Similarity (matrix)

In the mathematical subfield of linear algebra , similarity is an equivalence relation on the class of square matrices . Similar matrices describe the same linear self-mapping ( endomorphism ) when using different bases.

definition

Two- dimensional square matrices over the body are called similar to each other if there is a regular matrix such that ${\displaystyle n}$ ${\displaystyle A,B\in K^{n\times n}}$ ${\displaystyle K}$ ${\displaystyle S\in K^{n\times n}}$

${\displaystyle B=S^{-1}AS}$

or equivalent

${\displaystyle SB=AS}$

applies. The image

${\displaystyle A\mapsto B=S^{-1}AS}$

is called similarity mapping or similarity transformation . If a matrix is similar to a diagonal matrix , it is said to be diagonalizable ; if it is similar to an upper triangular matrix , it is called trigonalizable .

example

The two real matrices

${\displaystyle A={\begin{pmatrix}-3&2\\-1&0\end{pmatrix}}}$   and   ${\displaystyle B={\begin{pmatrix}2&3\\-4&-5\end{pmatrix}}}$

are similar to each other because with the regular matrix

${\displaystyle S={\begin{pmatrix}2&1\\3&2\end{pmatrix}}}$

applies

${\displaystyle S^{-1}AS={\begin{pmatrix}2&-1\\-3&2\end{pmatrix}}\cdot {\begin{pmatrix}-3&2\\-1&0\end{pmatrix}}\cdot {\begin{pmatrix}2&1\\3&2\end{pmatrix}}={\begin{pmatrix}2&-1\\-3&2\end{pmatrix}}\cdot {\begin{pmatrix}0&1\\-2&-1\end{pmatrix}}={\begin{pmatrix}2&3\\-4&-5\end{pmatrix}}=B}$.

The matrix is not unique, because each multiples with meets these identity. ${\displaystyle S}$ ${\displaystyle cS}$ ${\displaystyle c\neq 0}$

properties

Parameters

Two matrices that are similar to one another have the same characteristic polynomial , because it holds with the commutativity of the identity matrix , the determinant product theorem and the determinant of the inverse ${\displaystyle A,B\in K^{n\times n}}$ ${\displaystyle I\in K^{n\times n}}$

${\displaystyle {\begin{aligned}\chi _{B}(\lambda )&=\det(\lambda I-B)=\det(\lambda I-S^{-1}AS)=\det(S^{-1}\lambda IS-S^{-1}AS)=\\&=\det(S^{-1}(\lambda I-A)S)=\det(S^{-1})\det(\lambda I-A)\det(S)=\det(\lambda I-A)=\chi _{A}(\lambda ).\end{aligned}}}$

Therefore have matrices similar to one another

• the same eigenvalues (but not necessarily the same eigenvectors ),
• the same determinant and
• the same trace .

They also have matrices that are similar to one another

• the same rank ,
• the same minimal polynomial and
• the same Jordanian normal form .

characterization

Two complex matrices are similar to each other if and only if they have the same Jordanian normal form (except for the order of the Jordan blocks).

In general, according to the lemma of Frobenius two matrices and if and similar to each other if they have the same Frobenius normal form possess. This is the case if and only if their characteristic matrices and have the same Smith normal form . ${\displaystyle A}$ ${\displaystyle B}$ ${\displaystyle xI-A}$ ${\displaystyle xI-B}$

Equivalence classes

The similarity of matrices is an equivalence relation , i.e. reflexive , symmetric and transitive . One writes

${\displaystyle A\sim B}$,

when and are similar to each other and recorded to a matrix corresponding equivalence class by ${\displaystyle A}$ ${\displaystyle B}$ ${\displaystyle A\in K^{n\times n}}$

${\displaystyle [A]=\{B\in K^{n\times n}\mid B\sim A\}}$.

