# Finding matrix exponential

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?

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 $$e^A = \begin{pmatrix}e&2e&0&0\\ 0&e&0&0\\ 0&0&1/e&0\\ 0&0&1/e&1/e\end{pmatrix}$$

• 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) – Gottfried Helms Jan 21 '15 at 14:16

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 $$e^{\lambda I} e^N.$$

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.