Matrix exponential without series expansion

**bkarpuz** · March 28th 2011, 09:31 AM

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

**FernandoRevilla** · March 28th 2011, 09:39 AM

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

**bkarpuz** · March 28th 2011, 11:41 AM

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

**HallsofIvy** · March 28th 2011, 01:25 PM

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.)

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