# Eigenvalues and eigenvectors of matrices

## Calculators

The other online calculators calculate the determinant of 3x3, 4x4 and arbitrary quadratic arrays as well as the QR decomposition of matrices.

Here are the online computers:

## Calculation rules for determinants and matrices

#### Terms

An eigenvector of a map is a vector that changes only in magnitude but not in the direction through the map. The factor by which the amount changes is the associated eigenvalue . The set of eigenvectors to an eigenvalue is called eigenspace . The set of eigenvalues of a map is the spectrum of the map. The spectral radius is the eigenvalue with the largest amount. The dimension of the eigenspace is the geometric multiplicity of the eigenvalue. The algebraic multiplicity is defined by the multiplicity of zeros of the characteristic polynomial.

## 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 λi.

## Two-dimensional example part 2

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

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

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

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

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

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.

B0= 0;   cn= 1;

repeat

Bk= A Bk-1 + cn-k+1 I;    cn-k= - tr(A Bk) / k;

until k < Rang(A)

Matrix dimension N =

Number of digits =

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

QR-Algorithm

repeat

Am= QmRm;   QR-Zerlegung

Am+1= RmQm;

until Residuum < ε

Matrix dimension N =

Number of digits =

Maximal number of iterations =