Characteristic polynomial

The characteristic polynomial (CP) is a term from the mathematical branch of linear algebra . This polynomial , which is defined for square matrices and endomorphisms of finite-dimensional vector spaces , provides information about some properties of the matrix or linear mapping.

The equation in which the characteristic polynomial is set equal to zero is sometimes called a secular equation . Their solutions are the eigenvalues ​​of the matrix or the linear mapping. A matrix, inserted into its characteristic polynomial, gives the zero mapping ( Cayley-Hamilton theorem ).

definition

The characteristic polynomial of a quadratic matrix with entries from a body is defined by: ${\displaystyle \chi _{A}}$ ${\displaystyle n\times n}$ ${\displaystyle A}$ ${\displaystyle \mathbb {K} }$

${\displaystyle \chi _{A}(\lambda ):=\det(\lambda E_{n}-A)}$

Here denotes the -dimensional identity matrix and the determinant . The indefinite also stands for an element of . ${\displaystyle E_{n}}$ ${\displaystyle n}$ ${\displaystyle \det }$ ${\displaystyle \lambda }$ ${\displaystyle \mathbb {K} }$

The definition of the characteristic polynomial as is also common. For odd numbers it differs from the above definition in terms of the factor , that is, the polynomial is then no longer normalized . ${\displaystyle \det(A-\lambda E_{n})}$ ${\displaystyle n}$ ${\displaystyle -1}$

If one - dimensional - vector space and an endomorphism , then the characteristic polynomial is given by: ${\displaystyle V}$ ${\displaystyle n}$ ${\displaystyle \mathbb {K} }$ ${\displaystyle \varphi \colon V\to V}$ ${\displaystyle \chi _{\varphi }}$

${\displaystyle \chi _{\varphi }(\lambda )=\det(\lambda \cdot \mathrm {id} _{V}-\varphi )=\chi _{A}(\lambda ),}$

where is a representation matrix of the endomorphism with respect to a base. The characteristic polynomial of does not depend on the chosen basis. ${\displaystyle A}$ ${\displaystyle \varphi }$ ${\displaystyle \varphi }$

The characteristic polynomial is a normalized polynomial -th degree from the polynomial ring . The notation for the characteristic polynomial is very inconsistent, other variants are, for example, or from Bourbaki . ${\displaystyle n}$ ${\displaystyle \mathbb {K} [\lambda ]}$ ${\displaystyle \mathrm {CP} _{A}(\lambda )}$ ${\displaystyle \mathrm {Pc} _{A}(\lambda )}$

Relationship with eigenvalues

The characteristic polynomial plays an important role in determining the eigenvalues ​​of a matrix, because the eigenvalues ​​are exactly the zeros of the characteristic polynomial. Even if you always select a base and thus a representation matrix for the explicit calculation of the characteristic polynomial, the polynomial and the determinant do not depend on this choice.

To show that the eigenvalues ​​are the zeros of the characteristic polynomial, proceed as follows:

Let it be and a matrix over . Then the following equivalences apply: ${\displaystyle \lambda \in \mathbb {K} }$ ${\displaystyle A}$ ${\displaystyle n\times n}$ ${\displaystyle \mathbb {K} }$

${\displaystyle \lambda }$is an eigenvalue of . ${\displaystyle A}$

${\displaystyle \Leftrightarrow }$There is one with . ${\displaystyle x\in \mathbb {K} ^{n},x\neq 0}$ ${\displaystyle Ax=\lambda x}$

${\displaystyle \Leftrightarrow }$There is one with . ${\displaystyle x\in \mathbb {K} ^{n},x\neq 0}$ ${\displaystyle (\lambda E-A)x=0}$

${\displaystyle \Leftrightarrow \lambda E-A}$ is not invertible.

${\displaystyle \Leftrightarrow \det(\lambda E-A)=0}$

${\displaystyle \Leftrightarrow }$ ${\displaystyle \lambda }$is the root of the characteristic polynomial of . ${\displaystyle A}$

Formulas and algorithms

If you write the characteristic polynomial in the form

${\displaystyle \chi _{A}(\lambda )=\sum _{k=0}^{n}a_{k}\lambda ^{k}}$

so is always , with as the trace , and the determinant of . ${\displaystyle a_{n-1}=(-1)^{n-1}\cdot \operatorname {Tr} (A)}$ ${\displaystyle \operatorname {Tr} }$ ${\displaystyle a_{0}}$ ${\displaystyle A}$

