Recipe

Compute $e^{tA}$ using the Jordan-Chevalley decomposition

Use when: you need $e^{tA}$ for an operator $A$ and have (or can find) its Jordan-Chevalley decomposition $A=S+N$ , with $S,N$ commuting, $S$ diagonalizable and $N$ nilpotent

Goal: the closed-form matrix $e^{tA}=e^{tS}e^{tN}$ , with each factor computed by an already-known, easier method

Taught in The matrix exponential

Quick recipe

Find the Jordan-Chevalley decomposition $A=S+N$
Compute $e^{tS}$ using $S$ 's diagonalization
Compute $e^{tN}$ as a finite polynomial in $t$
Multiply $e^{tS}e^{tN}$ together

1
Split $A$ into its diagonalizable part $S$ and its nilpotent part $N$ , and confirm $SN=NS$ .

Example

For $A=\begin{pmatrix}3&1\\0&3\end{pmatrix}$ : $S=3I$ and $N=\begin{pmatrix}0&1\\0&0\end{pmatrix}$ , and $SN=NS=3N$ trivially since $S$ is a scalar multiple of $I$ .

Why: the identity $e^{tS}e^{tN}=e^{tA}$ used in the next steps is only valid when $S$ and $N$ genuinely commute.
2
Diagonalize $S$ (or note it is already diagonal), and exponentiate each eigenvalue separately.

Example

$S=3I$ is already diagonal with a single eigenvalue $3$ , so $e^{tS}=e^{3t}I$ .

⚡ Shortcut

If $A$ is already diagonalizable, then $N=0$ : skip straight to $e^{tA}=e^{tS}$ , with no nilpotent factor or multiplication step needed at all.

Why: this reuses section 6's diagonalizable-matrix construction directly, with no new work needed for this factor.
3
Expand the series $e^{tN}=I+tN+\frac{t^2N^2}{2}+\cdots$ , and stop once a power of $N$ reaches the zero matrix.

Example

$N^2=0$ , so $e^{tN}=I+tN=\begin{pmatrix}1&t\\0&1\end{pmatrix}$ .
- $N^2$ already reaches the zero matrix (a chain of length $2$ ) — the series stops after the linear term, as above.
- a higher power is needed to reach zero (a longer chain) — keep adding terms up through $t^{k-1}N^{k-1}/(k-1)!$ before it terminates.
Why: $N$ being nilpotent is exactly what turns an infinite series into a genuinely finite, exact polynomial in $t$ .
4
Multiply the two factors in either order, since $S$ and $N$ commute; the product is $e^{tA}$ exactly.

Example

$e^{tA}=e^{3t}I\cdot\begin{pmatrix}1&t\\0&1\end{pmatrix}=\begin{pmatrix}e^{3t}&te^{3t}\\0&e^{3t}\end{pmatrix}$ .

Why: this is exactly $e^{tS+tN}=e^{tA}$ , valid here because commuting factors are the one case where $e^Xe^Y=e^{X+Y}$ genuinely holds.

Common mistakes

Multiplying $e^{tS}$ and $e^{tN}$ in the wrong order, or applying $e^Xe^Y=e^{X+Y}$ to matrices that don't actually commute.
Stopping the $e^{tN}$ series one term too early or too late relative to how quickly $N$ 's own powers reach zero.
Forgetting the factorial in the denominator of each term of the $e^{tN}$ series.
Treating $S$ as if it always equals the diagonal entries of $A$ read off directly, instead of the genuine diagonalizable piece from the decomposition.
Reporting $e^{tS}$ alone as the final answer when $A$ genuinely has a nonzero nilpotent part.

Compute e^{tA} using the Jordan-Chevalley decomposition

Compute $e^{tA}$ using the Jordan-Chevalley decomposition