Jordan normal form

The Jordanian normal form is a term from the mathematical branch of linear algebra . It was named after Marie Ennemond Camille Jordan , who derived it in 1870 for finite bodies and in 1871 in connection with the solution of complex differential equation systems for complex matrices, but who was also known to Karl Weierstrass in 1868 for his treatment of bilinear forms in the complex. The Jordanian normal form is a simple representative of the equivalence class of matrices that are similar to a trigonalizable matrix . Trigonalization is synonymous with the fact that thecharacteristic polynomial of the matrix breaks down completely into linear factors . Matrices over an algebraically closed field are always trigonalizable and therefore always similar to a Jordanian normal form.

For each linear mapping of a finite-dimensional vector space, the characteristic polynomial of which is completely broken down into linear factors, a vector space basis can be selected so that the mapping matrix that describes the mapping with respect to this basis has Jordanian normal form.

The rational normal form or Frobenius normal form provides a standardized representative of the similarity class of this matrix for any matrix, even if it cannot be trigonalized .

definition

The Jordanian normal form of a square matrix over the complex numbers is a matrix in the following block diagonal form : ${\displaystyle n\times n}$ ${\displaystyle A}$ ${\displaystyle \mathbb {C} }$ ${\displaystyle J}$

${\displaystyle J={\begin{pmatrix}J_{1}&&0\\&\ddots &\\0&&J_{k}\end{pmatrix}}=Q^{-1}AQ.}$

The matrix is the matrix of the eigenvectors and principal vectors that make it up in columns. denotes the inverse matrix of . The matrices are called Jordan blocks or Jordan boxes and they are bidiagonal matrices with the following shape: ${\displaystyle Q}$ ${\displaystyle Q^{-1}}$ ${\displaystyle Q}$ ${\displaystyle J_{j}}$

${\displaystyle J_{j}={\begin{pmatrix}\lambda _{j}&1&&&0\\&\lambda _{j}&1&&\\&&\ddots {}&\ddots {}\\&&&\lambda _{j}&1\\0&&&&\lambda _{j}\end{pmatrix}}\in \mathbb {C} ^{s_{j}\times s_{j}}.}$

They are the eigenvalues of . For each eigenvalue there are many Jordan blocks corresponding to its geometric multiplicity . The geometric multiplicity is the dimension of the eigen-space to the eigen-value . The total dimension of all Jordan blocks of an eigenvalue corresponds to its algebraic multiplicity , i.e. H. its multiplicity in the characteristic polynomial. ${\displaystyle \lambda _{j}}$ ${\displaystyle A}$ ${\displaystyle \lambda _{j}}$ ${\displaystyle \lambda _{j}}$

In a Jordan block the so-called are Jordan chains "stored" (see Major Vector ). Is there e.g. B. only from a Jordan block with eigenvalue and denote a principal vector -th degree. There is an eigenvector for the eigenvalue and it holds and for . Then and for , that is, the mapping matrix with respect to the base is actually a Jordan block. ${\displaystyle A}$ ${\displaystyle \lambda }$ ${\displaystyle v_{l}}$ ${\displaystyle l}$ ${\displaystyle v_{1}}$ ${\displaystyle \lambda }$ ${\displaystyle (A-\lambda E)v_{1}=0}$ ${\displaystyle (A-\lambda E)v_{l}=v_{l-1}}$ ${\displaystyle l=2,\dots ,n}$ ${\displaystyle Av_{1}=\lambda v_{1}}$ ${\displaystyle Av_{l}=v_{l-1}+\lambda v_{l}}$ ${\displaystyle l=2,\dots ,n}$ ${\displaystyle (v_{1},\dotsc ,v_{n})}$

There is also an alternative representation of the Jordan blocks with 1 in the lower secondary diagonal .

In the special case of a diagonalizable matrix, the Jordanian normal form is a diagonal matrix .