For example, there is the equivalence class of up to a multiple of the identity matrix -like arrays of exactly one element , because regular for all matrices . ${\displaystyle cI}$ ${\displaystyle c\in K}$ ${\displaystyle I\in K^{n\times n}}$ ${\displaystyle \left[cI\right]=\{cI\}}$ ${\displaystyle S^{-1}(cI)S=cI}$ ${\displaystyle S\in K^{n\times n}}$

The similarity of matrices is a special case of the more generally defined equivalence on the class of matrices. ${\displaystyle (m\times n)}$

Calculation of the transformation matrix

method

Given two matrices that are similar to one another , a matrix with which applies can be determined as follows. First, the two matrices and are converted into the same Frobenius normal form (or, if possible, the same Jordan normal form) . Are the two similarity transformations used for this ${\displaystyle A,B\in K^{n\times n}}$ ${\displaystyle S}$ ${\displaystyle B=S^{-1}AS}$ ${\displaystyle A}$ ${\displaystyle B}$ ${\displaystyle F\in K^{n\times n}}$

${\displaystyle F=G^{-1}AG}$   and   ${\displaystyle F=H^{-1}BH}$

with regular matrices , it follows by equating ${\displaystyle G,H\in K^{n\times n}}$

${\displaystyle B=HG^{-1}AGH^{-1}=\left(GH^{-1}\right)^{-1}A\left(GH^{-1}\right)}$.

The transformation matrix we are looking for is therefore

${\displaystyle S=GH^{-1}}$.

example

Let the two matrices and be given as in the example above. The characteristic polynomials of the two matrices result in ${\displaystyle (2\times 2)}$ ${\displaystyle A}$ ${\displaystyle B}$

${\displaystyle \chi _{A}(\lambda )=\det(\lambda I-A)=(\lambda +3)\lambda +2=(\lambda +2)(\lambda +1)}$

and

${\displaystyle \chi _{B}(\lambda )=\det(\lambda I-B)=(\lambda -2)(\lambda +5)+12=(\lambda +2)(\lambda +1)}$.

The two characteristic polynomials therefore agree, with the eigenvalues ​​being and . Because the characteristic polynomial completely breaks down into real linear factors, the same Jordan normal form can be set up for both matrices, which in this case is the diagonal form ${\displaystyle \lambda _{1}=-2}$ ${\displaystyle \lambda _{2}=-1}$

${\displaystyle F={\begin{pmatrix}-2&0\\0&-1\end{pmatrix}}}$

Has. The transformation matrices into the Jordan normal form have the form and , with eigenvectors for the eigenvalue and eigenvectors for the eigenvalue . For there are two eigenvectors by solving and as ${\displaystyle G=(v_{1}\mid v_{2})}$ ${\displaystyle H=(w_{1}\mid w_{2})}$ ${\displaystyle v_{1},w_{1}}$ ${\displaystyle \lambda _{1}=-2}$ ${\displaystyle v_{2},w_{2}}$ ${\displaystyle \lambda _{2}=-1}$ ${\displaystyle A}$ ${\displaystyle (-2I-A)v_{1}=0}$ ${\displaystyle (-I-A)v_{2}=0}$

${\displaystyle v_{1}={\begin{pmatrix}2\\1\end{pmatrix}}}$   and   . ${\displaystyle v_{2}={\begin{pmatrix}1\\1\end{pmatrix}}}$

Correspondingly, for two eigenvectors, by solving and as ${\displaystyle B}$ ${\displaystyle (2I+B)w_{1}=0}$ ${\displaystyle (I+B)w_{2}=0}$

${\displaystyle w_{1}={\begin{pmatrix}3\\-4\end{pmatrix}}}$   and   . ${\displaystyle w_{2}={\begin{pmatrix}1\\-1\end{pmatrix}}}$

The two transformation matrices into the Jordan normal form are accordingly ${\displaystyle F}$

${\displaystyle G={\begin{pmatrix}2&1\\1&1\end{pmatrix}}}$   and   , ${\displaystyle H={\begin{pmatrix}3&1\\-4&-1\end{pmatrix}}}$

and the similarity transformation matrix we are looking for is thus

${\displaystyle S=GH^{-1}={\begin{pmatrix}2&1\\1&1\end{pmatrix}}\cdot {\begin{pmatrix}-1&-1\\4&3\end{pmatrix}}={\begin{pmatrix}2&1\\3&2\end{pmatrix}}}$.

See also

• Congruence (matrix)

literature

• Gerd Fischer : Linear Algebra. 18th edition. Springer Spectrum, 2014, ISBN 978-3-8348-0996-4 .