Solving Eigenvectors and Eigenvalues by hand

Given a matrix of vectors like:

$$ A = \begin{bmatrix} 1 & 5 \\ 2 & 4 \end{bmatrix} $$

1. Subtract $\lambda$ along the diagonal, representing $A - \lambda I$: $$ A = \begin{bmatrix} 1 - \lambda& 5 \\ 2 & 4 - \lambda \end{bmatrix} $$

2. Find the determinant:

$$ = (1 - \lambda)(4 - \lambda) - 10 \\ = 4 - 5\lambda + \lambda^2 - 10 \\ = -6 -5\lambda + lambda^2 \\ = (\lambda - 6)(\lambda + 1) $$

1. The eigenvalues are the values of $\lambda$ that make the determinant 0, so in this case they are 6 and -1.

Now, finding the Eigenvectors, we know that $\lambda$ is either 6 or -1, so we plug these in:

$$ (A - \lambda I) = \begin{bmatrix} 1 & 5 \\ 2 & 4 \end{bmatrix} - \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix} \vec v = \begin{bmatrix} -5 & 5 \\ 2 & -2 \end{bmatrix} \vec v = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $$

Gives us $\vec v = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$

Or, $$ (A - \lambda I) = \begin{bmatrix} 1 & 5 \\ 2 & 4 \end{bmatrix} - \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \vec v = \begin{bmatrix} 2 & 5 \\ 2 & 5 \end{bmatrix} \vec v = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $$

Gives us $\vec v = \begin{bmatrix} 5 \\ -2 \end{bmatrix}$

ICA Question 2: Find the eigenvalues and the eigenvectors of the given matrix:

$$ A = \begin{bmatrix} -1 & 3 \\ 0 & 2 \end{bmatrix} $$

import numpy as np

A = np.array([[-1, 3], [0, 2]])

# column 0 corresponds to eigenvalue 0 
np.linalg.eig(A)

(array([-1.,  2.]),
 array([[1.        , 0.70710678],
        [0.        , 0.70710678]]))