Shape of the transformation matrix

There are principal vectors of the respective th stage, wherein the dimension of the -th Jordan block . ${\displaystyle v_{j,1},\ldots ,v_{j,l},\ldots ,v_{j,s_{j}}}$ ${\displaystyle l}$ ${\displaystyle s_{j}}$ ${\displaystyle j}$ ${\displaystyle j=1,\dotsc ,k}$

Then is defined by ${\displaystyle Q}$

${\displaystyle Q:=(v_{1,1}|\ldots |v_{1,s_{1}}|\ldots |v_{k,1}|\ldots |v_{k,s_{k}})}$

a transformation matrix that uses the Jordan normal form of . ${\displaystyle Q^{-1}AQ=J}$ ${\displaystyle J}$ ${\displaystyle A}$

In words: The columns of are the eigenvectors with the associated main vectors in the order of the associated Jordan blocks. However, it is not clearly determined. ${\displaystyle Q}$ ${\displaystyle Q}$

Algorithm for determining a complex Jordanian normal form

For Jordan normal form of an endomorphism of a -dimensional -Vektorraums to choose a basis of a vector space , and calculates the Jordan normal form of the imaging array of respect to the base . ${\displaystyle u\colon V\to V}$ ${\displaystyle n}$ ${\displaystyle \mathbb {C} }$ ${\displaystyle V}$ ${\displaystyle B=(b_{1},\ldots ,b_{n})}$ ${\displaystyle V}$ ${\displaystyle A=M_{B}(u)}$ ${\displaystyle u}$ ${\displaystyle B}$

The following is therefore set and the complex Jordanian normal form of a square matrix is determined. The identity matrix is denoted by. ${\displaystyle V=\mathbb {C} ^{n}}$ ${\displaystyle A\in \mathbb {C} ^{n\times n}}$ ${\displaystyle E_{n}}$

Determination of the eigenvalues

With the help of the characteristic polynomial

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

the pairwise different eigenvalues ​​are calculated from its zeros

${\displaystyle \lambda _{1},\ldots ,\lambda _{k}\in \mathbb {C} \,.}$

The eigenvalues ​​are therefore not listed here according to their multiplicity.

Determination of the size of the Jordan blocks

To do this, the dimensions of the generalized eigenspaces must first be determined. That is, one calculates the numbers for all of them ${\displaystyle 1\leq j\leq k}$

${\displaystyle a_{s}:=\dim \operatorname {Kern} (A-\lambda _{j}E_{n})^{s},\quad s\in \mathbb {N} _{0}\,.}$

In particular, is always and is precisely the geometric multiplicity of the eigenvalue . The dimension of the core can be calculated from the rank with the aid of the dimension set , which can be determined, for example, with the Gaussian algorithm . ${\displaystyle a_{0}=0}$ ${\displaystyle a_{1}}$ ${\displaystyle \lambda _{j}}$

The sequence of is increasing monotonically and becomes stationary from a certain value , at the latest when the algebraic multiplicity of the eigenvalue in the characteristic polynomial is reached . The number of Jordan blocks of size to the eigenvalue can then be calculated using the formula ${\displaystyle a_{s}}$ ${\displaystyle s}$ ${\displaystyle s}$ ${\displaystyle \lambda _{j}}$

${\displaystyle 2a_{s}-a_{s-1}-a_{s+1}}$

to calculate. It also gives the total number of Jordan blocks belonging to this eigenvalue. ${\displaystyle a_{1}}$

Complex Jordanian normal form

The Jordan blocks obtained are written in a matrix and the complex Jordanian normal form of a matrix is ​​obtained. If all blocks have the size 1, there is the special case of a diagonal matrix and can therefore be diagonalized. ${\displaystyle A}$

The minimal polynomial of is obtained from , where denotes the size of the largest Jordan block for the eigenvalue . ${\displaystyle g\in \mathbb {C} [X]}$ ${\displaystyle A}$ ${\displaystyle g=\prod _{j=1}^{k}(X-\lambda _{j})^{m_{j}}}$ ${\displaystyle m_{j}}$ ${\displaystyle \lambda _{j}}$

The Jordanian normal form is uniquely determined except for the order of the Jordan blocks. If all eigenvalues are in, two matrices that have the same Jordanian normal form are similar to each other. ${\displaystyle \mathbb {K} }$

example

Consider the matrix which is defined as follows ${\displaystyle A\in \mathbb {C} ^{5\times 5}}$

${\displaystyle A:={\begin{pmatrix}25&-16&30&-44&-12\\13&-7&18&-26&-6\\-18&12&-21&36&12\\-9&6&-12&21&6\\11&-8&15&-22&-3\end{pmatrix}}.}$

Its characteristic polynomial is . Thus this matrix has exactly one eigenvalue, namely 3. With the abbreviation the following are determined: ${\displaystyle \chi _{A}=(X-3)^{5}}$ ${\displaystyle B:=A-3E_{5}}$ ${\displaystyle a_{s}}$

It applies . So is . ${\displaystyle \mathrm {rg} (B)=2}$ ${\displaystyle a_{1}=\dim(V)-\mathrm {rg} (B)=5-2=3}$

Furthermore, the zero matrix is valid, and therefore and the sequence becomes stationary from this point on. ${\displaystyle B^{2}}$ ${\displaystyle \mathrm {rg} (B^{2})=0}$ ${\displaystyle a_{2}=5-0=5}$ ${\displaystyle a_{s}}$

So it follows:

There are Jordan blocks, and of them ${\displaystyle a_{1}=3}$

${\displaystyle 2a_{1}-a_{0}-a_{2}=6-0-5=1}$ Jordan block with size 1 and

${\displaystyle 2a_{2}-a_{1}-a_{3}=10-3-5=2}$ Jordan blocks with size 2.

