You can give an equivalent definition of the matrix exponential. or any matrix function $f$, using the Jordan form. If $A = VJV^{-1}$, and $J$ is the direct sum of diagonal blocks of the form $$ J_i = \begin{bmatrix} \lambda_i & 1\\ & \lambda_i & \ddots\\ & & \ddots & 1\\ & & & \lambda_i \end{bmatrix}, $$ then $f(A) = Vf(J)V^{-1}$, where $f(J)$ is the direct sum of Toeplitz triangular blocks of the form $$ J_i = \begin{bmatrix} f(\lambda_i) & f'(\lambda_i) & \frac{f''(\lambda_i)}{2} & \dots & \frac{f^{(k)}(\lambda_i)}{k!}\\ & \ddots & \ddots & & \vdots\\ & & \ddots &\\ & & & f(\lambda_i)& f'(\lambda_i)\\ & & & & f(\lambda_i) \end{bmatrix}. $$ For the exponential, the derivatives are trivial, so any field for which you can define both the scalar $\exp$ and the Jordan form works. Algebraically closed is the only condition needed for the Jordan form, if I am not missing anything. [EDIT: Also the fractions $\frac{1}{k!}$ are a problem in non-zero characteristic, unfortunately. Note, though, that they are needed only in cases where $A$ has a Jordan block of size larger than $\operatorname{char}(\mathbb{F})$, so at least for some matrices you can still apply the definition.]

Also, at this point, I imagine that you can extend the definition also to a non-algebraically-closed field $\mathbb{F}$: work in its algebraic closure $\overline{\mathbb{F}}$, and note that $f(A)$ must have entries in $\mathbb{F}$ since it does not depend on the choice of the closure.

Alternatively, a matrix function $f(A)$ can also be defined using Hermite interpolation: take a polynomial $p(x)$ such that $p^{(j)}(\lambda_i) = f^{(j)}(\lambda_i)$, for each eigenvalue $i$ and each $j=0,1,\dots, m_a(\lambda_i)$ (algebraic multiplicity of $\lambda_i$); then $f(A) = p(A)$. This definition works out of the box for the exponential.

Note that all these definitions coincide with the limit of the series in cases where it is defined.

For more information on these techniques, you can check Chapter 1 in Higham's excellent book Functions of matrices.

You can give an equivalent definition of the matrix exponential. or any matrix function $f$, using the Jordan form. If $A = VJV^{-1}$, and $J$ is the direct sum of diagonal blocks of the form $$ J_i = \begin{bmatrix} \lambda_i & 1\\ & \lambda_i & \ddots\\ & & \ddots & 1\\ & & & \lambda_i \end{bmatrix}, $$ then $f(A) = Vf(J)V^{-1}$, where $f(J)$ is the direct sum of Toeplitz triangular blocks of the form $$ J_i = \begin{bmatrix} f(\lambda_i) & f'(\lambda_i) & \frac{f''(\lambda_i)}{2} & \dots & \frac{f^{(k)}(\lambda_i)}{k!}\\ & \ddots & \ddots & & \vdots\\ & & \ddots &\\ & & & f(\lambda_i)& f'(\lambda_i)\\ & & & & f(\lambda_i) \end{bmatrix}. $$ For the exponential, the derivatives are trivial, so any field for which you can define both the scalar $\exp$ and the Jordan form works. Algebraically closed is the only condition needed for the Jordan form, if I am not missing anything. [EDIT: Also the fractions $\frac{1}{k!}$ are a problem in non-zero characteristic, unfortunately. Note, though, that they are needed only in cases where $A$ has a Jordan block of size larger than $\operatorname{char}(\mathbb{F})$, so at least for some matrices you can still apply the definition.]

Also, at this point, I imagine that you can extend the definition also to a non-algebraically-closed field $\mathbb{F}$: work in its algebraic closure $\overline{\mathbb{F}}$, and note that $f(A)$ must have entries in $\mathbb{F}$ since it does not depend on the choice of the closure.

Alternatively, a matrix function $f(A)$ can also be defined using Hermite interpolation: take a polynomial $p(x)$ such that $p^{(j)}(\lambda_i) = f^{(j)}(\lambda_i)$, for each eigenvalue $i$ and each $j=0,1,\dots, m_a(\lambda_i)$ (algebraic multiplicity of $\lambda_i$); then $f(A) = p(A)$. This definition works out of the box for the exponential.

Note that all these definitions coincide with the limit of the series in cases where it is defined.

For more information on these techniques, you can check Chapter 1 in Higham's excellent book Functions of matrices.