The characteristic polynomial has a particularly simple form especially for matrices ${\displaystyle (2\times 2)}$

${\displaystyle \chi _{A}(\lambda )=\lambda ^{2}-\operatorname {spur} (A)\cdot \lambda +\det(A).}$

The form for matrices is: ${\displaystyle (3\times 3)}$

${\displaystyle \chi _{A}(\lambda )=-\lambda ^{3}+\operatorname {spur} (A)\cdot \lambda ^{2}-\left(\det(A_{1})+\det(A_{2})+\det(A_{3})\right)\cdot \lambda +\det(A).}$

Here is the matrix that is obtained by deleting the -th row and the -th column (a minor ). ${\displaystyle A_{i}}$ ${\displaystyle (2\times 2)}$ ${\displaystyle i}$ ${\displaystyle i}$

The coefficients of can be determined with the aid of suitable methods, e.g. B. the algorithm of Faddejew-Leverrier or the algorithm of Samuelson-Berkowitz , also determine systematically. ${\displaystyle \chi _{A}\;(\lambda )}$

properties

• The characteristic polynomials of two similar matrices are the same. However, the reverse is generally not correct.
• The matrix and its transpose have the same characteristic polynomial. ${\displaystyle A}$
• According to Cayley-Hamilton's theorem , a matrix is ​​the root of its characteristic polynomial:

  ${\displaystyle \chi _{A}\left(A\right)=0}$.

• The minimal polynomial of a linear mapping divides its characteristic polynomial.
• If there is a matrix and a matrix then holds . ${\displaystyle A}$ ${\displaystyle m\times n}$ ${\displaystyle B}$ ${\displaystyle n\times m}$ ${\displaystyle \chi _{AB}(\lambda )\,\lambda ^{n}=\chi _{BA}(\lambda )\,\lambda ^{m}}$

Proof:

From the matrix equations

${\displaystyle {\begin{pmatrix}\lambda E_{m}&-A\\0&E_{n}\end{pmatrix}}\,{\begin{pmatrix}E_{m}&A\\B&\lambda E_{n}\end{pmatrix}}={\begin{pmatrix}\lambda E_{m}-AB&0\\B&\lambda E_{n}\end{pmatrix}}}$

${\displaystyle {\begin{pmatrix}\lambda E_{m}&0\\-B&E_{n}\end{pmatrix}}\,{\begin{pmatrix}E_{m}&A\\B&\lambda E_{n}\end{pmatrix}}={\begin{pmatrix}\lambda E_{m}&\lambda A\\0&\lambda E_{n}-BA\end{pmatrix}}}$

as well as the rule

${\displaystyle \det {\begin{pmatrix}T&0\\S&W\end{pmatrix}}=\det(T)\,\det(W)}$

follows

${\displaystyle \det(\lambda E_{m}-AB)\,\lambda ^{n}=\det {\begin{pmatrix}E_{m}&A\\B&\lambda E_{n}\end{pmatrix}}\,\lambda ^{m}=\det(\lambda E_{n}-BA)\,\lambda ^{m}}$.

example

Find the characteristic polynomial of the matrix

${\displaystyle A={\begin{pmatrix}1&0&1\\2&2&1\\4&2&1\end{pmatrix}}.}$

According to the definition above, one calculates as follows:

${\displaystyle {\begin{aligned}\chi _{A}(\lambda )&=\det(\lambda E-A)\\&=\det {\begin{pmatrix}\lambda -1&0&-1\\-2&\lambda -2&-1\\-4&-2&\lambda -1\end{pmatrix}}\\&=\lambda ^{3}-4\lambda ^{2}-\lambda +4\\&=(\lambda -1)(\lambda +1)(\lambda -4).\end{aligned}}}$

Thus 1, −1 and 4 are the zeros of the characteristic polynomial and thus also the eigenvalues ​​of the matrix . Since every zero has the multiplicity 1, the characteristic polynomial in this example is also the minimal polynomial. ${\displaystyle \chi _{A}(\lambda )}$ ${\displaystyle A}$

literature

• Oliver Deiser, Caroline Lasser: First Aid in Linear Algebra: Overview and basic knowledge with many illustrations and examples . Springer, 2015, ISBN 978-3-642-41627-9 , p. 204 ff

Web links

Wikiversity: Introduction to the characteristic polynomial on Wikiversity  course materials

• Online tool for calculating the characteristic polynomial
• Characteristic polynomial in an online script from the University of Göttingen