Question 3.26: Analyze the rational canonical form
Analyze the rational canonical form
A = \left [ \begin{matrix} 0 & 1 & 0 & 0 & \vdots & & & & & \\ 0 & 0 & 1 & 0 & \vdots & & & & & \\ 0 & 0 & 0 & 1 & \vdots & & & & & \\ -1 & -2 & -3 & -2 & \vdots & & & & & \\\ldots & \ldots & \ldots & \ldots &\vdots & \ldots & \ldots & \ldots & & \\ & & & & \vdots & 0 & 1 & \vdots & & \\ & & & & \vdots & -1 & -1 & \vdots & & \\ & & & &\vdots &\ldots &\ldots & \vdots & \ldots &\ldots \\ & & & & & & & \vdots & 0 & 1 \\ & & & & & & & \vdots & -1 & 1 \end{matrix} \right ] _{8 \times 8}
= \left[\begin{matrix} R_{1} & & 0 \\ \ & R_{2} & \\ 0 & & R_{3} \end{matrix} \right] .
Analysis The characteristic polynomial is
det \left(A-tI_{8}\right) = det\left(R_{1}-tI_{4}\right) det\left(R_{2}-tI_{2}\right) det\left(R_{3}-tI_{2}\right)
= \left(t^{2}+t+1\right) ^{2} \cdot \left(t^{2}+t+1\right)\cdot \left(t^{2}-t+1\right)
= \left(t^{2}+t+1\right) ^{3}\left(t^{2}-t+1\right).
For computations concerned with block matrices, refer to Ex. <C> of Sec. 2.7.5. Let N =\left\{\overrightarrow{e_{1}}, \overrightarrow{e_{2}}, \ldots , \overrightarrow{e_{8}} \right\} be the natural basis for R^{8}.
For R_{1} and R_{2}:
\overrightarrow{e_{1}}A=\overrightarrow{e_{2}},
\overrightarrow{e_{2}}A=\overrightarrow{e_{3}}=\overrightarrow{e_{1}}A^{2} ,
\overrightarrow{e_{3}}A=\overrightarrow{e_{4}}=\overrightarrow{e_{1}}A^{3} ,
\overrightarrow{e_{4}}A=-\overrightarrow{e_{1}}-2\overrightarrow{e_{2}}-3 \overrightarrow{e_{3}} -2 \overrightarrow{e_{4}}
= -\overrightarrow{e_{1}}-2\overrightarrow{e_{1}}A -3\overrightarrow{e_{1}}A^{2} -2 \overrightarrow{e_{1}}A^{3}=\overrightarrow{e_{1}}A^{4}
\Rightarrow \overrightarrow{e_{i}} \left(A^{4}+2A^{3}+3A^{2}+I_{8}\right)=\overrightarrow{e_{i}} \left(A^{2}+A+I_{8}\right)^{2}=\overrightarrow{0} for 1≤ i ≤ 4;
\overrightarrow{e_{5}}A=\overrightarrow{e_{6}},
\overrightarrow{e_{6}}A=-\overrightarrow{e_{5}}A-\overrightarrow{e_{6}}=-\overrightarrow{e_{5}}-\overrightarrow{e_{5}}A=\overrightarrow{e_{5}}\left(-I_{8}-A\right)=\overrightarrow{e_{5}}A^{2}
\Rightarrow \overrightarrow{e_{i}} \left(A^{2}+A+I_{8}\right)=\overrightarrow{0} fori = 5, 6.
Also, compute the ranks as follows:
A^{2}= \left[\begin{matrix} R^{2}_{1} & & 0 \\ \ & R^{2}_{2} & \\ 0 & & R^{2}_{3} \end{matrix} \right]
= \left [ \begin{matrix} 0 & 0 & 1 & 0 & \vdots & & & & & \\ 0 & 0 & 0 & 1 & \vdots & & & & & \\ -1 & -2 & -3 & -2 & \vdots & & & & & \\ 2 & 3 & 4 & 1 & \vdots & & & & & \\\ldots & \ldots & \ldots & \ldots &\vdots & \ldots & \ldots & \ldots & & \\ & & & & \vdots & -1 & -1 & \vdots & & \\ & & & & \vdots & 1 & 0 & \vdots & & \\ & & & &\vdots &\ldots &\ldots & \vdots & \ldots &\ldots \\ & & & & & & & \vdots & -1 & 1 \\ & & & & & & & \vdots & -1 & 0 \end{matrix} \right ]
\Rightarrow A^{2}+A+I_{8}= \left [ \begin{matrix} 1 & 1 & 1 & 0 & \vdots & & & & & \\ 0 & 1 & 1 & 1 & \vdots & & & & & \\ -1 & -2 & -2 & -1 & \vdots & & & & & \\ 1 & 1 & 1 & 0 & \vdots & & & & & \\\ldots & \ldots & \ldots & \ldots &\vdots & \ldots & \ldots & \ldots & & \\ & & & & \vdots & 0 & 0 & \vdots & & \\ & & & & \vdots & 0 & 0 & \vdots & & \\ & & & &\vdots &\ldots &\ldots & \vdots & \ldots &\ldots \\ & & & & & & & \vdots & 0 & 2 \\ & & & & & & & \vdots & -2 & 2 \end{matrix} \right ]
\Rightarrow \left(A^{2}+A+I_{8} \right)=4;
\Rightarrow \left(A^{2}+A+I_{8} \right)^{2}=\left [ \begin{matrix} O_{4 \times 4} & & & & \\ & O_{2 \times 2} & & & \\ & & \vdots & \ldots & \ldots \\ & & \vdots & -4 & 4 \\ & & \vdots & -4 & 0 \end{matrix} \right ] {8 \times 8}
\Rightarrow r\left(A^{2}+A+I_{8} \right)^{k}=2 for k ≥ 2.