This is the Jordanian normal form of . The minimal polynomial of is . ${\displaystyle {\begin{pmatrix}3&1&0&0&0\\0&3&0&0&0\\0&0&3&1&0\\0&0&0&3&0\\0&0&0&0&3\end{pmatrix}}}$ ${\displaystyle A}$ ${\displaystyle A}$ ${\displaystyle (X-3)^{2}}$

Determination of a basic transformation to the complex Jordanian normal form

Now a basic transformation matrix is ​​to be determined which ${\displaystyle P\in {\rm {GL}}(n;\mathbb {C} )}$

${\displaystyle J=P^{-1}AP}$

Fulfills. It is known that it is not clearly determined by this equation. The standard procedure uses previous knowledge of the complex Jordanian normal form . ${\displaystyle J}$

A standard procedure

A common procedure for obtaining a basis transformation is as follows: First determine (as with the naive approach above) the Jordan normal form . Then in particular all eigenvalues ​​have already been calculated as well as the kernels for all in which the dimension of the largest Jordan block denotes the eigenvalue . Subsequently, work for determining a regular matrix with the blocks in succession. It should be noted that with Jordan blocks with the same eigenvalue one always proceeds from the largest block to the smallest block. ${\displaystyle J}$ ${\displaystyle \lambda }$ ${\displaystyle {\rm {Kern}}(A-\lambda I)^{k}}$ ${\displaystyle 1\leq k\leq m(\lambda )}$ ${\displaystyle m(\lambda )\in \mathbb {N} }$ ${\displaystyle \lambda }$ ${\displaystyle P}$ ${\displaystyle J=P^{-1}AP}$

For each block of size and eigenvalue are columns of the base transformation matrix according to a certain scheme determined. If the block occupies the columns , the vectors are also inserted into the columns (from left to right) . The vectors are now determined as follows: ${\displaystyle s}$ ${\displaystyle \lambda }$ ${\displaystyle s}$ ${\displaystyle v^{1},\ldots ,v^{s}}$ ${\displaystyle J}$ ${\displaystyle m,\ldots ,m+s-1}$ ${\displaystyle v^{1},\ldots ,v^{s}}$ ${\displaystyle P}$ ${\displaystyle m,\ldots ,m+s-1}$ ${\displaystyle v^{1},\ldots ,v^{s}}$

• One can choose arbitrarily, in which denotes the set of previously calculated columns (ie basis vectors) of the level from previously processed Jordan blocks for the same eigenvalue (if any). This relatively free choice in particular shows that the basic transformation cannot be unique. If is simply an eigenvector to the eigenvalue ${\displaystyle v^{s}\in {\rm {Kern}}(A-\lambda I)^{s}\setminus [{\rm {Span}}({\rm {Kern}}(A-\lambda I)^{s-1}\cup M)]}$ ${\displaystyle M}$ ${\displaystyle s}$ ${\displaystyle \lambda }$ ${\displaystyle s=1,}$ ${\displaystyle v^{1}}$ ${\displaystyle \lambda .}$
• After choosing the above vector, there is no longer any freedom of choice for the other basic vectors: You have to set successively for all of them. ${\displaystyle v^{j}:=(A-\lambda I)v^{j+1}}$ ${\displaystyle j=s-1,\ldots ,1}$

After all Jordan blocks have been processed in the above way, all columns of have been filled in at the end . The following applies: is regular and satisfied , and its columns form a basis with respect to which the representation possesses. ${\displaystyle P}$ ${\displaystyle P}$ ${\displaystyle P^{-1}AP=J}$ ${\displaystyle A}$ ${\displaystyle J}$

If the alternative representation of the Jordan blocks is selected, i. H. with 1 in the lower secondary diagonal, only the order of the basis vectors per Jordan block has to be reversed.

