the normalized v2 = (1/√3, 1/√3, 1/√3).
Step 3: Form the orthogonal matrix P
The orthogonal matrix P is formed by placing the normalized eigenvectors as its columns:
P = (1/√6, 1/√3 -2/√6, 1/√3 1/√6, 1/√3)
This is the orthogonal matrix P that orthogonally diagonalizes A.
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