Is the relation $e^{At} = Pe^{Jt} P^{-1}$ valid for a Jordan form?

Question

Imagine you have a system of equations of the form: $$ \mathbf {\dot{x}} = A \mathbf{x} $$ And $A$ cannot be diagonalized, so you find it's Jordan form: $$ A = P J P^{-1} \\ \; \\ J = \begin{pmatrix} \lambda_1 & 1 & 0 & \dots & 0 \\ 0 & \lambda_2 & 1 & \dots & 0 \\ 0 & 0 & \lambda_3 & \dots & 0 \\ \vdots &\vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0& \dots & \lambda_n \\ \end{pmatrix} $$ The solution of the differential equation is: $$ \mathbf{x}(t) = e^{At}\mathbf{x_0} $$ My question is: If the matrix $A$ was diagonalizable, then: $$ e^{At} = P \begin{pmatrix} e^{\lambda_1t} & 0 & 0 & \dots & 0 \\ 0 & e^{\lambda_2 t} & 0 & \dots & 0 \\ 0 & 0 & e^{\lambda_3 t} & \dots & 0 \\ \vdots &\vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0& \dots & e^{\lambda_n t} \\ \end{pmatrix} P^{-1} $$ because: $$ e^{At} = \sum_{n = 0}^{\infty} \frac{(At)^n}{n!} = \sum_{n = 0}^{\infty} \frac{(PDP^{-1})^n}{n!}t^n = P \left ( \sum_{n = 0}^{\infty} \frac{D^n}{n!}t^n \right )P^{-1} = P e^{Dt} P^{-1} $$ but, is this last relation: $$ \sum_{n = 0}^{\infty} \frac{(PJP^{-1})^n}{n!}t^n = P \left ( \sum_{n = 0}^{\infty} \frac{J^n}{n!}t^n \right )P^{-1} = P e^{Jt} P^{-1} $$ still apply for a Jordan form? I know that when you sum all the terms involved in $e^{Jt}$, polynomials of $t$ will appear apart from the exponential related to the eigenvalue: $e^{\lambda_k t}$. But, is this still valid?

Answer 1

Yes, although there is a bit of checking you need to do first. First, your Jordan form is slightly wrong. $J$ should be a block diagonal matrix

$$ J = \begin{bmatrix} J_1 & \; & \; \\ \; & \ddots & \; \\ \; & \; & J_p\end{bmatrix} $$

where each block $J_i$ is given by

$$ J_i = \begin{bmatrix} \lambda_i & 1 & \; & \; \\ \; & \lambda_i & \ddots & \; \\ \; & \; & \ddots & 1 \\ \; & \; & \; & \lambda_i \end{bmatrix}. $$ In particular, we only have ones above the diagonals within blocks corresponding to a single eigenvalue each. The matrix exponential of this block diagonal matrix is given by

$$ e^{tJ} = \begin{bmatrix} e^{tJ_1} & \; & \; \\ \; & \ddots & \; \\ \; & \; & e^{tJ_p}\end{bmatrix}. $$ Now we just have to compute the matrix exponential of a single Jordan block.

We decompose the Jordan block $J_i$ into diagonal and off-diagonal terms, $J_i = D_i + N_i$, where $D_i = \lambda_i I$ and

$$ N_i = \begin{bmatrix} 0 & 1 & \; & \; \\ \; & 0 & \ddots & \; \\ \; & \; & \ddots & 1 \\ \; & \; & \; & 0 \end{bmatrix}. $$ These matrices commute, so we can use the fact that $e^{A+B} = e^Ae^B$ for commuting matrices $A$ and $B$. Notice that the Jordan form you wrote does not satisfy this property since it is essential that the diagonal terms are the same for them to commute. Also notice that $N_i$ is nilpotent, so $N_i^k = 0_{k\times k}$, where $k$ is the size of the Jordan block. We can now calculate the matrix exponential.

$$ \begin{aligned} e^{tJ_i} &= e^{tD_i + tN_i} = e^{tD_i}e^{tN_i}\\ &= e^{\lambda_i t}I_{k\times k} \cdot \left(\sum_{j=0}^k \frac{t^jN_i^j} {j!}\right) \\ &= e^{\lambda_i t}\left(\sum_{j=0}^k \frac{t^jN_i^j} {j!}\right), \end{aligned} $$ which is a vector of polynomials in $t$ times $e^{\lambda_i t}$, which is what we would expect. One can then apply the $P$ and $P^{-1}$ from both sides to obtain the final result.