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\begin{document}
\centerline{Math 322 lecture for Friday, Week 4}
\bigskip

\centerline{\sc exponentiation of jordan matrix}
\medskip

To solve the linear system~$x'=Ax$, we need to compute~$e^{At}$.
If~$P^{-1}AP=J$ where~$J$ is the Jordan form of~$A$,
then~$e^{At}=Pe^{Jt}P^{-1}$.  Then, to exponentiate~$J$, we must exponentiate
each of its blocks.  If
\begin{center}
  \begin{tikzpicture}
    \node at (0,0) {$\displaystyle
    J:=
    \left(\begin{array}{cccccc}
	J_{k_1}(\lambda_1)&&&&&\\
	  &J_{k_2}(\lambda_2)&&&&\\
	  &&J_{k_3}(\lambda_3)&&&\\
	&&&\ddots&&\\
        &&&&&\\
	&&&&&J_{k_{\ell}}(\lambda_{\ell})
 \end{array} \right)$,};
 \node at (-1.2,-0.6) {\huge 0};
 \node at (2.6,1) {\huge 0};
  \end{tikzpicture}
\end{center}
then
\begin{center}
  \begin{tikzpicture}
    \node at (0,0) {$\displaystyle
    e^{Jt}:=
    \left(\begin{array}{cccccc}
	e^{J_{k_1}(\lambda_1)t}&&&&&\\
	  &e^{J_{k_2}(\lambda_2)t}&&&&\\
	  &&e^{J_{k_3}(\lambda_3)t}&&&\\
	&&&\ddots&&\\
        &&&&&\\
	&&&&&e^{J_{k_{\ell}}(\lambda_{\ell}t)}
 \end{array} \right).$}; \node at (-1.2,-0.6) {\huge 0};
 \node at (2.6,1) {\huge 0};
  \end{tikzpicture}
\end{center}
Thus, we are reduced to exponentiating Jordan
blocks, which we talk about here, starting with an example.  Let~$\lambda\in
F$ and consider the Jordan block
\[
  J_4(\lambda)=
    \left(\begin{array}{cccc}
	\lambda&1&0&0\\
	0&\lambda&1&0\\
	0&0&\lambda&1\\
	0&0&0&\lambda
    \end{array} \right)
    = 
    \left(\begin{array}{cccc}
	\lambda&0&0&0\\
	0&\lambda&0&0\\
	0&0&\lambda&0\\
	0&0&0&\lambda
    \end{array} \right)
    +
    \left(\begin{array}{cccc}
	0&1&0&0\\
	0&0&1&0\\
	0&0&0&1\\
	0&0&0&0
    \end{array} \right)
    =\lambda I_4 + N_4
  \]
where
\[
N_4 = 
    \left(\begin{array}{cccc}
	0&1&0&0\\
	0&0&1&0\\
	0&0&0&1\\
	0&0&0&0
    \end{array} \right).
\]
Since~$\lambda I_4$ and~$N_4$ commute,
\[
  e^{J_4(\lambda)t}=e^{(\lambda I_4+N_4)t}=e^{\lambda tI_4}e^{tN_4}.
\]
As usual, 
\[
  e^{\lambda tI_4}
 = 
    \left(\begin{array}{cccc}
	e^{\lambda t}&0&0&0\\
	0&e^{\lambda t}&0&0\\
	0&0&e^{\lambda t}&0\\
	0&0&0&e^{\lambda t}
    \end{array} \right)
    =e^{\lambda t}I_4.
\]
So we are left with computing~$e^{tN_4}$:
\[
  e^{N_4t}=I_4+tN_4+\frac{t^2}{2!}N_4^2+\frac{t^3}{3!}N_4^3+\frac{t^4}{4!}N_4^4+\frac{t^5}{5!}N_4^5+\cdots
\]
Consider the powers of~$N_4$:
\begin{align*}
N_4^2
&= 
\left(\begin{array}{cccc}
    0&1&0&0\\
    0&0&1&0\\
    0&0&0&1\\
    0&0&0&0
\end{array} \right)
\left(\begin{array}{cccc}
    0&1&0&0\\
    0&0&1&0\\
    0&0&0&1\\
    0&0&0&0
\end{array} \right)
=
\left(\begin{array}{cccc}
    0&0&1&0\\
    0&0&0&1\\
    0&0&0&0\\
    0&0&0&0
\end{array} \right)\\[10pt]
N_4^3&= 
\left(\begin{array}{cccc}
    0&1&0&0\\
    0&0&1&0\\
    0&0&0&1\\
    0&0&0&0
\end{array} \right)
\left(\begin{array}{cccc}
    0&0&1&0\\
    0&0&0&1\\
    0&0&0&0\\
    0&0&0&0
\end{array} \right)
=
\left(\begin{array}{cccc}
    0&0&0&1\\
    0&0&0&0\\
    0&0&0&0\\
    0&0&0&0
\end{array} \right)\\[10pt]
N_4^4&= 0.
\end{align*}
All higher powers of~$N_4$ are~$0$.  Notice how as we take powers, the diagonal
of~$1$s climbs up to the right along successively higher diagonals. 

