# Matrix exponential using the Jordan form

How do I calculate the matrix exponential $\Bbb e^{At}$ for $A = \left( \begin{matrix} 1 & 0 & 0 \\ 0 & 2 & 3 \\ 0 & 0 & 2 \end{matrix} \right)$ using the Jordan form of $A$? I already computed that the similarity transform to turn $A$ into the Jordan form is given by $X = \left( \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & \dfrac 1 3 \end{matrix} \right)$.

I believe that $\Bbb e^{At}$ should be given by $X \Bbb e^{ct} \left( \begin{matrix} 1 & 0 & 0 \\ 0 & 2 & t \\ 0 & 0 & 2 \end{matrix} \right) X^{-1}$ but I'm not sure what the constant $c$ should be. I've tried both of the eigenvalues of $A$ ($1$ and $2$) but neither of them works.

Let $A$ be any matrix and $A=SJS^{-1}$ its Jordan decomposition. Then, by the definition of the exponential function for matrices, we obtain:

$$\exp(At)=\exp(SJS^{-1}t) = \sum_{i=0}^{\infty} \frac{(SJS^{-1})^i t^i}{i!} = S\sum_{i=0}^{\infty} \frac{J^it^i}{i!}S^{-1}=S\exp(Jt)S^{-1}$$

since the $S$ cancel in the powers. Now, given $J=\bigoplus_j J_j$ (decomposition of $J$ into Jordan blocks), you have, since the Jordan blocks commute:

$$\exp(Jt)=\bigoplus_j \exp{J_jt}$$ In particular, if $A$ is diagonalizable, then $\exp(J)$ is also diagonal and the diagonal entries are just $\exp(J_jt)$, with $J_j$ the eigenvalues. Otherwise, you need to calculate the exponential function of the Jordan blocks. This already tells you that there is no "constant c" the way you want to have it. The constant changes with the different Jordan blocks.

Your Jordan blocks will look like $J:=\lambda I + N$ with a diagonal part and an nilpotent part $N$. Since the two parts commute (check this!) we have:

$$\exp(Jt)=\exp(\lambda I t)\exp(Nt)=\exp(\lambda t)\exp{Nt}$$

which might help for calculations.

• (multiples of) the identity commute with all other elements, including $N$. $\exp{Nt}$ can be calculated by expanding it into a series which will be a finite summation because $N$ is nilpotent. – WorldSEnder Oct 9 '18 at 15:11