These relations together provide the following information:
1. Let p_{1}\left(t\right)=t^{2}+t+1. The set
K_{p_{1}}= \left\{\overrightarrow{x} \in R^{8} \mid \overrightarrow{x} \left(A^{2}+A+I_{8} \right)^{3}=\overrightarrow{0}\right\}=Ker p_{1}\left(A\right)^{3}
= \ll \overrightarrow{e_{1}}, \ldots , \overrightarrow{e_{4}},\overrightarrow{e_{5}},\overrightarrow{e_{6}}\gg = \ll \overrightarrow{e_{1}}, \overrightarrow{e_{2}} , \overrightarrow{e_{3}},\overrightarrow{e_{4}}\gg \oplus \ll \overrightarrow{e_{5}}, \overrightarrow{e_{6}}\gg
is an invariant subspace of R^{8} of dimension equal to
3·2 = (the algebraic multiplicity 3 of p_{1}\left(t\right) ) . (the degree of p_{1}\left(t\right) ) =6.
2.K_{p_{1}} contains an invariant subspace
\ll \overrightarrow{e_{1}}, \overrightarrow{e_{2}} , \overrightarrow{e_{3}},\overrightarrow{e_{4}}\gg = \ll \overrightarrow{e_{1}},\overrightarrow{e_{1}}A,\overrightarrow{e_{1}}A^{2},\overrightarrow{e_{1}}A^{3}\gg=C\left(\overrightarrow{e_{1}}\right)
which is an A-cycle of length 4 = degree of p_{1}\left(t\right)^{2}.Note that p_{1}\left(A\right)^{2} is an annihilator of C_{A}\left(\overrightarrow{e_{1}}\right), simply denoted by C\left(\overrightarrow{e_{1}}\right), of least degree, i.e.
p_{1}\left(A\right)^{2} \mid_{C\left(\overrightarrow{e_{1}}\right)}=O _{4 \times 4}.
3.K_{p_{1}} contains another invariant subspace
\ll \overrightarrow{e_{5}},\overrightarrow{e_{6}} \gg= \ll \overrightarrow{e_{5}},\overrightarrow{e_{5}}A \gg = C\left(\overrightarrow{e_{5}}\right)
which is an A-cycle of length 2 = degree of p_{1}\left(t\right). Note that p_{1}\left(A\right) s an annihilator of C_{A}\left(\overrightarrow{e_{5}}\right) of least degree, i.e.
p_{1}\left(A\right) \mid_{C_{A}\left(\overrightarrow{e_{5}}\right)}=O _{2 \times 2}.
4. Solve
\overrightarrow{x} p_{1}\left(A\right)= \overrightarrow{0} for \overrightarrow{x} \in R^{8}
\Rightarrow \overrightarrow{x} =x_{3} \left(1,1,1,0,0,0,0,0\right)+ x_{4} \left(-1,0,0,1,0,0,0,0\right)+\left(0,0,0,0,x_{5},x_{6},0,0\right)
\Rightarrow Ker p_{1}\left(A\right)= \ll \overrightarrow{v_{1}},\overrightarrow{v_{2}},\overrightarrow{e_{5}},\overrightarrow{e_{6}}\gg where
\overrightarrow{v_{1}} =\left(1,1,1,0,0,0,0,0\right) and \overrightarrow{v_{2}} =\left(-1,0,0,1,0,0,0,0\right).