EDIT: note that the second method has the same requirements on denominators as the first one, since the formulas to define Hermite interpolants use $\frac{f^{(k)}(\lambda_i)}{k!}$ internally; see e.g. the divided differences method on Wikipedia.

You can give an equivalent definition of the matrix exponential. or any matrix function $f$, using the Jordan form. If $A = VJV^{-1}$, and $J$ is the direct sum of diagonal blocks of the form $$ J_i = \begin{bmatrix} \lambda_i & 1\\ & \lambda_i & \ddots\\ & & \ddots & 1\\ & & & \lambda_i \end{bmatrix}, $$ then $f(A) = Vf(J)V^{-1}$, where $f(J)$ is the direct sum of Toeplitz triangular blocks of the form $$ J_i = \begin{bmatrix} f(\lambda_i) & f'(\lambda_i) & \frac{f''(\lambda_i)}{2} & \dots & \frac{f^{(k)}(\lambda_i)}{k!}\\ & \ddots & \ddots & & \vdots\\ & & \ddots &\\ & & & f(\lambda_i)& f'(\lambda_i)\\ & & & & f(\lambda_i) \end{bmatrix}. $$ For the exponential, the derivatives are trivial, so any field for which you can define the scalar $\exp$ and write the Jordan form works; Algebraically closed is the only condition needed for the Jordan form, if I am not missing anything. Also at this point I imagine you can extend the definition to any field $\mathbb{F}$ for which $\exp$ is defined as well: work in its algebraic closure $\overline{\mathbb{F}}$, and note that $f(A)$ must have entries in $\mathbb{F}$ since it does not depend on the choice of the closure.

Alternatively, a matrix function $f(A)$ can also be defined using Hermite interpolation: take a polynomial $p(x)$ such that $p^{(j)}(\lambda_i) = f^{(j)}(\lambda_i)$, for each eigenvalue $i$ and each $j=0,1,\dots, m_a(\lambda_i)$ (algebraic multiplicity of $\lambda_i$); then $f(A) = p(A)$. This works out of the box for the exponential.

All these definitions coincide with the limit of the series in cases where it is defined.

For more information on these techniques, you can check Chapter 1 in Higham's excellent book Functions of matrices.

You can give an equivalent definition of the matrix exponential. or any matrix function $f$, using the Jordan form. If $A = VJV^{-1}$, and $J$ is the direct sum of diagonal blocks of the form $$ J_i = \begin{bmatrix} \lambda_i & 1\\ & \lambda_i & \ddots\\ & & \ddots & 1\\ & & & \lambda_i \end{bmatrix}, $$ then $f(A) = Vf(J)V^{-1}$, where $f(J)$ is the direct sum of Toeplitz triangular blocks of the form $$ J_i = \begin{bmatrix} f(\lambda_i) & f'(\lambda_i) & \frac{f''(\lambda_i)}{2} & \dots & \frac{f^{(k)}(\lambda_i)}{k!}\\ & \ddots & \ddots & & \vdots\\ & & \ddots &\\ & & & f(\lambda_i)& f'(\lambda_i)\\ & & & & f(\lambda_i) \end{bmatrix}. $$ For the exponential, the derivatives are trivial, so any field for which you can define both the scalar $\exp$ and the Jordan form works. Algebraically closed is the only condition needed for the Jordan form, if I am not missing anything. [EDIT: Also the fractions $\frac{1}{k!}$ are a problem in non-zero characteristic, unfortunately. Note, though, that they are needed only in cases where $A$ has a Jordan block of size larger than $\operatorname{char}(\mathbb{F})$, so at least for some matrices you can still apply the definition.]

Also, at this point, I imagine that you can extend the definition also to a non-algebraically-closed field $\mathbb{F}$: work in its algebraic closure $\overline{\mathbb{F}}$, and note that $f(A)$ must have entries in $\mathbb{F}$ since it does not depend on the choice of the closure.

Alternatively, a matrix function $f(A)$ can also be defined using Hermite interpolation: take a polynomial $p(x)$ such that $p^{(j)}(\lambda_i) = f^{(j)}(\lambda_i)$, for each eigenvalue $i$ and each $j=0,1,\dots, m_a(\lambda_i)$ (algebraic multiplicity of $\lambda_i$); then $f(A) = p(A)$. This definition works out of the box for the exponential.

Note that all these definitions coincide with the limit of the series in cases where it is defined.

For more information on these techniques, you can check Chapter 1 in Higham's excellent book Functions of matrices.