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Is the matrix exponential independent of the norm?

Mathematics Asked by user736690 on February 12, 2021

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)

One Answer

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!

Answered by Qiaochu Yuan on February 12, 2021

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