Math853:6-7-13

3104 days ago by jimlb

A=Matrix(QQ,[[3,3,0,0,0,-1,0,2],[-3,4,1,-1,-1,0,1,-1],[0,6,3,0,0,-2,0,-4],[-2,4,0,1,-1,0,2,-5],[-3,2,1,-1,2,0,1,-2],[-1,1,0,-1,-1,3,1,-1],[-5,10,1,-3,-2,-1,6,-10],[-3,2,1,-1,-1,0,1,1]]);A 
       
f=A.characteristic_polynomial();f 
       
f.parent() 
       
f.factor() 
       
alpha=3+2*sqrt(3); 
       
beta=3-2*sqrt(3); 
       
I=matrix.identity(QQ,8);I 
       
B=A-3*I;B 
       
B.nullity() 
       
B.kernel() 
       

This shows we have two vectors, say $v_1$ and $v_2$ in $ker(T-3I)$.

B^2 
       
(B^2).nullity() 
       

This gives two vectors in $ker(T-3I)^2$ that are not in $ker(T-3I)$; call these $v_3$ and $v_4$.

B^3 
       
(B^3).nullity() 
       

Since we are now up to dimension 5, which is the power of $(x-3)$ appearing in the characteristic polynomial, we have all the vectors.