Returning to the calculation,
\begin{align*}
  e^{J_4(\lambda t)}
  &= 
  e^{\lambda t}\left(
  I_4+tN_4+\frac{t^2}{2!}N_4^2+\frac{t^3}{3!}N_4^3\right)\\[10pt]
  &= 
  e^{\lambda t}
  \left(\begin{array}{cccc}
      1&t&\frac{t^{2}}{2!}&\frac{t^3}{3!}\\[4pt]
      0&1&t&\frac{t^{2}}{2!}\\[4pt]
      0&0&1&t\\[4pt]
      0&0&0&1\\[4pt]
  \end{array} \right)
  =
  \left(\begin{array}{cccc}
      e^{\lambda t}&te^{\lambda t}&\frac{t^{2}}{2!}e^{\lambda t}&\frac{t^3}{3!}e^{\lambda t}\\[4pt]
      0&e^{\lambda t}&te^{\lambda t}&\frac{t^{2}}{2!}e^{\lambda t}\\[4pt]
      0&0&e^{\lambda t}&te^{\lambda t}\\[4pt]
      0&0&0&e^{\lambda t}\\[4pt]
  \end{array} \right).
\end{align*}
For instance, the solution to~$x'=J_4(\lambda)x$ with initial
condition~$x_0=(4,3,2,1)$ is
\begin{align*}
  x(t)&= e^{J_4(\lambda)t}x_0\\[10pt]
  &= 
  e^{\lambda t}
  \left(\begin{array}{cccc}
      1&t&\frac{t^{2}}{2!}&\frac{t^3}{3!}\\[4pt]
      0&1&t&\frac{t^{2}}{2!}\\[4pt]
      0&0&1&t\\[4pt]
      0&0&0&1\\[4pt]
  \end{array} \right)
  \left(\begin{array}{c}
      4\\3\\2\\1
  \end{array} \right)\\[10pt]
  &= e^{\lambda t}
  \left(\begin{array}{cc}
      4+3t+2\frac{t^2}{2!}+\frac{t^3}{3!}\\
      3+2t+\frac{t^2}{3!}\\
      2+t\\
      1
  \end{array} \right),
\end{align*}
or
\[
  x(t)=e^{\lambda t}\left( 4+3t+2\frac{t^2}{2!}+\frac{t^3}{3!},\ 
  3+2t+\frac{t^2}{3!},\ 2+t,\ 1\right).
\]
Now consider a general Jordan block:
\[
  J_k(\lambda)=\lambda I_k+ N_k
\]
where~$N_k$ is the matrix with~$1$s along the superdiagonal.  As before, taking
powers of~$N_k$ causes the diagonal of~$1$ to march up to the right, and we get
$N_k^k=0$.  A matrix~$N$ such that~$N^k=0$ is called {\em nilpotent}.  The
minimum~$k$ such that~$N^k=0$ is the {\em degree} of nilpotency.  Thus,~$N_k$ is
nilpotent of degree~$k$.  We have
\begin{align*}
  e^{J_k(\lambda)t}&= e^{(\lambda I_k+N_k)t}=e^{\lambda t I_k}e^{N_kt}\\
  &= e^{\lambda t}\left( I_k+tN_k+\frac{t^2}{2}N_k^2+\frac{t^3}{3!}N_k^3+\dots
  +\frac{t^{k-1}}{(k-1)!}N_k^{k-1}\right)\\[10pt]
  &= 
  e^{\lambda t}
  \left(\begin{array}{cccccc}
      1&t&\frac{t^2}{2!}&\dots&\dots&\frac{t^{k-1}}{(k-1)!}\\[4pt]
      0& 1&t&\dots&\dots&\frac{t^{k-2}}{(k-2)!}\\[4pt]
      0& 0& 1&\dots&\dots&\frac{t^{k-3}}{(k-3)!}\\[4pt]
      &\ddots & &\vdots&\vdots&\vdots\\[4pt]
      0&\dots&\dots&0&1&t\\[4pt]
      0&\dots&\dots&\dots&0&1\\[4pt]
  \end{array} \right).
\end{align*}
{\bf Note.}  If the real part of~$\lambda$ is negative, notice how
\[
  \lim_{t\to\infty}e^{J_k(\lambda)t}=0.
\]

