Matrix Exponential¶

Definition¶

For a square matrix $A\in\mathbb{R}^{n\times n}$ (or $\mathbb{C}^{n\times n}$), the matrix exponential is

where $A^0=I$ and powers are matrix products. For a real variable $t$, $\displaystyle \mathrm{e}^{At}=\sum_{k\geq0}\frac{t^k}{k!}A^k$$\sum _{k\ge 0}\frac{t^k}{k!}{A}^k$. The series converges entrywise for every square matrix.

Interpretation¶

The definition copies the Taylor series of the scalar exponential function, $\displaystyle \mathrm{e}^{a}=\sum a^k/k!$$\sum a^k/k!$. Its purpose is the same as in the scalar case: $\frac{\mathrm{d}}{\mathrm{d}t}\mathrm{e}^{At}=A\mathrm{e}^{At}$$A{\mathrm{e}}^{At}$, so $\mathrm{e}^{At}\boldsymbol{x}_0$ solves the linear vector ODE $\boldsymbol{x}'(t)=A\boldsymbol{x}(t)$ with $\boldsymbol{x}(0)=\boldsymbol{x}_0$${\mathbf{x}}_0$.

The series is rarely summed directly. One computes $\mathrm{e}^{At}$ through diagonalization or the Jordan normal form, because $(PDP^{-1})^k=PD^kP^{-1}$$P{D}^k{P}^{-1}$: the exponential respects a similarity transformation.

The main pitfall: $\mathrm{e}^{A}$ is not the matrix of entrywise exponentials $(\mathrm{e}^{a_{ij}})$. For a diagonal matrix the rule is "exponentiate the diagonal entries": the off-diagonal zeros stay $0$ rather than becoming $\mathrm{e}^0=1$$1$. For a non-diagonal matrix the off-diagonal entries of $\mathrm{e}^{A}$ have no entrywise formula (see the examples). Also, $\mathrm{e}^{A+B}=\mathrm{e}^{A}\mathrm{e}^{B}$ can fail when $AB\neq BA$$BA$.

Key Results¶

• $\mathrm{e}^{O}=I$$I$, $\ \frac{\mathrm{d}}{\mathrm{d}t}\mathrm{e}^{At}=A\mathrm{e}^{At}=\mathrm{e}^{At}A$${\mathrm{e}}^{At}A$, and $\mathrm{e}^{A(s+t)}=\mathrm{e}^{As}\mathrm{e}^{At}$${\mathrm{e}}^{As}{\mathrm{e}}^{At}$.
• If $AB=BA$$BA$, then $\mathrm{e}^{A+B}=\mathrm{e}^{A}\mathrm{e}^{B}$${\mathrm{e}}^A{\mathrm{e}}^B$.
• $\mathrm{e}^{A}$ is always invertible with $(\mathrm{e}^{A})^{-1}=\mathrm{e}^{-A}$${\mathrm{e}}^{-A}$, and $\det \mathrm{e}^{A}=\mathrm{e}^{\operatorname{tr}A}$ (determinant, trace).
• If $A=PDP^{-1}$ with $D=\operatorname{diag}(\lambda_1,\dots,\lambda_n)$$\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)$, then $\mathrm{e}^{At}=P\operatorname{diag}(\mathrm{e}^{\lambda_1t},\dots,\mathrm{e}^{\lambda_nt})P^{-1}$$P\mathrm{diag}({\mathrm{e}}^{{\lambda }_{1}t},\dots ,{\mathrm{e}}^{{\lambda }_{n}t})P^{-1}$; the eigenvalues of $\mathrm{e}^{At}$ are $\mathrm{e}^{\lambda_it}$.
• If $N^m=O$ (nilpotent), the series stops: $\displaystyle \mathrm{e}^{Nt}=\sum_{k<m}\frac{t^k}{k!}N^k$$\sum _{k<m}\frac{t^k}{k!}{N}^k$. For a Jordan block $\lambda I+N$ this gives $\mathrm{e}^{(\lambda I+N)t}=\mathrm{e}^{\lambda t}\mathrm{e}^{Nt}$${\mathrm{e}}^{\lambda t}{\mathrm{e}}^{Nt}$.

Examples¶

Diagonal and scalar. $\mathrm{e}^{\operatorname{diag}(2,-1)t}=\operatorname{diag}(\mathrm{e}^{2t},\mathrm{e}^{-t})$$\mathrm{diag}({\mathrm{e}}^{2t},{\mathrm{e}}^{-t})$, and $\mathrm{e}^{\lambda It}=\mathrm{e}^{\lambda t}I$${\mathrm{e}}^{\lambda t}I$.

Jordan block. $A=\begin{pmatrix}2&1\\0&2\end{pmatrix}=2I+N$ with $N=\begin{pmatrix}0&1\\0&0\end{pmatrix}$$\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$, $N^2=O$$O$. Since $2I$ and $N$ commute,

Check: $\det \mathrm{e}^{At}=\mathrm{e}^{4t}=\mathrm{e}^{\operatorname{tr}(At)}$${\mathrm{e}}^{\mathrm{tr}(At)}$.

Diagonalizable. $A=\begin{pmatrix}1&2\\2&1\end{pmatrix}=P\operatorname{diag}(3,-1)P^{-1}$ with $P=\begin{pmatrix}1&1\\1&-1\end{pmatrix}$$\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)$, $P^{-1}=\tfrac{1}{2}\begin{pmatrix}1&1\\1&-1\end{pmatrix}$$\frac{1}{2}\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)$. Then

which equals $I$ at $t=0$$0$.

The figure shows the route taken in this example. Rather than summing the series for $\mathrm{e}^{At}$ directly (dashed), one passes to eigenvector coordinates with $P^{-1}$, where the exponential acts as $\operatorname{diag}(\mathrm{e}^{3t},\mathrm{e}^{-t})$, and returns with $P$. Both routes give the same $\boldsymbol{x}(t)=\mathrm{e}^{At}\boldsymbol{x}_0$ because $\mathrm{e}^{At}=P\operatorname{diag}(\mathrm{e}^{3t},\mathrm{e}^{-t})P^{-1}$$P\mathrm{diag}({\mathrm{e}}^{3t},{\mathrm{e}}^{-t})P^{-1}$.

Rotation. For $A=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$$\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$, $A^2=-I$$-I$, and splitting the series into even and odd powers gives $\mathrm{e}^{At}=\begin{pmatrix}\cos t&-\sin t\\\sin t&\cos t\end{pmatrix}$$\left(\begin{array}{cc}\cos t& -\sin t\\ \sin t& \cos t\end{array}\right)$, the matrix analogue of Euler's formula.

Prerequisites¶

• Matrix Product
• Taylor Series
• Exponential Function
• Diagonalization
• Jordan Normal Form

Related Concepts¶

• Linear Vector ODE
• General Solution of a Linear ODE
• System of Differential Equations
• Similarity Transformation

Sources¶

• Math V: Advanced Linear Algebra · Week 4

Aliases: exponential of a matrix, e^{At} Keywords: exp(A), e^{At}=Pe^{Dt}P^{-1}, exponential series, nilpotent matrix, det e^A = e^{tr A} Symbols: complex numbers, element of, infinity, lambda, real numbers, sum

Last updated: 22 October 2026 19:04