example

As an illustrative example, consider the matrix

${\displaystyle A:={\begin{pmatrix}25&-16&30&-44&-12\\13&-7&18&-26&-6\\-18&12&-21&36&12\\-9&6&-12&21&6\\11&-8&15&-22&-3\end{pmatrix}}}$

as above. It applies

${\displaystyle {\rm {Kern}}(A-3I)={\rm {Span}}\left(\left\{{\begin{pmatrix}-2\\1\\2\\0\\0\\\end{pmatrix}},{\begin{pmatrix}2\\0\\0\\1\\0\\\end{pmatrix}},{\begin{pmatrix}2\\2\\0\\0\\1\\\end{pmatrix}}\right\}\right)}$and . ${\displaystyle {\rm {Kern}}(A-3I)^{2}=\mathbb {C} ^{5}}$

Its normal Jordan form is

${\displaystyle J={\begin{pmatrix}3&1&0&0&0\\0&3&0&0&0\\0&0&3&1&0\\0&0&0&3&0\\0&0&0&0&3\end{pmatrix}}}$.

Start with the first Jordan block of dimension 2. To do this, choose

${\displaystyle v^{2}\in {\rm {Kern}}(A-3I)^{2}\setminus {\rm {Kern}}(A-3I)^{1}}$

any, for example . Then choose. From this you get . Now move on to the second Jordan block, size 2. Choose now ${\displaystyle v^{2}:={\begin{pmatrix}1\\0\\0\\0\\0\\\end{pmatrix}}}$ ${\displaystyle v^{1}:=(A-3I)v^{2}={\begin{pmatrix}22\\13\\-18\\-9\\11\\\end{pmatrix}}}$ ${\displaystyle P:={\begin{pmatrix}22&1&*&*&*\\13&0&*&*&*\\-18&0&*&*&*\\-9&0&*&*&*\\11&0&*&*&*\end{pmatrix}}}$

${\displaystyle w^{2}\in {\rm {Kern}}(A-3I)^{2}\setminus {\rm {Span}}({\rm {Kern}}(A-3I)^{1}\cup \{v^{2}\})}$

any, for example . Then is , and you end up with . Finally, it is the turn of the last Jordan block (size 1). Choose for this ${\displaystyle w^{2}:={\begin{pmatrix}0\\1\\0\\0\\0\\\end{pmatrix}}}$ ${\displaystyle w^{1}:=(A-3I)w^{2}={\begin{pmatrix}-16\\-10\\12\\6\\-8\\\end{pmatrix}}}$ ${\displaystyle P={\begin{pmatrix}22&1&-16&0&*\\13&0&-10&1&*\\-18&0&12&0&*\\-9&0&6&0&*\\11&0&-8&0&*\end{pmatrix}}}$

${\displaystyle x^{1}\in {\rm {Kern}}(A-3I)^{1}\setminus {\rm {Span}}(\{v^{1},w^{1}\})}$

any, for example . Then there is a regular matrix with . ${\displaystyle x^{1}:={\begin{pmatrix}2\\0\\0\\1\\0\\\end{pmatrix}}}$ ${\displaystyle P={\begin{pmatrix}22&1&-16&0&2\\13&0&-10&1&0\\-18&0&12&0&0\\-9&0&6&0&1\\11&0&-8&0&0\end{pmatrix}}}$ ${\displaystyle J=P^{-1}AP}$

Real Jordanian normal form

If one considers real matrices, their characteristic polynomial generally no longer breaks down completely into linear factors, but only into irreducible factors, which in this case are always linear or quadratic factors. The question of a normal form now arises if only real basis transformations are allowed.

A quadratic irreducible factor with is defined as a Jordan block ${\displaystyle (\lambda -a_{j})^{2}+b_{j}^{2}}$ ${\displaystyle b_{j}>0}$

${\displaystyle J_{j}={\begin{pmatrix}a_{j}&b_{j}&1&0&&&&0\\-b_{j}&a_{j}&0&1&&&&\\&&a_{j}&b_{j}&1&0&&\\&&-b_{j}&a_{j}&0&1&&\\&&&\ddots {}&\ddots {}&\ddots {}&1&0\\&&&&\ddots {}&\ddots {}&0&1\\&&&&&\ddots {}&a_{j}&b_{j}\\0&&&&&&-b_{j}&a_{j}\\\end{pmatrix}}.}$

We call the number of rows (or columns) the size of this block. Then one designates

