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\newtheorem{thm}{Theorem}

\begin{document}

\noindent
Math 325

\noindent
Spring, 2001

\begin{center}
{\bf Jordan Canonical Form}
\end{center}

  A matrix 
\begin{math}
  B
\end{math}
is a \textbf{Jordan block}
if it is either of the form 
\begin{eqnarray}
  B=\lambda I_{1\times 1}
\end{eqnarray}
where 
\begin{math}
I_{1\times 1}
\end{math}
is the 
\begin{math}
1\times 1
\end{math} 
identity matrix 
or of the form 
\begin{eqnarray}
\label{eqn:block}
  B = \left[ 
  \begin{array}{cccccc}
     \lambda & 1 & 0 & \dots & 0 & 0 \\
     0 & \lambda & 1 & \dots & 0 & 0 \\
    \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
     0 & 0 & 0 & \dots & \lambda & 1\\
    0 & 0 & 0 & \dots & 0 & \lambda
  \end{array}
   \right]
\end{eqnarray}
Notice that a matrix of the form (\ref{eqn:block}) has zeros below the
diagonal, the same number 
\begin{math}
  \lambda
\end{math}
in each entry on the diagonal, a 
\begin{math}
  1
\end{math}
in each entry just above the diagonal and zeros every place else above
the diagonal.  

\begin{thm}
  If 
  \begin{math}
    A
  \end{math}
is a complex 
\begin{math}
  n \times n
\end{math}
matrix, there is an invertible matrix 
\begin{math}
  U
\end{math}
such that 
\begin{eqnarray}
  U^{-1}A U = J = \left[
    \begin{array}{cccc}
     B_1 & 0 & \dots & 0\\
     0 & B_2 & \dots & 0\\
     \vdots & \vdots & \ddots & \vdots\\
     0 & 0 & \dots & B_k 
    \end{array}
  \right]
\end{eqnarray}
where each matrix 
\begin{math}
  B_i,\ i=1\dots k,
\end{math}
is a Jordan block.
\end{thm}

The matrix 
\begin{math}
  J
\end{math}
is called the \textbf{Jordan canonical form} of 
\begin{math}
  A.
\end{math}
The entries on the diagonal of 
\begin{math}
  J
\end{math}
are the eigenvalues of 
\begin{math}
  A.
\end{math}
 If 
\begin{math}
  A
\end{math}
is diagonalizable, 
\begin{math}
  J
\end{math}
is a diagonal matrix, which we usually call 
\begin{math}
  D.
\end{math}
So, the interest of this theorem is that it gives a similarity
transformation from 
\begin{math}
  A
\end{math}
to a matrix of a simple form for non-diagonalizable matrices 
\begin{math}
  A.
\end{math}

A \textbf{generalized eigenvector} of an 
\begin{math}
  n\times n
\end{math}
matrix 
\begin{math}
  A
\end{math}
corresponding to the eigenvalue 
\begin{math}
  \lambda 
\end{math}
is a  nonzero vector 
\begin{math}
  \mathbf \eta
\end{math}
which solves the equation 
\begin{eqnarray}
\label{eq:generalized}
  (A - \lambda I)^k \eta = 0.
\end{eqnarray}
For 
\begin{math}
  k=1
\end{math}
 a solution 
\begin{math}
  \eta _1
\end{math}
to (\ref{eq:generalized}) is
just 
an eigenvector.  To obtain generalized eigenvectors for 
\begin{math}
  k \ge 2,
\end{math}
let 
\begin{math}
  \eta _2 
\end{math}
solve 
\begin{eqnarray}
  (A-\lambda I) \eta _2 = \eta _1, \qquad 
 (A-\lambda I) \eta _3 = \eta _2,
\end{eqnarray}
etc.  


How do you find the matrix 
\begin{math}
  U?
\end{math}
If  the eigenvalues 
\begin{math}
  \lambda _i
\end{math}
of the Jordan  blocks
\begin{math}
  B_i
\end{math}
are distinct, you can let 
\begin{math}
  U_i
\end{math}
be the matrix whose 
\begin{math}
  j
\end{math}th 
column is 
\begin{math}
  \eta _j ^{(i)}
\end{math}
where 
\begin{math}
  \eta _j ^{(i)}
\end{math}
is the 
\begin{math}
  j
\end{math}th
generalized eigenvector of 
\begin{math}
  A
\end{math}
with eigenvalue 
\begin{math}
  \lambda _i,
\end{math}
\begin{displaymath}
  (A-\lambda_i I )^j \eta _j ^{(i)} =0
\end{displaymath}
but 
\begin{displaymath}
  (A-\lambda_i I )^{j-1} \eta _j ^{(i)} \ne 0.
\end{displaymath}
Then 
\begin{eqnarray}
  U = [
  \begin{array}{cccc}
   U_1& U_2 &\dots& U_k
  \end{array}
].
\end{eqnarray}
In general, the columns of 
\begin{math}
  U
\end{math}
will be generalized eigenvectors of 
\begin{math}
  A
\end{math},
but the ones corresponding to two different Jordan blocks with the
same eigenvalue will be harder to find.
\end{document}
