Eigen Values, Eigen Vectors

1 minute read

Overview

If we have a matrix \(A \in \mathbb{R}^n\), then its eigen vectors are defined:

\[Av = \lambda v\]

Where:

\(\lambda\) is a scalar
\(v\) is a vector

That is, \(v\) are vectors that are only scaled by \(A\), but do not change direction.

\(\lambda\) are the eigen values of \(A\).

Calculating eigen vectors and eigen values

To find the eigen values:

\[det(\lambda I_n - A) = 0\]

To find the eigen vectors (once we know the eigen values):

\[(\lambda I_n - A) v = \boldsymbol{0}\]

2x2 matrix example

\[A = \left[ \begin{array}{ccc} 1 & 1 \\ 4 & -2 \end{array} \right]\]

Finding eigen values

Take the determinent of \((\lambda I_n - A)\) and set to zero:

\[det\left(\left[ \begin{array}{ccc} \lambda - 1 & -1 \\ -4 & \lambda + 2 \end{array} \right]\right) = 0\] \[(\lambda - 1)(\lambda + 2) - 4 = 0\]

Has solutions \(\lambda = 2, -3\)

Finding eigen vectors

For \(\lambda = 2\):

\[\left[ \begin{array}{ccc} 1 & -1 \\ -4 & 4 \end{array} \right] v = \boldsymbol{0}\]

Let:

\[v = \left[ \begin{array}{ccc} v_1 \\ v_2 \end{array} \right]\]

Then:

\[v_1 - v_2 = 0\] \[\therefore v = \left[ \begin{array}{ccc} 1 \\ 1 \end{array} \right]\]

For \(\lambda = -3\):

\[\left[ \begin{array}{ccc} -4 & -1 \\ -4 & -1 \end{array} \right] v = \boldsymbol{0}\]

Then:

\[4 v_1 + v_2 = 0\] \[\therefore v = \left[ \begin{array}{ccc} 1 \\ -4 \end{array} \right]\]

Python

import numpy as np

A = np.array([[1, 1],[4, -2]])
w, v = np.linalg.eig(A)

print(f"Eigen values = {w[0], w[1]}") 
print(f"Eigen vectors = {v[:,0]/v[0,0], v[:,1]/v[0,1]}") 

With output:

Eigen values = (2.0, -3.0)
Eigen vectors = (array([1., 1.]), array([ 1., -4.]))

Octave/Matlab

A = [1, 1; 4, -2];
[v, w] = eigs(A);
disp("Eigen values")
disp(w)
disp("Eigen vectors")
disp(v(:, 1)/v(1, 1))
disp(v(:, 2)/v(1, 2))