Eigen Values, Eigen Vectors

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))



With output:

Eigen values
Diagonal Matrix
-3   0
0   2

Eigen vectors
1
-4
1
1



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