Jordan decomposition example of matrix

Matrix Jordan decompose ： Standard + Transformation matrix

〇、 subject

The subject comes from 《 Computer Science Computing 》 The second edition , Editor Zhang Hongwei , Golden sun , Shi Jilin , Dong Bo . book P86 The first 12 topic .

O matrix $A=\left[\begin{array}{cccc}4& -1& -1& 0\\ 4& 0& -2& 0\\ 0& 0& 2& 0\\ 0& 0& 6& 1\end{array}\right]$ Of Jordan decompose .

One 、 Find its Jordan Standard

Calculation $det(\lambda I-A)=\left|\begin{array}{cccc}\lambda -4& 1& 1& 0\\ -4& \lambda & 2& 0\\ 0& 0& \lambda -2& 0\\ 0& 0& -6& \lambda -1\end{array}\right|$.

The eigenvalue of the solution is ${\lambda }_1=1$（ A multiplicity of algebra ）,${\lambda }_2=2$（ Triple algebraic multiplicity 、 Order ）.

Calculation $rank({\lambda }_{2}I-A)=2$, obtain ${\lambda }_2$ The geometric multiplicity of is 2, Namely its Jordan The number of blocks is 2, Because the order is 3, You can get two of them Jordan The block must be $1+2$ The format of .
Available Jordan The standard type is ：$J=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 2& 0& 0\\ 0& 0& 2& 1\\ 0& 0& 0& 2\end{array}\right]$.

Two 、 Find its transformation matrix T

By definition $A\cdot (T_1,T_2,T_3,T_4)=(T_1,T_2,T_3,T_4)\cdot J$

Two 、 One 、 The first eigenvalue

The following solution , Superscript indicates the block serial number , Subscript indicates the number in the block .

about ${\lambda }_1=1$, Find its linearly independent eigenvector .
$A\cdot t^1={\lambda }_1\cdot t^1$, One vector of the solution is $t^1=(0,0,0,1{)}^T$, Because it is a geometric multiplicity , You can do it directly Jordan Head of chain .

Two 、 Two 、 The second eigenvalue

Empathy , solve equations $(A-{\lambda }_{2}I)\cdot t^2=0$

$\left[\begin{array}{cccc}2& -1& -1& 0\\ 4& -2& -2& 0\\ 0& 0& 0& 0\\ 0& 0& 6& 1\end{array}\right]\cdot t^2=0$

Solution ：$t_1^3=t_1^2=(1,1,1,6{)}^T,t_2^3=t_2^2=(0,0,1,6{)}^T$

Previously known ${\lambda }_2=2$ It's divided into two parts , The order of a piece ( chain length ) by 1, The order of a piece is 2.
Here we can directly get the order 1 Get the head of the chain ,$t^2=(1,1,1,6{)}^T$

For chain length 2 Chain , In order to ensure that the second ring can be launched from the head of the chain , That is, the following equation has a solution .$(ybychainThe\; first{t}_1^3,zbyThe\; firstTwoRing{t}_2^3)$
$$(A-{\lambda }_{2}I)\cdot z=y$$

Make $y=k_1\cdot t_1^3+k_2\cdot t_2^3=(k_1+k_2,2k_1-k_2,k_2,6k_2{)}^T$

The condition of its solution is $r(A-{\lambda }_{2}I)=r((A-{\lambda }_{2}I)|y)$.

Obtainable $k_2=0,k_1=1$, namely $y=(1,2,0,0{)}^T$, Substituting into the original equation , You can get $z=(1,1,0,0{)}^T$.

The transformation matrix obtained by synthesis is ：$T=\left[\begin{array}{cccc}0& 1& 1& 1\\ 0& 1& 2& 1\\ 0& 1& 0& 0\\ 1& 6& 0& 0\end{array}\right]$.

3、 ... and 、 Scientific checking

Use an online calculator Cloud operator , verification $A=TJ{T}^{-1}$.

Correct checking calculation , The following is its $T^{-1}$ value .

$T^{-1}=\left[\begin{array}{cccc}0& 0& -6& 1\\ 0& 0& 1& 0\\ -1& 1& 0& 0\\ 2& -1& -1& 0\end{array}\right]$