5. Solve
\overrightarrow{x} p_{1}\left(A\right)^{2}= \overrightarrow{0}
\Rightarrow Ker p_{1}\left(A\right)^{2}= \ll \overrightarrow{e_{1}},\overrightarrow{e_{2}},\overrightarrow{e_{3}},\overrightarrow{e_{4}},\overrightarrow{e_{5}},\overrightarrow{e_{6}}\gg.
Hence
Ker p_{1}\left(A\right)^{2}= \ll \overrightarrow{e_{1}},\overrightarrow{e_{2}},\overrightarrow{e_{3}},\overrightarrow{e_{4}}\gg \oplus \ll \overrightarrow{e_{5}},\overrightarrow{e_{6}}\gg
= \ll \overrightarrow{v_{1}},\overrightarrow{v_{2}},\overrightarrow{e_{3}},\overrightarrow{e_{4}}\gg \oplus \ll \overrightarrow{e_{5}},\overrightarrow{e_{6}}\gg.
6. Take any vector \overrightarrow{v} \in Ker p_{1}\left(A\right)^{2} but not in Ker p_{1}\left(A\right), say \overrightarrow{v} = \overrightarrow{e_{3}}+\overrightarrow{e_{4}}.
Then
\overrightarrow{v}A = \left(-1,-2,-3,-1,0,\dots , 0\right),
\overrightarrow{v}A ^{2}= \left(1,1,1,-1,0,\dots , 0\right),
\overrightarrow{v}A^{4} = \left(-3,-5,-6,-2,0,\dots , 0\right)= -\overrightarrow{v} -2\overrightarrow{v}A-3\overrightarrow{v}A ^{2}-2\overrightarrow{v}A ^{3}.
Also, B_{\overrightarrow{v}}=\left\{\overrightarrow{v},\overrightarrow{v}A,\overrightarrow{v}A ^{2},\overrightarrow{v}A ^{3}\right\} is a basis for Ker p_{1}\left(A\right)^{2}. In B_{\overrightarrow{v}},
\left[A \mid_{Ker p_{1}\left(A\right)^{2}}\right]_{B_{\overrightarrow{v}}}=R_{1}.
The matrix R_{1}is called the companion matrix of p_{1}\left(t\right)^{2} = t^{4}+2t^{3}+3t^{2}-2t+1. Take any vector \overrightarrow{u} \in Ker p_{1}\left(A\right), say \overrightarrow{u} =\overrightarrow{v_{1}}. Then
\overrightarrow{u}A= \left(0,1,1,1,0,0,0,0\right),
\overrightarrow{u}A^{2}= \left(-1,2,-2,-1,0,0,0,0\right)= -\overrightarrow{u}-\overrightarrow{u}A
and B_{\overrightarrow{u}}= \left\{\overrightarrow{u} ,\overrightarrow{u}A\right\} is a basis for Ker p_{1}\left(A\right). In B_{\overrightarrow{u}},
\left[A \mid_{Ker p_{1}\left(A\right)}\right]_{B_{\overrightarrow{u}}}=R_{2}.
The matrix R_{2} is called the companion matrix of p_{1}\left(t\right) = t^{2}+t+1.
7. It can be show that B_{\overrightarrow{v}} \cup B_{\overrightarrow{u}} is linearly independent. Hence B_{\overrightarrow{v}} \cup B_{\overrightarrow{u}} is a basis for Ker p_{1}\left(A\right)^{3}=Ker p_{1}\left(A\right)^{2}. Notice that
Where the 2 in \frac{1}{2} is the degree of p_{1}\left(t\right). Also,
dim Ker p_{1}\left(A\right)^{3}= deg p_{1}\left(t\right)^{2} + deg p_{1}\left(t\right)
= 2· 2 + 2 · 1
= 2· 3
= (the degree of p_{1}\left(t\right). ) · (the number of dots)
Combining together, we get
\left[A \mid_{Ker p_{1}\left(A\right)^{3}}\right]_{B_{\overrightarrow{v}} \cup B_{\overrightarrow{u}}}=\begin{bmatrix} R_{1} & 0 \\ 0 & R_{2} \end{bmatrix}_{6 \times 6} . \left(*_{1}\right)
For R_{3}:
\overrightarrow{e_{7}}A=\overrightarrow{e_{8}},
\overrightarrow{e_{8}}A=\left(0,\ldots,0,-1,1\right)=-\overrightarrow{e_{7}}+\overrightarrow{e_{8}}=-\overrightarrow{e_{7}}+\overrightarrow{e_{7}}A=\overrightarrow{e_{7}}A^{2}
\Rightarrow \overrightarrow{e_{i}} \left(A^{2}-A+I_{8}\right)=\overrightarrow{0} fori = 7, 8.