${\displaystyle J={\begin{pmatrix}J_{1}&&0\\&\ddots {}&\\0&&J_{k}\end{pmatrix}}=P^{-1}AP}$

as a real Jordanian normal form. To find it and a suitable real matrix , one can proceed as follows: ${\displaystyle P\in \mathbb {R} ^{n\times n}}$

• Find the characteristic polynomial and factor it into irreducible factors. It turns out

${\displaystyle \chi (\lambda )=\prod _{j=1}^{k}\left(\lambda -\lambda _{j}\right)^{\mu _{j}}\cdot \prod _{j=1}^{l}\left(\left(\lambda -a_{j}\right)^{2}+b_{j}^{2}\right)^{\nu _{j}}}$,

where pairwise denote different eigenvalues ​​with multiplicity . Next were in it , , and pairwise different. ${\displaystyle \lambda _{1},\ldots ,\lambda _{k}\in \mathbb {R} }$ ${\displaystyle \mu _{j}\in \mathbb {N} }$ ${\displaystyle a_{1},\ldots ,a_{l}\in \mathbb {R} }$ ${\displaystyle b_{1},\ldots ,b_{l}>0}$ ${\displaystyle \nu _{1},\ldots ,\nu _{l}\in \mathbb {N} }$ ${\displaystyle (a_{1},b_{1}),\ldots ,(a_{l},b_{l})}$

• Determine for each ${\displaystyle j\in \{1,\ldots ,k\}}$

${\displaystyle K_{j,m}:={\rm {Kern}}_{\mathbb {R} }(A-\lambda _{j}E)^{m}}$for , ${\displaystyle m=1,2,\ldots ,m_{j}}$

where is the smallest natural number with . Determine analogously for each ${\displaystyle m_{j}\leq \mu _{j}}$ ${\displaystyle \dim _{\mathbb {R} }K_{j,m_{j}}=\mu _{j}}$ ${\displaystyle j\in \{1,\ldots ,l\}}$

${\displaystyle K^{j,m}:={\rm {Kern}}_{\mathbb {R} }\left(\left(A-a_{j}E\right)^{2}+b_{j}^{2}E\right)^{m}}$for , ${\displaystyle m=1,2,\ldots ,n_{j}}$

where is the smallest natural number with . ${\displaystyle n_{j}\leq \nu _{j}}$ ${\displaystyle \dim _{\mathbb {R} }K^{j,n_{j}}=2\nu _{j}}$

We also bet . ${\displaystyle K_{j,0}:=K^{j,0}:=\{0\}}$

• Now set up the Jordanian normal form. It applies here
  • ${\displaystyle \ \dim _{\mathbb {R} }K_{j,m}-\dim _{\mathbb {R} }K_{j,m-1}}$is the number of Jordan blocks for the eigenvalue whose size is greater than or equal . ${\displaystyle \lambda _{j}}$ ${\displaystyle m}$
  • ${\displaystyle {\frac {1}{2}}\left(\dim _{\mathbb {R} }K^{j,m}-\dim _{\mathbb {R} }K^{j,m-1}\right)}$is the number of Jordan blocks for the factor whose size is greater or equal . ${\displaystyle (\lambda -a_{j})^{2}+b_{j}^{2}}$ ${\displaystyle 2m}$

In addition, the sum of the Jordan block sizes is the eigenvalue and the sum of the Jordan block sizes is the factor . From this information one can clearly determine the Jordanian normal form . ${\displaystyle \mu _{j}}$ ${\displaystyle \lambda _{j}}$ ${\displaystyle 2\nu _{j}}$ ${\displaystyle (\lambda -a_{j})^{2}+b_{j}^{2}}$ ${\displaystyle J}$

• Then determine the basis transformation matrix , that is, one looks for a real invertible matrix such that . ${\displaystyle P}$ ${\displaystyle P\in \mathbb {R} ^{n\times n}}$ ${\displaystyle J=P^{-1}AP}$

One method to get a base transformation is as follows:

