Square Root of a 2x2 Matrix (by diagonalization)

[krozer]

Homework Statement

Show that the -1 -2
4 -1


2x2 matrix has one square root.

Homework Equations

det(A-λI) to find Eigenvalues
(A-λI)v=0 to find Eigenvectors
A^1/2 = V D^1/2 V^-1 to find the square root of A where V is the created matrix with the eigenvectors of A, and D is the diagonalized matrix A

The Attempt at a Solution

I know I can use diagonalization to find the square root:
Using the formula its eigenvalues are -1+(-32)^1/2 and -1-(-32)^1/2
But I don't know how to find the eigenvectors given that (-32)^1/2 it's a complex number.

 

[mighty2000]

To find the eigenvectors, we can use the fact that for a 2x2 matrix, the eigenvectors are given by the columns of the matrix V in the formula A1/2 = V D1/2 V-1.

First, let's find the eigenvalues of the matrix A. We can use the characteristic polynomial det(A-λI) = 0 to find the eigenvalues. In this case, the characteristic polynomial is λ^2 + 2λ + 3 = 0, which has the solutions λ = -1+(-32)1/2 and λ = -1-(-32)1/2.

Next, we can find the eigenvectors for each eigenvalue. For the eigenvalue λ = -1+(-32)1/2, we can substitute this into the equation (A-λI)v=0 to get the following system of equations:

3x + 2y = 0
2x + 3y = 0

Solving this system, we get the eigenvector v1 = [2, -3].

Similarly, for the eigenvalue λ = -1-(-32)1/2, we get the eigenvector v2 = [2, 3].

Now, we can form the matrix V with these eigenvectors as its columns:
V = [2, 2; -3, 3].

Next, we can find the diagonalized matrix D by substituting the eigenvalues into the diagonal matrix D = diag(λ1, λ2):
D = diag(-1+(-32)1/2, -1-(-32)1/2) = [-1+(-32)1/2, 0; 0, -1-(-32)1/2].

Finally, we can use the formula A1/2 = V D1/2 V-1 to find the square root of A. Substituting the matrices we found, we get:
A1/2 = V D1/2 V-1 = [2, 2; -3, 3] [-1+(-32)1/2, 0; 0, -1-(-32)1/2]1/2 [2, -3; 2, 3] = [1, 0; 0,