Working exclusively over the reals, we will need to exponentiate Jordan blocks
corresponding to pairs of conjugate eigenvalues.  Let
\[
M := 
\left(\begin{array}{cc}
    a&-b\\b&a
\end{array} \right)
\]
and consider a real Jordan block for~$\lambda=a+bi$ with~$b\neq 0$:
\[
J :=
  \left(\begin{array}{cccccc}
      M&I_2&0&\dots&\dots&0\\[4pt]
      0& M&I_2&\dots&\dots&0\\[4pt]
      0& 0& M&\dots&\dots&0\\[4pt]
      &\ddots & &\vdots&\vdots&\vdots\\[4pt]
      0&\dots&\dots&0&M&I_2\\[4pt]
      0&\dots&\dots&\dots&0&M\\[4pt]
  \end{array} \right).
\]
To exponentiate, let
\[
R:=
\left(\begin{array}{cc}
    \cos(bt)&-\sin(bt)\\
    \sin(bt)&\cos(bt)
\end{array} \right).
\]
So
\[
  e^{Mt}=e^{at}R.
\]
By an argument that is essentially the same as just given above, we get the
matrix of~$2\times 2$ blocks
\[
  e^{Jt}= 
  e^{at}
  \left(\begin{array}{cccccc}
      R&tR&\frac{t^2}{2!}R&\dots&\dots&\frac{t^{k-1}}{(k-1)!}R\\[4pt]
      0& R&tR&\dots&\dots&\frac{t^{k-2}}{(k-2)!}R\\[4pt]
      0& 0& R&\dots&\dots&\frac{t^{k-3}}{(k-3)!}R\\[4pt]
      &\ddots & &\vdots&\vdots&\vdots\\[4pt]
      0&\dots&\dots&0&R&tR\\[4pt]
      0&\dots&\dots&\dots&0&R\\[4pt]
  \end{array} \right).
\]
Again, notice that if~$\mathrm{Re}(\lambda)=a<0$, then
\[
  \lim_{t\to\infty}e^{Jt}=0.
\]
{\bf Algorithm for computing the Jordan form.}  Our book has a careful
discussion of an algorithm for computing the Jordan form of a matrix~$A$.  We will
not go into the details (unless there is demand for it!). Here, we'll give over
a couple of points, though.  To start the algorithm, compute the eigenvalues
of the matrix by finding the zeros of the characteristic polynomial.  We would
like to know the number of Jordan blocks for each eigenvalue and their sizes.
The key to this is as follows: Let~$\lambda$ be an eigenvalue, and consider the
sequence of integers
\[
  \delta_{\ell}:=\delta_{\ell}(\lambda):=\dim\ker(A-\lambda I)^{\ell}
\]
for~$\ell=0,1,2,\dots$.  {\bf These~$\delta_{\ell}$ are invariant with respect to
conjugation, so we might as well imagine that~$A$ is in Jordan form already and
work block-by-block.} For a Jordan block~$J_k(\mu)$
with~$\mu\neq\lambda$,
\[
\ker(J_k(\mu)-\lambda I)^{\ell}=0
\]
for all~$\ell$ since each diagonal entry of each power is nonzero.  So
the~$\delta_{\ell}(\lambda)$ for any block like this are all~$0$.  Now consider
each Jordan block of the form~$J_k(\lambda)$.  We have
\[
  \ker(J_k(\lambda)-\lambda I)^{\ell}=\ker N_k^{\ell}
\]
where~$N_k$ is the nilpotent matrix from earlier.  Thinking about the form
of~$N_k^{\ell}$ is it easy to see that the~$\delta_{\ell}$ sequence  for blocks
like these is
\[
  \delta_{\ell}=
  \begin{cases}
    \ell&\text{for~$0\leq\ell\leq k$},\\
    k&\text{for~$\ell>k$.}
  \end{cases}
\]
See Figure~\ref{04frifig} for the case where~$k=4$.


