# Exponential of jordan form involving distinct eigenvalues

1. Oct 28, 2011

### JamesGoh

For a Jordan form (which is a direct sum of individual jordan blocks) that has distinct eigenvalues along the diagonal, how would the exponential of the Jordan form be calculated ?

As far as Im aware the formula for calculating the exponential of a jordan block can only involve one eigenvalue, so how would this work for the case of a matrix having distinct eigenvalues ?

2. Oct 28, 2011

### quasar987

If J is a jordan matrix with block diagonal B(λ1),...,B(λn), use the notation J=diag(B(λi)). Notice that Jk=diag(B(λi)k). By definition,

$$e^J=\sum_{k=0}^{\infty}\frac{J^k}{k!}=\sum_{k=0}^{\infty}\frac{\mathrm{diag}(B(\lambda_i)^k)}{k!}= \mathrm{diag} \left( \sum_{k=0}^{\infty}\frac{B( \lambda_i)^k}{k!}\right)=\mathrm{diag}(e^{B( \lambda_i)})$$

More generally, the exponential of a block diagonal matrix is the block diagonal matrix of exponentials.

3. Oct 29, 2011

### HallsofIvy

To take a simple example,
$$\begin{bmatrix}1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 2\end{bmatrix}$$
is a Jordan matrix for a matrix having eigenvalues 1 and 2 (each with geometric multiplicity 2).
You can think of it as made of the two "elementary" Jordan matrices
$$\begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}$$
and
$$\begin{bmatrix} 2 & 1 \\ 0 & 2\end{bmatrix}$$

The exponential of the first is, of course,
$$\begin{bmatrix}e & e \\ 0 & e\end{bmatrix}$$
and of the second
$$\begin{bmatrix}e^2 & e^2 \\ 0 & e^2\end{bmatrix}$$
so it should be no surprise that the exponential of the original matrix is
$$\begin{bmatrix}e & e & 0 & 0 \\ 0 & e & 0 & 0 \\ 0 & 0 & e^2 & e^2 \\ 0 & 0 & 0 & e^2\end{bmatrix}$$

4. Oct 29, 2011

### JamesGoh

I thought that you can only produce an exponential of a matrix if it's in diagaonalisable form only ?

How did you get the top right terms of both the individual Jordan matrices to be exponents ?

5. Oct 29, 2011

### micromass

Not at all, finding the exponential of a Jordan canonical form is easy.
Indeed, in the Jordan canonical form, we are dealing with matrices of the form D+N, where D is diagonal and where N is nilpotent. If we have that DN=ND, then we have that

$$e^{D+N}=e^De^N$$

The exponential of D can be easily find since it is diagonal. The exponential of N is also easy. Indeed, we know that $N^k=0$ for a certain k, so

$$e^N=I+N+\frac{N^2}{2}+...+\frac{N^{k-1}}{(k-1)!}$$

which is a finite sum!!

In the case $$A=\begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix}=\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}+\begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix}=D+N$$

We see that

$$e^D=\begin{pmatrix} e & 0\\ 0 & e \end{pmatrix}$$

Furthermore, we know that $N^2=0$, so

$$e^N=I+N=\begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix}$$

We get that

$$e^A=e^De^N=\begin{pmatrix} e & 0\\ 0 & e \end{pmatrix}\begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix}=\begin{pmatrix} e & e\\ 0 & e \end{pmatrix}$$