The definition of the matrix exponential uses powers, multiplication by scalars, sums and a limit. The former 3 are given by the vector space the matrices are in. A limit would require making that space of matrices into a normed space, i believe. Since there are different possible norms for a space of matrices, it is not clear that the same sequences converge for each norm, which could mean that the matrix exponential depends on norm.

While writing this i remembered that a space of matrices is isomorphic (as a vector space) to $\mathbb{F}^k$ for some $k \in \mathbb{N}$ and some field $\mathbb{F}$ and all norms in $\mathbb{F}^k$ are equivalent. I believe isomorphism as a vector space would imply that all norms are equivalent since norms are defined in terms of the vector space. Is this correct?

(Edit: My last claim might be wrong if what someone commented is true, but its at least true for the real and complex fields)

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    $\begingroup$ Yes. Convergence the series for exponential does not depend on the norm. $\endgroup$ Commented Sep 30, 2020 at 8:49
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    $\begingroup$ Yes, in finite-dimensional spaces, all norms are equivalent. $\endgroup$
    – user436658
    Commented Sep 30, 2020 at 9:00
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    $\begingroup$ All norms on $\mathbb F^k$ are equivalent in the case that $\mathbb F$ is complete. So $\mathbb R$ and $\mathbb C$ are OK. It is an interesting exercise to find an example two norms on $\mathbb Q^2$ that are not equivalent. $\endgroup$
    – GEdgar
    Commented Oct 1, 2020 at 0:07

1 Answer 1


The limit only depends on the underlying topology, and all norms on a finite-dimensional vector space generate the same topology, which after picking a basis is the topology of pointwise convergence of coefficients. So the matrix exponential is independent of the choice of norm, but it's good of you to have spotted the possibility that it might not be!

  • $\begingroup$ Do non-equivalent norms over a finite-dimensional vector space (as can happen over non-complete fields: mathoverflow.net/q/265192/150499) still generate the same topology? $\endgroup$
    – user736690
    Commented Dec 6, 2020 at 13:46
  • $\begingroup$ @Carla: do you ask that the norm take values in the field? Does the field itself have an order or a (real-valued) norm? I don’t know what the standard definition is here. $\endgroup$ Commented Dec 6, 2020 at 16:45

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