Jordan Form and Mathematica
Jordan Form and Mathematica
In[56]:=
?JordanDecomposition
JordanDecomposition[A] returns { S, J }, where A = S . J .
Inverse[S] and J is the Jordan canonical form of A. All
entries of A must have infinite precision.
In[57]:=
MatrixForm[A]
MatrixForm[JA]
Out[57]=
1 1 -1 0 0 2 0 -1 3
Out[58]=
1 1 0 0 1 0 0 0 2
In[59]:=
{s1,j1}=JordanDecomposition[A]
Out[59]=
{{{1, 0, 0}, {0, 2, 1}, {0, 1, 1}},
{{1, 1, 0}, {0, 1, 0}, {0, 0, 2}}}
In[60]:=
MatrixForm[Inverse[s1].A.s1]
MatrixForm[j1]
Out[60]=
1 1 0 0 1 0 0 0 2
Out[61]=
1 1 0 0 1 0 0 0 2
In[62]:=
MatrixForm[B]
MatrixForm[JB]
Out[62]=
2 -1 2 0 0 3 -1 0 0 1 1 0 0 1 -3 5
Out[63]=
2 1 0 0 0 2 1 0 0 0 2 0 0 0 0 5
In[64]:=
{s2,j2}=JordanDecomposition[B]
Out[64]=
2 5
{{{1, 0, 0, 0}, {0, 1, 2, 0}, {0, 1, 1, 0}, {0, -, -, 1}},
3 9
{{2, 1, 0, 0}, {0, 2, 1, 0}, {0, 0, 2, 0}, {0, 0, 0, 5}}}
In[65]:=
MatrixForm[Inverse[s2].B.s2]
MatrixForm[j2]
Out[65]=
2 1 0 0 0 2 1 0 0 0 2 0 0 0 0 5
Out[66]=
2 1 0 0 0 2 1 0 0 0 2 0 0 0 0 5
In[67]:=
MatrixForm[a]
MatrixForm[Ja]
Out[67]=
2 -4 2 2 -2 0 1 3 -2 -2 3 3 -2 -6 3 7
Out[68]=
4 1 0 0 0 4 0 0 0 0 2 0 0 0 0 2
In[69]:=
{s3,j3}=JordanDecomposition[a]
Out[69]=
{{{2, 0, 0, 1}, {1, 1, 1, -1}, {0, 2, 1, -1},
{2, 0, 1, 0}}, {{2, 0, 0, 0}, {0, 2, 0, 0},
{0, 0, 4, 1}, {0, 0, 0, 4}}}
In[70]:=
MatrixForm[Inverse[s3].a.s3]
MatrixForm[j3]
Out[70]=
2 0 0 0 0 2 0 0 0 0 4 1 0 0 0 4
Out[71]=
2 0 0 0 0 2 0 0 0 0 4 1 0 0 0 4
Why is this slightly different? It is simply because of the ordering of the columns in S (or s3 here). Let's rearrange s3.
In[72]:=
MatrixForm[s3]
Out[72]=
2 0 0 1 1 1 1 -1 0 2 1 -1 2 0 1 0
In[73]:=
newS={{0,1,2,0},{1,-1,1,1},{1,-1,0,2},{1,0,2,0}};
MatrixForm[newS]
Out[73]=
0 1 2 0 1 -1 1 1 1 -1 0 2 1 0 2 0
In[74]:=
MatrixForm[Inverse[newS].a.newS]
Out[74]=
4 1 0 0 0 4 0 0 0 0 2 0 0 0 0 2
Up to Jordan Canonical Form