Let A be a 3 × 3 real symmetric matrix and have eigenvalues λ1 = 2, λ2 = λ3 = 1. The eigenvectors corresponding to λ2 = λ3 = 1 are respectively given by v2=(1,0,1), v3=(1,2.-1),Find the eigenvector of A corresponding to λ1 = 2 and the matrix A
Question
Let A be a 3 × 3 real symmetric matrix and have eigenvalues λ1 = 2, λ2 = λ3 = 1. The eigenvectors corresponding to λ2 = λ3 = 1 are respectively given by v2=(1,0,1), v3=(1,2.-1),Find the eigenvector of A corresponding to λ1 = 2 and the matrix A
Solution
Given that A is a real symmetric matrix, we know that its eigenvectors corresponding to distinct eigenvalues are orthogonal.
We have the eigenvalues λ1 = 2 and λ2 = λ3 = 1, with corresponding eigenvectors v2 = (1,0,1) and v3 = (1,2,-1).
We can find the eigenvector corresponding to λ1 = 2 by taking the cross product of v2 and v3, since the cross product of two vectors gives a vector that is orthogonal to both.
v1 = v2 x v3 = (1,0,1) x (1,2,-1) = (-2, 0, -2)
So, the eigenvector corresponding to λ1 = 2 is v1 = (-2, 0, -2).
Now, we can find the matrix A by using the formula A = PDP^-1, where D is the diagonal matrix of eigenvalues and P is the matrix of eigenvectors.
D = diag([2, 1, 1]) = [[2, 0, 0], [0, 1, 0], [0, 0, 1]]
P = [v1, v2, v3] = [[-2, 1, 1], [0, 0, 2], [-2, 1, -1]]
P^-1 is the inverse of P.
P^-1 = [[-1/2, 0, -1/2], [1, 0, 1], [1, 1, -1]]
Now, we can compute A = PDP^-1.
A = [[-2, 1, 1], [0, 0, 2], [-2, 1, -1]]
- [[2, 0, 0], [0, 1, 0], [0, 0, 1]]
- [[-1/2, 0, -1/2], [1, 0, 1], [1, 1, -1]]
After performing the matrix multiplications, we get:
A = [[1, 2, 1], [1, 0, 1], [2, 2, 1]]
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