## Definition

For a square matrix A, each vector v is an eigenvector if the following conditions are met:

$Av=\lambda v$

$\text{with}\phantom{\rule{1em}{0ex}}\lambda \phantom{\rule{0.5em}{0ex}}\in \phantom{\rule{0.5em}{0ex}}\mathbb{C}\phantom{\rule{1em}{0ex}}\text{und}\phantom{\rule{1em}{0ex}}v\phantom{\rule{0.5em}{0ex}}\ne \phantom{\rule{0.5em}{0ex}}0$

The factor λ is the eigenvalue belonging to the eigenvector v. The eigenvalues can be real or complex.

## Vivid interpretation

In general, a matrix A maps a vector a into another vector b. The eigenvectors v are distinguished by the fact that they are represented by the image only by a factor? stretched or compressed.

## Two-dimensional example

Using this example, the eigenvalue and eigenvector are explained clearly in the plane.

$A=\left(\begin{array}{cc}3& -1\\ 1& 1\end{array}\right)$

This matrix maps the vector (1,1) as follows.

$\left(\begin{array}{cc}3& -1\\ 1& 1\end{array}\right)\left(\begin{array}{c}1\\ 1\end{array}\right)=\left(\begin{array}{c}2\\ 2\end{array}\right)=2\left(\begin{array}{c}1\\ 1\end{array}\right)$

That the vector is an eigenvector of A, which is stretched by the factor of 2. Thus, the eigenvalue belongs to this eigenvector 2.

For another vector, e.g., (2,3), this is not true.

$\left(\begin{array}{cc}3& -1\\ 1& 1\end{array}\right)\left(\begin{array}{c}2\\ 3\end{array}\right)=\left(\begin{array}{c}3\\ 5\end{array}\right)$

However, the matrix does not just have an eigenvector. For example, the vector (2,2) is an eigenvector.

$\left(\begin{array}{cc}3& -1\\ 1& 1\end{array}\right)\left(\begin{array}{c}2\\ 2\end{array}\right)=\left(\begin{array}{c}4\\ 4\end{array}\right)=2\left(\begin{array}{c}2\\ 2\end{array}\right)$

## Calculation

The equation

$Av=\lambda v$

can be transformed into the homogeneous equation system

$\left(A-\lambda E\right)v=0$

The system of equations has a non-trivial solution if and only if the determinant disappears. That if applicable

$\mathrm{det}\left(A-\lambda E\right)=0$

The polynomial is called the characteristic polynomial of A and the equation is the characteristic equation of A. If λ_{ i } is an eigenvalue of A then the solutions of the characteristic equation are the eigenvectors of A to the eigenvalue λ_{i}.

## Two-dimensional example part 2

In the second part of the example eigenvalue and eigenvector are calculated. First, the characteristic polynomial is determined.

$\mathrm{det}\left(A-\lambda E\right)=$

$\left|\left(\begin{array}{cc}3& -1\\ 1& 1\end{array}\right)-\mathrm{\lambda}\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\right|=$

$\left|\begin{array}{cc}3-\lambda & -1\\ 1& 1-\lambda \end{array}\right|=$

${\lambda}^{2}-4\lambda +4$

The zeros of the characteristic polynomial are the eigenvalues of A.

${\lambda}^{2}-4\lambda +4=0$

So the eigenvalue of A is:

${\lambda}_{i}=2$

The solutions of the characteristic equation provide the eigenvectors.

$\left(A-{\lambda}_{i}E\right)v=$

$\left(\left(\begin{array}{cc}3& -1\\ 1& 1\end{array}\right)-2\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\right)v=$

$\left(\begin{array}{cc}1& -1\\ 1& -1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=0$

This results in the following two equations.

$x-y=0$

$x-y=0$

That means all vectors v_{ i } where both components are equal, are eigenvectors to eigenvalue λ_{ i } = 2.

${v}_{i}=\left(\begin{array}{c}x\\ x\end{array}\right)$

### Calculator for the calculation of the characteristic polynomial p(λ) of matrix A

Calculation according to the algorithm of Faddejew-Leverrier.

B_{0}= 0; c_{n}= 1;

repeat

B_{k}= A B_{k-1} + c_{n-k+1} I; c_{n-k}= - tr(A B_{k}) / k;

until k < Rang(A)

### Input matrix elements: a_{11}, a_{12}, ...

### Calculator for calculating the eigenvalues using the QR decomposition

QR-Algorithm

repeat

A_{m}= Q_{m}R_{m}; QR-Zerlegung

A_{m+1}= R_{m}Q_{m};

until Residuum < ε

### Input matrix elements: a_{11}, a_{12}, ...