Also,
A^{2}-A+I_{8}= = \left [ \begin{matrix} 0 & -1 & 1 & 0 & \vdots & & & & & \\ 0 & 0 & -1 & 0 & \vdots & & & & & \\ -1 & -2 & -3 & -3 & \vdots & & & & & \\ 3 & 5 & 7 & 3 & \vdots & & & & & \\\ldots & \ldots & \ldots & \ldots &\vdots & \ldots & \ldots & \ldots & & \\ & & & & \vdots & -1 & 2 & \vdots & & \\ & & & & \vdots & 2 & 1 & \vdots & & \\ & & & &\ldots &\ldots &\ldots & \vdots & \ldots &\ldots \\ & & & & & & & \vdots & 0 & 0 \\ & & & & & & & \vdots & 0 & 0 \end{matrix} \right ]
\Rightarrow r\left(A^{2}-A+I_{8}\right)=6
\Rightarrow r\left(A^{2}-A+I_{8}\right)^{k}=2 for k ≥ 2.
These facts provide the following information:
1. Set p_{2}\left(t\right) = t^{2}-t+1. Then
Ker p_{2}\left(A\right) = \left\{\overrightarrow{x} \in R^{8} \mid p_{2}\left(A\right) =\overrightarrow{0} \right\}
is an invariant subspace of dimension 2, the degree of p_{2}\left(t\right).
2. Ker p_{2}\left(A\right) = \ll \overrightarrow{e_{7}},\overrightarrow{e_{8}}\gg=\ll \overrightarrow{e_{7}},\overrightarrow{e_{7}}A\gg= C\left(\overrightarrow{e_{7}}\right) is an A-cycle of length 2 which is annihilated by p_{2}\left(t\right), i.e.
p_{2}\left(A\right) \mid_{C\left(\overrightarrow{e_{7}}\right)}=O_{2 \times 2}.
3. Solve
\overrightarrow{x} p_{2}\left(A\right)=\overrightarrow{0} for \overrightarrow{x} \in R^{8}
\Rightarrow Ker p_{2}\left(A\right)= \ll \overrightarrow{e_{7}},\overrightarrow{e_{8}}\gg as it should be by 2.
4. Take any nonzero vector \overrightarrow{w} \in Ker p_{2}\left(A\right), say \overrightarrow{w}=\alpha \overrightarrow{e_{7}}+\beta \overrightarrow{e_{8}}. Then
\overrightarrow{w} A = \alpha \overrightarrow{e_{7}}A+\beta \overrightarrow{e_{8}}A= \alpha \overrightarrow{e_{7}}A+\beta \overrightarrow{e_{8}}A^{2} =-\beta \overrightarrow{e_{7}}+\left(\alpha+\beta \right)\overrightarrow{e_{8}}
\Rightarrow \overrightarrow{w} A^{2} = -\beta \overrightarrow{e_{7}}A+\left(\alpha+\beta \right)\overrightarrow{e_{8}}A = -\beta \overrightarrow{e_{7}}A+\left(\alpha+\beta \right) \left(-\overrightarrow{e_{7}}+\overrightarrow{e_{7}}A\right)
= – \left(\alpha+\beta \right) \overrightarrow{e_{7}}+\alpha\overrightarrow{e_{7}}A
= -\overrightarrow{w}+\overrightarrow{w} A
Then B_{\overrightarrow{w}}=\left\{\overrightarrow{w},\overrightarrow{w} A\right\} is a basis for Ker p_{2}\left(A\right).The matrix
\left[A \mid _{Ker p_{2}\left(A\right)}\right]_{B_{\overrightarrow{w}}}=R_{3} \left(*_{2}\right)
is called the companion matrix of p_{2}\left(t\right)=t^{2}-t+1.
5. Notice that
Putting \left(*_{1}\right) and \left(*_{2}\right) together, let
B= B_{\overrightarrow{v}} \cup B_{\overrightarrow{u}} \cup B_{\overrightarrow{w}} = \left\{\overrightarrow{v},\overrightarrow{v}A,\overrightarrow{v}A^{2},\overrightarrow{v}A^{3},\overrightarrow{u},\overrightarrow{u}A,\overrightarrow{w},\overrightarrow{w}A\right\} .
B is basis for R^{8} and is called a rational canonical basis of A. In B,
\left[A\right]_{B}=PAP^{-1}= \left[\begin{matrix} R_{1} & & 0 \\ & R_{2} & 0 \\ 0 & & R_{3} \end{matrix} \right] , where P = \left[\begin{matrix} \overrightarrow{v} \\ \overrightarrow{v}A \\ \vdots \\ \overrightarrow{w}A \end{matrix} \right] _{8 \times 8}
is called the rational canonical form of A.