• Work through the blocks one after the other. It should be noted that with Jordan blocks for the same irreducible factor one always proceeds from the largest block to the smallest block. For each block of size are columns of the base transformation matrix determined according to a certain scheme. If the block occupies the columns , the vectors are also inserted into the columns (from left to right) . The vectors are now determined as follows: ${\displaystyle t}$ ${\displaystyle t}$ ${\displaystyle v^{1},\ldots ,v^{t}}$ ${\displaystyle J}$ ${\displaystyle m,\ldots ,m+t-1}$ ${\displaystyle v^{1},\ldots ,v^{t}}$ ${\displaystyle P}$ ${\displaystyle m,\ldots ,m+t-1}$ ${\displaystyle v^{1},\ldots ,v^{t}}$
  • For a Jordan block of the size of the eigenvalue, one can arbitrarily choose , in which the set of previously calculated columns (i.e. basis vectors) of the level from previously processed Jordan blocks for the same eigenvalue (if any) is designated. Then you bet successively for all . ${\displaystyle m}$ ${\displaystyle \lambda _{j}}$ ${\displaystyle v^{m}\in K_{j,m}\setminus [{\rm {Span}}_{\mathbb {R} }(K_{j,m-1}\cup M)]}$ ${\displaystyle M}$ ${\displaystyle m}$ ${\displaystyle \lambda }$ ${\displaystyle v^{t-1}:=(A-\lambda _{j}E)v^{t}}$ ${\displaystyle t=m,\ldots ,2}$
  • For a Jordan block of the size of the irreducible factor, choose a vector , whereby the main vectors of the steps already calculated for the same irreducible factor are made up. ${\displaystyle 2m}$ ${\displaystyle \left(\left(\lambda -a_{j}\right)^{2}+b_{j}^{2}\right)}$ ${\displaystyle v^{2m}\in K^{j,m}\setminus [{\rm {Span}}_{\mathbb {R} }(K^{j,m-1}\cup M)]}$ ${\displaystyle M}$ ${\displaystyle 2m,2m-1}$ ${\displaystyle \left(\left(\lambda -a_{j}\right)^{2}+b_{j}^{2}\right)}$

Then you set for successively ${\displaystyle t=2m,\ldots ,2}$ ${\displaystyle v^{t-1}:=\left\{{\begin{array}{ll}{\frac {1}{b_{j}}}(A-a_{j}E)v^{t}\ ,&{\textrm {falls}}\ t\ {\textrm {gerade}}\ ,\\(A-a_{j}E)v^{t}+b_{j}v^{t+1}\ ,&{\textrm {falls}}\ t\ {\textrm {ungerade}}\ .\\\end{array}}\right.}$

Finally, as usual, you put the vectors together. ${\displaystyle P}$ ${\displaystyle v^{1},\ldots ,v^{2m}}$

• After you have processed all Jordan blocks in the above way, all columns are filled by at the end . The following applies: is regular and satisfied , and its columns form a basis with respect to which the representation possesses. ${\displaystyle P}$ ${\displaystyle P}$ ${\displaystyle P^{-1}AP=J}$ ${\displaystyle A}$ ${\displaystyle J}$

example

Consider the matrix which is defined as follows ${\displaystyle B\in M_{5}(\mathbb {R} )}$

${\displaystyle B:={\begin{pmatrix}6&-2&6&1&1\\1&-1&2&1&-2\\-2&0&-1&0&-1\\-1&0&-2&2&-1\\-4&4&-6&-2&3\end{pmatrix}}.}$

Its characteristic polynomial is where irreducible is over . Now we calculate the Jordanian normal form: ${\displaystyle \chi (\lambda )=((\lambda -2)^{2}+1)^{2}(\lambda -1)}$ ${\displaystyle (\lambda -2)^{2}+1}$ ${\displaystyle \mathbb {R} }$

${\displaystyle {\rm {Kern}}_{\mathbb {R} }(B-1E)={\rm {Span}}_{\mathbb {R} }\left(\left\{{\begin{pmatrix}1\\-1\\-1\\-1\\0\\\end{pmatrix}}\right\}\right)}$.

This core has dimension 1. So there is only one Jordan block that size . On the other hand, the sum of the Jordan block sizes must be 1 (the power of ), so that there is exactly one Jordan block with eigenvalue 1, and it has size 1. Further ${\displaystyle \geq 1}$ ${\displaystyle \lambda -1}$

${\displaystyle {\rm {Kern}}_{\mathbb {R} }\left(\left(B-2E\right)^{2}+1^{2}E\right)={\rm {Span}}_{\mathbb {R} }\left(\left\{{\begin{pmatrix}2\\0\\-1\\-1\\-1\\\end{pmatrix}},{\begin{pmatrix}0\\1\\0\\1\\-1\\\end{pmatrix}}\right\}\right)}$

the dimension 2, so that there is consequently only Jordan block of the size . Since the sum of the Jordan block sizes must be 4 (twice the power of ), it follows that this one Jordan block is size 4. We also calculate ${\displaystyle {\tfrac {1}{2}}\cdot 2=1}$ ${\displaystyle \geq 2}$ ${\displaystyle (\lambda -2)^{2}+1}$

${\displaystyle {\rm {Kern}}_{\mathbb {R} }\left(\left(B-2E\right)^{2}+1^{2}E\right)^{2}={\rm {Span}}_{\mathbb {R} }\left(\left\{{\begin{pmatrix}1\\0\\0\\0\\-1\\\end{pmatrix}},{\begin{pmatrix}0\\1\\0\\0\\-1\\\end{pmatrix}},{\begin{pmatrix}0\\0\\1\\0\\-1\\\end{pmatrix}},{\begin{pmatrix}0\\0\\0\\1\\0\\\end{pmatrix}}\right\}\right)}$.

