8
$\begingroup$

I am trying to compute the matrix exponential for $$A=\left( \begin{array}{ccc} 1 & 2 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 1 & -1 \end{array} \right) $$ But I am stuck. There are a couple of methods I know of, but none of them seem to be working.

First, I tried to see if the matrix was nilpotent. It is not. I then tried to split the matrix into the identity and hoped the remaining matrix was nilpotent. It is not.

Next, I tried to do it by solving for the fundamental matrix, but the characteristic polynomial is $(x^2-1)^2$, which implies there are two eigenvalues $1$ and $-1$ with multiplicity $2$ each. I was unable to find the corresponding eigenvectors since $(A-I)^2 \neq 0$ and $(A+I)^2 \neq 0$.

What am I missing?

$\endgroup$
13
$\begingroup$

You have a matrix composed by two 2x2 diagonals blocks. You can compute the exponential of the blocks separaterly. The blocks themselves are of the form $I+N$ and $-I+M$ where $N$ and $M$ are nilpotent ($N^2=0$, $M^2=0$). So: $$ e^{N+I} =e^Ne^I = (I+N)e^I, \qquad e^{M-I} = e^M e^{-I}=(I+M)e^{-I}. $$

Matrix exponential can be computed blockwise because the exponential is a sum of powers, and both sums and products can be computed blockwise. The exponential of a square free matrix $N$ is $I+N$ since all higher powers: $N^2$, $N^3$... in the sum: $e^N = I + N + N^2/2 + N^3/3! + ...$ are null. Clearly $I$ commutes with every matrix, hence $\exp(N+I) = e^Ne^I$. The same is true for $-I$ which is a multiple of $I$.

Specifically: $$ \exp\begin{pmatrix}1&2\\0&1\end{pmatrix} = \begin{pmatrix}1&2\\0&1\end{pmatrix}\begin{pmatrix}e&0\\0&e\end{pmatrix} =\begin{pmatrix}e&2e\\0&e\end{pmatrix} $$ while $$ \exp\begin{pmatrix}-1&0\\1&-1\end{pmatrix} = \begin{pmatrix}1&0\\1&1\end{pmatrix}\begin{pmatrix}1/e&0\\0&1/e\end{pmatrix} =\begin{pmatrix}1/e & 0\\1/e & 1/e\end{pmatrix} $$ Hence $$ e^A = \begin{pmatrix}e&2e&0&0\\ 0&e&0&0\\ 0&0&1/e&0\\ 0&0&1/e&1/e\end{pmatrix} $$

$\endgroup$
  • $\begingroup$ One should possibly add in the introductory remark, that the block-composition is one of the diagonal form, so the blocks do not interfere when -for instance- the power series of A is evaluated (which we thus do not need to do actually, but only that blocks on the diagonal, as you've done it) $\endgroup$ – Gottfried Helms Jan 21 '15 at 14:16
1
$\begingroup$

You can write the matrix in the Jordan form $$ A = P J P^{-1}$$ and calculate the exponential as $$ e^A = P e^J P^{-1}.$$

The $e^J$ can be calculated easily by noting that each Jordan block can be written as $\lambda I + N$, where $N$ is a nilpotent matrix and the exponential of each block is $$ e^{\lambda I} e^N.$$

$\endgroup$
1
$\begingroup$

You can deduce from the information you've gathered that the Jordan normal form $J$ of the matrix is $J_2(1) \oplus J_2(-1)$, where $J_k(\lambda)$ is the $k \times k$ Jordan block of eigenvalue $\lambda$, so there is a real matrix $P$ such that

$$A = PJP^{-1} = P (J_2(1) \oplus J_2(-1)) P^{-1}.$$

This is useful, as substituting into the power series expansion for $\exp A$ gives that $$\exp A = P (\exp J) P^{-1} = P \exp (J_2(1) \oplus J_2(-1)) P^{-1}.$$

Better yet, again using the power series formula gives (more or less trivially) that $$\exp (B \oplus C) = \exp B \oplus \exp C$$ for square matrices $B, C$, which reduces the problem to computing $\exp J_2(1)$ and $\exp J_2(-1)$.

An easy argument (yet again using the power series expression for $\exp $) gives that $$\exp J_2(\lambda) := \begin{pmatrix}e^{\lambda} & e^{\lambda}\\ 0 &e^{\lambda}\end{pmatrix}.$$

Similar formulae exist for $\exp J_k(\lambda)$ for general $k$; see, e.g., the bottom of page 6 of http://www.ing.unitn.it/~bertolaz/2-teaching/2012-2013/AA-2012-2013-DYSY/lucidi/Exponential.pdf, which also gives some justification for the above argument and some comments about computing matrix exponentials explicitly in general.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.