\begin{figure}[!ht]
\[
\begin{array}{cccc}
k&(A-\lambda I)^k&\text{\bf basis for kernel}&\text{\bf dimension}\\\hline
\rule[+0pt]{0pt}{40pt}1&  
\left(\begin{array}{cccc}
   0&1&0&0\\
   0&0&1&0\\
   0&0&0&1\\
   0&0&0&0
\end{array} \right)
  & e_1&1
  \\[40pt]
2&  
\left(\begin{array}{cccc}
   0&0&1&0\\
   0&0&0&1\\
   0&0&0&0\\
   0&0&0&0
\end{array} \right)
& e_1,e_2&2\\
  \\[10pt]
3&  
\left(\begin{array}{cccc}
   0&0&0&1\\
   0&0&0&0\\
   0&0&0&0\\
   0&0&0&0
\end{array} \right)
& e_1,e_2,e_3&3\\
  \\[10pt]
4&  
\left(\begin{array}{cccc}
   0&0&0&0\\
   0&0&0&0\\
   0&0&0&0\\
   0&0&0&0
\end{array} \right)
& e_1,e_2,e_3,e_4&4.
\end{array}
\]
\caption{The case where~$A=J_4(\lambda)$.}\label{04frifig}
\end{figure}
The~$\delta_{\ell}(\lambda)$-sequence for~$A$ is the sum of
the~$\delta_{\ell}(\lambda)$-sequences for each of its Jordan blocks.
For instance,~$\delta_1(\lambda)$ for~$A$ is the number of its Jordan blocks
for~$\lambda$---we've just seen that each of these contributes its~$\delta_1=1$
to the count.  With just a little more thought (see our text), letting~$\nu_k$ be
the number of~$k\times k$ Jordan blocks for~$\lambda$ for the~$n\times n$
matrix~$A$, we get
\[
\nu_k=
\begin{cases}
  2\delta_1-\delta_2&\text{for~$k=1$},\\
  2\delta_{k}-\delta_{k+1}-\delta_{k-1}&\text{for~$1<k<n$},\\
  \delta_n-\delta_{n-1}&\text{for~$k=n$}.
\end{cases}
\]
The point is that the numbers of Jordan blocks of each size for each eigenvalue
are determined by the~$\delta$-sequences, i.e, by the sequence of dimensions of
the kernels, $\ker(A-\lambda I)^{\ell}$.

To actually conjugate~$A$ to Jordan form, for each
eigenvalue~$\lambda$, we consider the tower of subspaces
\[
  \ker(A-\lambda I)\subseteq\ker(A-\lambda I)^2\subseteq\ker(A-\lambda
  I)^3\subseteq\dots
\]
Starting at the leftmost kernel in this tower of subsets, we could successively
build bases for these kernels, adding vectors as we move to the right, as we
could see earlier in the case where~$A=J_4(\lambda)$.
Appropriately chosen, these vectors are called {\em generalized eigenvectors}.
We use them as columns of a matrix~$P$ so that~$P^{-1}AP$ is the Jordan form
for~$A$

Let's consider the case where 
\[
  A=
  J_4(\lambda)=
    \left(\begin{array}{cccc}
	\lambda&1&0&0\\
	0&\lambda&1&0\\
	0&0&\lambda&1\\
	0&0&0&\lambda
    \end{array} \right).
\]
Notice that we have
\[
  Ae_1=\lambda e_1,\ Ae_2=e_1+\lambda e_2,\ Ae_3= e_2+\lambda e_3,\ Ae_4=e_3+\lambda
  e_4.
\]
Therefore,
\begin{align*}
(A-\lambda I)e_1&= 0\\
(A-\lambda I)e_2&= e_1\\
(A-\lambda I)e_3&= e_2\\
(A-\lambda I)e_4&= e_3,
\end{align*}
and~$(A-\lambda I)^{i+1}e_i=0$ for~$i=2,3,4$.
So if~$A$ is not in Jordan form already, we will look for
vectors~$v_1,\dots,v_4$ that behave like the~$e_i$, above.  We need to
solve~$(A-\lambda I)v_i=v_{i-1}$ starting with~$v_1$ an eigenvector with
eigenvalue~$\lambda$. These~$v_i$ will be columns in the matrix~$P$.
\end{document}