Thus is the real Jordanian normal form of . ${\displaystyle J:={\begin{pmatrix}1&&&&\\&2&1&1&0\\&-1&2&0&1\\&&&2&1\\&&&-1&2\end{pmatrix}}}$ ${\displaystyle B}$

For comparison, the complex Jordanian normal form is . ${\displaystyle J_{\mathbb {C} }:={\begin{pmatrix}1&&&&\\&2+i&1&&\\&&2+i&&\\&&&2-i&1\\&&&&2-i\end{pmatrix}}}$

To calculate a basic transformation matrix, start with the first real eigenvalue and then with the (first) Jordan block of dimension 1. Choose

${\displaystyle u^{1}\in {\rm {Kern}}_{\mathbb {R} }(B-1E)^{1}\setminus {\rm {Kern}}_{\mathbb {R} }(B-1E)^{0}}$

any, for example . From this you get . ${\displaystyle u^{1}:={\begin{pmatrix}1\\-1\\-1\\-1\\0\\\end{pmatrix}}}$ ${\displaystyle P:={\begin{pmatrix}1&*&*&*&*\\-1&*&*&*&*\\-1&*&*&*&*\\-1&*&*&*&*\\0&*&*&*&*\end{pmatrix}}}$

Now move on to the first irreducible factor (complex eigenvalue) and then to the Jordan block of size 4. To do this, choose

${\displaystyle v^{4}\in {\rm {Kern}}_{\mathbb {R} }((B-2E)^{2}+1^{2}E)^{2}\setminus {\rm {Kern}}_{\mathbb {R} }((B-2E)^{2}+1^{2}E)^{1}}$

any, for example . Then , and to choose. This yields: . is a regular matrix with . ${\displaystyle v^{4}:={\begin{pmatrix}0\\0\\0\\1\\0\\\end{pmatrix}}}$ ${\displaystyle v^{3}:={\frac {1}{1}}(B-2E)v^{4}={\begin{pmatrix}1\\1\\0\\0\\-2\\\end{pmatrix}}}$ ${\displaystyle v^{2}:=(B-2E)v^{3}+bv^{4}={\begin{pmatrix}0\\2\\0\\2\\-2\\\end{pmatrix}}}$ ${\displaystyle v^{1}:={\frac {1}{1}}(B-2E)v^{2}={\begin{pmatrix}-4\\0\\2\\2\\2\\\end{pmatrix}}}$ ${\displaystyle P:={\begin{pmatrix}1&-4&0&1&0\\-1&0&2&1&0\\-1&2&0&0&0\\-1&2&2&0&1\\0&2&-2&-2&0\end{pmatrix}}}$ ${\displaystyle P}$ ${\displaystyle J=P^{-1}BP}$

Jordan normal form in general solids

The Jordanian normal form can be further generalized to general bodies . In this context it is often referred to as the Weierstraß normal form (or Frobenius normal form ). This allows an unambiguous matrix representation of endomorphisms of finite-dimensional vector spaces, in which all similar endomorphisms can be represented by an unambiguous matrix. In this way, similar linear maps can be identified. The lemma Frobenius characterized matrices similar to each other through the elementary divisors their characteristic matrices and provides the Frobenius normal form as the normal form of the vector space under the operation of a polynomial ring.

Due to the representation in the Weierstrasse normal form, the structure of the minimal polynomial is immediately recognizable and the characteristic polynomial can be easily calculated.

