# Math853:6-7-13

## 3269 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
 $\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrrrrr} 3 & 3 & 0 & 0 & 0 & -1 & 0 & 2 \\ -3 & 4 & 1 & -1 & -1 & 0 & 1 & -1 \\ 0 & 6 & 3 & 0 & 0 & -2 & 0 & -4 \\ -2 & 4 & 0 & 1 & -1 & 0 & 2 & -5 \\ -3 & 2 & 1 & -1 & 2 & 0 & 1 & -2 \\ -1 & 1 & 0 & -1 & -1 & 3 & 1 & -1 \\ -5 & 10 & 1 & -3 & -2 & -1 & 6 & -10 \\ -3 & 2 & 1 & -1 & -1 & 0 & 1 & 1 \end{array}\right)$
f=A.characteristic_polynomial();f
 $\newcommand{\Bold}[1]{\mathbf{#1}}x^{8} - 23 x^{7} + 243 x^{6} - 1527 x^{5} + 6165 x^{4} - 16173 x^{3} + 26649 x^{2} - 25029 x + 10206$
f.parent()
 $\newcommand{\Bold}[1]{\mathbf{#1}}\Bold{Q}[x]$
f.factor()
 $\newcommand{\Bold}[1]{\mathbf{#1}}(x - 2) \cdot (x - 3)^{5} \cdot (x^{2} - 6 x + 21)$
alpha=3+2*sqrt(3);
beta=3-2*sqrt(3);
I=matrix.identity(QQ,8);I
 $\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrrrrr} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{array}\right)$
B=A-3*I;B
 $\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrrrrr} 0 & 3 & 0 & 0 & 0 & -1 & 0 & 2 \\ -3 & 1 & 1 & -1 & -1 & 0 & 1 & -1 \\ 0 & 6 & 0 & 0 & 0 & -2 & 0 & -4 \\ -2 & 4 & 0 & -2 & -1 & 0 & 2 & -5 \\ -3 & 2 & 1 & -1 & -1 & 0 & 1 & -2 \\ -1 & 1 & 0 & -1 & -1 & 0 & 1 & -1 \\ -5 & 10 & 1 & -3 & -2 & -1 & 3 & -10 \\ -3 & 2 & 1 & -1 & -1 & 0 & 1 & -2 \end{array}\right)$
B.nullity()
 $\newcommand{\Bold}[1]{\mathbf{#1}}2$
B.kernel()
 $\newcommand{\Bold}[1]{\mathbf{#1}}\mathrm{RowSpan}_{\Bold{Q}}\left(\begin{array}{rrrrrrrr} 0 & 1 & -\frac{1}{2} & -1 & 0 & 0 & 1 & -2 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & -1 \end{array}\right)$

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

B^2
 $\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrrrrr} -14 & 6 & 5 & -4 & -4 & 0 & 4 & -6 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & -12 \\ -4 & -4 & 2 & 0 & 0 & 0 & 0 & 4 \\ 0 & -2 & 0 & 0 & 0 & 0 & 0 & -6 \\ 0 & -1 & 0 & 0 & 0 & 0 & 0 & -11 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & -4 \\ -2 & -6 & 1 & 0 & 0 & 0 & 0 & -14 \\ 0 & -1 & 0 & 0 & 0 & 0 & 0 & -11 \end{array}\right)$
(B^2).nullity()
 $\newcommand{\Bold}[1]{\mathbf{#1}}4$

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
 $\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrrrrrr} 0 & -2 & 0 & 0 & 0 & 0 & 0 & -54 \\ 36 & -24 & -12 & 12 & 12 & 0 & -12 & 24 \\ 0 & 4 & 0 & 0 & 0 & 0 & 0 & -20 \\ 24 & -14 & -8 & 8 & 8 & 0 & -8 & 14 \\ 36 & -23 & -12 & 12 & 12 & 0 & -12 & 23 \\ 12 & -8 & -4 & 4 & 4 & 0 & -4 & 8 \\ 60 & -34 & -20 & 20 & 20 & 0 & -20 & 26 \\ 36 & -23 & -12 & 12 & 12 & 0 & -12 & 23 \end{array}\right)$
(B^3).nullity()
 $\newcommand{\Bold}[1]{\mathbf{#1}}5$

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.