In class we solved for the eigenvalues and eigenvectors of t

In class, we solved for the eigenvalues and eigenvectors of the matrix. A = [-5 2 2 -2] The eigenvalues are lambda_1 = -1 and lambda_2 = -6. lambda_1 has a corresponding eigenvector of [1 2]^T. lambda_2 has a corresponding eigenvector of [1 -0.5]^T. You can use Matlab to calculate eigenstuff also. Enter the matrix A into MatLab. Use the command e = eig(A) to calculate the eigenvalues. Do they agree with our answers from class? Explain. Use the command [V, D] = eig(A) where the answer for V is the eigenvector matrix and the answer for D is the diagonalized eigenvalue matrix. Do these values agree with our answers from class? Explain.

Solution

>> A = [-5 2;2 -2]

A =

-5 2
2 -2

>> eig(A)

ans =

-6
-1

A) Eigen values of given matrix is same as calculated .Because eigen values are unique to given matrix

>> [V,D] = eig(A)

V =

-0.8944 -0.4472
0.4472 -0.8944

D =

-6 0
0 -1

B) D matrix is equalent with hand calculation , where as V matrix is not unique because eigen vector is not unique. For a non singular matrix eigen vectors corresponding to eigen values are different.