Use in systems of linear differential equations of the first order with constant coefficients

A linear differential equation system (of equations) of the first order with constant coefficients is given ${\displaystyle n}$

${\displaystyle y'=A\cdot y+g(x)}$

by a matrix and a continuous function . It is known that the unique solution to the initial value problem ${\displaystyle A\in \mathbb {C} ^{n\times n}}$ ${\displaystyle g\colon \mathbb {R} \rightarrow \mathbb {C} ^{n}}$

${\displaystyle y(x_{0})=y_{0}\in \mathbb {C} ^{n}}$

is given by

${\displaystyle y(x)=e^{(x-x_{0})A}\cdot y_{0}+\int _{x_{0}}^{x}e^{(x-t)A}g(t){\rm {d}}t\ }$,

wherein

${\displaystyle \exp(B):=e^{B}:=\sum _{k=0}^{\infty }{\frac {1}{k!}}B^{k}}$ For ${\displaystyle B\in \mathbb {C} ^{n\times n}}$

denotes the matrix exponential function . Note:

• The matrix exponential function of a complex Jordan block can be calculated explicitly:

${\displaystyle \exp \left(t\cdot {\begin{pmatrix}\lambda &1&0&\cdots &0\\0&\lambda &1&\ddots &0\\\vdots &\ddots &\ddots &\ddots &\vdots \\0&\cdots &0&\lambda &1\\0&\cdots &\cdots &0&\lambda \end{pmatrix}}\right)=e^{t\lambda }\cdot {\begin{pmatrix}1&{\frac {t^{1}}{1!}}&{\frac {t^{2}}{2!}}&\cdots &{\frac {t^{n-1}}{(n-1)!}}\\0&1&{\frac {t^{1}}{1!}}&\cdots &{\frac {t^{n-2}}{(n-2)!}}\\\vdots &\ddots &\ddots &\ddots &\vdots \\0&\cdots &0&1&{\frac {t^{1}}{1!}}\\0&\cdots &\cdots &0&1\end{pmatrix}}}$.

• The matrix exponential function of a complex Jordan normal form can be calculated explicitly using: ${\displaystyle J={\rm {diag}}(J_{1},\ldots ,J_{m})}$

${\displaystyle \exp(t\cdot {\rm {diag}}(J_{1},\ldots ,J_{m}))={\rm {diag}}(\exp(tJ_{1}),\ldots ,\exp(tJ_{m}))}$.

• The matrix exponential function of a matrix whose complex Jordan normal form is known together with a base transformation matrix , that is , can be calculated explicitly using: ${\displaystyle A}$ ${\displaystyle J}$ ${\displaystyle P\in \mathbb {C} ^{n\times n}}$ ${\displaystyle A=PJP^{-1}}$

${\displaystyle \exp(tA)=\exp(t\cdot PJP^{-1})=P\cdot \exp(tJ)\cdot P^{-1}}$.

In other words: If you know a representation with the complex Jordanian normal form , you can calculate explicitly for each , so that to determine ${\displaystyle A=PJP^{-1}}$ ${\displaystyle J}$ ${\displaystyle \exp(tA)}$ ${\displaystyle t\in \mathbb {R} }$

${\displaystyle y(x)=e^{(x-x_{0})A}\cdot y_{0}+\int _{x_{0}}^{x}e^{(x-t)A}g(t){\rm {d}}t\ }$

only the integration problem remains to be solved, which is completely eliminated in the homogeneous case . ${\displaystyle g=0}$

See also

• Diagonalization is a special case of the Jordanian normal form.
• The Jordan normal form is a special case of the Weierstrass normal form .
• The existence of the Jordanian normal form provides the existence of the (additive) Jordan-Chevalley decomposition of an endomorphism.
• Since the existence of zeros of the characteristic polynomial is decisive for the existence of a Jordanian normal form, the real normal form, as described here, can be more general for affine self-maps of the two- dimensional affine space over a Euclidean space and an affine space with any finite dimension over a real closed body to be determined.

literature

• Herbert Amann: Ordinary differential equations . 2nd Edition. De Gruyter, Berlin 1995, ISBN 3-11-014582-0 .
• Gilbert Strang : Linear Algebra . 1st edition. Springer-Verlag, Berlin 2003, ISBN 3-540-43949-8 . (Literature on eigenvalues ​​and eigenvectors as well as matrix calculation).

Web links

Wikiversity: Lecture on Jordan normal form  - course materials

• Daniel Winkler: Cooking with Jordan. (PDF file; 264 kB)
• Jordan matrix. In: Encyclopaedia of Mathematics . (English)
• Jordan canonical form theorem. In: PlanetMath . (English)
• The Real Jordan Form. In: Number Theory Web. (English; PDF file; 110 kB)
• The yellow arithmetic book / additions: Jordan form of matrices (PDF file; 81 kB)

Individual evidence

1. ^ Wilhelm von Alten u. a., 4000 Years of Algebra, Springer 2008, p. 409