% m211oh89.tex (prev.: *oh85.*) - The Jordan Canonical Form: Examples 
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\vspace*{-20pt}
\begin{center}\tcr{\rmH The Jordan Canonical Form -- Examples} \end{center} 
\LARGE 

\newcommand{\ueq}{\vec{u}_{\scr eq}}
\newcommand{\vu}[1]{\vec{u}_{\scr #1}}
\newcommand{\vv}[1]{\vec{v}_{\scr #1}}
\newcommand{\rank}{\mbox{rank}}
\newcommand{\mult}{\mbox{mult}}

\begin{description}  
\item[\tcr{Example 1:}] \tcg{Given} \mc{A = \left(\begin{array}{rr}
0 & 1 \\ -1 & -2 \end{array}\right),} \tcg{find} its \tcr{JCF} and
\vspace{-10pt} \mc{P}. 

Here: \exprow{1.2}
 \mc{\left.\begin{array}{l}
\ch_A(t) = (t+1)^2 \\ \nu_{A}(-1) = 1\end{array}\right\} }
\exprow{1.0}
\mc{\Rightarrow J = \left(\begin{array}{rr}
-1 & 1 \\ 0 & -1 \end{array}\right).}

We want to find an \tcg{invertible} matrix \mc{P} such that
\tcb{\beq
P^{\scr-1}AP \; = \; J\; \mbox{or, equivalently,}\; AP \; = \; PJ.
\eeq}
Write \mc{P = (\vv{1}|\vv{2})}. Then, since \mc{AP = (A\vv{1}|A\vv{2})} and \mc{PJ = 
(-\vv{1}|\vv{1}-\vv{2})}, we see that equation (1) \tcr{is equivalent} to:
\tcb{$$
\begin{array}{rlcl}
1) & A\vv{1} = -\vv{1} & \mbox{or} & (A + I)\vv{1} = \vec{0} \\
2) & A\vv{2} = \vv{1} - \vv{2} & \mbox{or} & (A + I)\vv{2} = \vv{1} \\
3) & \vv{1}, \vv{2}\; \mbox{are lin.\ indep.} & & (\Leftrightarrow P 
\mbox{ is invertible)}
\end{array} \vspace{-25pt}
$$}
\item[\tcg{Observations:}] a) Equations 1) and 2) imply \\[.3cm]
\hspace*{50pt}\tcb{4)} \mc{(A + I)^2\vv{2} = \vec{0}.}

b) If we \tcr{pick} a vector \mc{\vv{2} \in \C^2} which satisfies 4) and \tcr{define} \mc{\vv{1}} by equation 2), then \mc{\vv{1}, \vv{2}} satisfy 1) and 2).

\tcr{However:} we have to pick \mc{\vv{2}} \tcr{carefully} such that condition 3) is also satisfied.

c) \tcg{It turns out:} it is \tcr{enough to require} that 
\tcb{$$
\vv{1}:= (A+I)\vv{2} \ne \vec{0},\, \mbox{i.e.}\, \vv{2} \notin 
E_A(-1) = \{c(1,-1)^t\}.
$$}  
Take \mc{\vv{2} = \left(\begin{array}{r} 1  \\ 0 \end{array}\right) \Rightarrow \vv{1} = \left(\begin{array}{r} 1 \\ -1 
\end{array}\right)}, so \mc{P = 
\left(\begin{array}{rr} 1 & 1 \\ -1 & 0 \end{array}\right).}
\end{description} 
\newpage

\vspace*{-30pt}
\begin{description}  
\item[\tcr{Example 2:}] If \mc{A = \left(\begin{array}{rrr}
1 & 2 & -1 \\ 0 & 2 & 0 \\ 1 & -2 & 3 \end{array}\right),}
\tcg{find} its \tcr{JCF} and \vspace{-10pt} \mc{P}.
%\vspace{-15pt} 

Here: \exprow{1.2}
 \mc{\left.\begin{array}{l}
\ch_A(t) = (t-2)^3 \\ \nu_{A}(2) = 2\end{array}\right\} }
\exprow{1.0}
\mc{\Rightarrow J = \left(\begin{array}{rrr}
2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array}\right).}

\tcg{Find} an \tcg{invertible} matrix \mc{P = (\vv{1}|\vv{2}|\vv{3})} such that \mc{AP  =  PJ}.
Since \mc{AP = (A\vv{1}|A\vv{2}| A\vv{3})} and \mc{PJ = 
(2\vv{1}|\vv{1}+2\vv{2}|2\vv{3})}, this is equivalent to: %\vspace{-10pt}
\tcb{$$
\begin{array}{rlcl}
1) & A\vv{1} = 2\vv{1} & \mbox{or} & (A - 2I)\vv{1} = \vec{0} \\
2) & A\vv{2} = \vv{1} + 2\vv{2} & \mbox{or} & (A - 2I)\vv{2} = \vv{1} \\
3) & A\vv{3} = 2\vv{3} & \mbox{or} & (A - 2I)\vv{3} = \vec{0} \\
4) & \vv{1}, \vv{2}, \vv{3}\, \mbox{are lin.\ indep.} & & (\Leftrightarrow P 
\mbox{ is invertible)}
\end{array} \vspace{-20pt}
$$}
\item[\tcg{Again:}] Equations 1) and 2) imply: \tcb{5)} \mc{(A - 2I)^2\vv{2} = \vec{0}.}
\item[\tcg{Thus:}] \tcr{pick} a vector \mc{\vv{2} \in E_A^{\scr2}(2)} \tcr{not in} \mc{E_A(2)};

\tcr{define} \mc{\vv{1} = (A - 2I)\vv{2}};

\tcr{pick} \mc{\vv{3} \in E_A(2)}, \tcr{linearly independent} from \mc{\vv{1}, \vv{2}}. \vspace{-5pt}
  
\item[\tcg{Here:}] \mc{E_A(2) = \{c_{\scr1}\left(\begin{array}{r} 1 \\ 0 \\ 
-1 \end{array}\right) 
+ c_{\scr2}\left(\begin{array}{r}2\\1\\0\end{array}\right)\},\,  E_A^{\scr2}(2) = \C^3. } \vspace{-20pt}
%\mbox{since} A - 2I)^2 = \vec{0} \end{array}\right\} }\exprow{1.0}
\item[\tcg{Take}] \mc{\vv{2} = \left(\begin{array}{r} 1  \\ 0 \\ 0 \end{array}\right) \Rightarrow \vv{1} = \left(\begin{array}{r} -1 \\ 0 \\ 1 
\end{array}\right)}; \mbox{also take } \mc{\vv{3} = \left(\begin{array}{r} 2 \\ 1 \\ 0 \end{array}\right).}
\item[\tcr{Then:}] \mc{P = (\vv{1} | \vv{2} | \vv{3}) = 
\left(\begin{array}{rrr} -1 & 1 & 2 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array}\right).}
\end{description} 

\end{document}                          