Thread: Matrix exponential without series expansion

1. Matrix exponential without series expansion

Dear MHF members,

my problem is explained below.
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Problem. Compute the matrix exponential of the matrix
$A:=
\left(
\begin{array}{rrrr}
1 & 0 & -1 & 1 \\
0 & 1 & 1 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 1 & 0 \\
\end{array}
\right).$

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I can compute the matrix exponential by using the series expansion
$\mathrm{e}^{At}=\mathrm{I}+\displaystyle\sum_{\ell =1}^{\infty}\displaystyle\frac{1}{\ell!}A^{\ell}t^ {\ell},$
where
$A^{n}=
\left(
\begin{array}{rrrr}
1 & 0 & -1 & 1 \\
0 & 1 & n & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 1 & 0 \\
\end{array}
\right)$
for $n\in\mathbb{N}$.

However, I could not figure out how to compute it by using the eigenvalues.
I would be glad if you show me the solution in this way.

Thanks.
bkarpuz

2. You can find the canonical form of Jordan $J$ for $A$ and a non singular matrix $P$ such that $P^{-1}AP=J$ . Then, $e^{tA}=Pe^{tJ}P^{-1}$ .

3. Originally Posted by FernandoRevilla
You can find the canonical form of Jordan $J$ for $A$ and a non singular matrix $P$ such that $P^{-1}AP=J$ . Then, $e^{tA}=Pe^{tJ}P^{-1}$.
Actually, this is the part I am having problems since it has repeated eigenvalues.
Do you mind explaining me?

Thanks.
bkarpuz

4. Yes, and so the matrix may not be diagonalizable. But the crucial part is not the number of eigenvalues but the number of independent eigenvectors. Here, 0 is a single eigenvalue and 1 is triple eigenvalue. An eigenvector corresponding to eigenvalue 0 (any multiple of <1, 0, 0, -1> is such an eigenvalue) is independent of any other eigenvalues but the eigenvalue 1 may have 3, 2, or only 1 independent eigenvector. In fact, looking for eigenvectors corresponding to eigenvalue 1 means looking for solutions to
$\begin{pmatrix}1 & 0 & -1 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0\end{pmatrix}\begin{pmatrix}x \\ y \\ z \\ w\end{pmatrix}=\begin{pmatrix}x- z+ w\\ y+ z \\ z \\ z\end{pmatrix}= \begin{pmatrix}x \\ y \\ z \\ w\end{pmatrix}$
So we must have x- z+ w= x, y+ z= y, z= z, and z= w y+ z= y gives immediately z= 0 so then w= 0. x- z+ w= x becomes x= x and y+ z= 7 becomes y= y both of which are satisfied by all x and y. That is, any eigenvector corresponding to eigenvalue 1 is of the form <x, y, 0, 0>= <x, 0, 0, 0>+ <0, y, 0, 0>= x<1, 0, 0, 0>+ y<0, 1, 0, 0>. That is, <1, 0, 0, 0> and <0, 1, 0, 0> are independent eigenvectors corresponding to eigenvalue 1. Because there are only 3 independent eigenvectors and not 4, we cannot "diagoalize" this matrix, but we can write it in "Jordan Normal form' as Fernando Revilla suggested. To do that, we need to find a "generalized eigenvector"- a vector, v, such that $(A- \lambda I)v\ne 0$ but $(A- \lambda I)^2v= 0$. That, of course, is the same as saying $(A- \lambda)((A- \lambda)v))= 0$ which, in turn, means that $(A- \lambda I)v$ is an eigenvector. That is, we must have
$\begin{pmatrix}1-1 & 0 & -1 & 1 \\ 0 & 1-1 & 1 & 0 \\ 0 & 0 & 1-1 & 0 \\ 0 & 0 & 1 & 1-1\end{pmatrix}\begin{pmatrix}x \\ y \\ z \\ w\end{pmatrix}$
must be equal to either $\begin{pmatrix}1 \\ 0 \\ 0 \\ 0\end{pmatrix}$ or $\begin{pmatrix}0 \\ 1 \\ 0 \\ 0 \end{pmatrix}$.

The first of those is impossible but the second is satisfied by $\begin{pmatrix}0 \\ 1 \\ 1 \\ 1\end{pmatrix}$.

Constructing a matrix, P, having those vectors as columns, say,
$P= \begin{pmatrix}1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1 \end{pmatrix}$
then $P^{-1}AP$ will be the Jordan Normal form.

(That's the idea. The calculations were done quickly and I don't guarentee them.)