Hi I am very stuck on this problem and appreciate any form of help I can get:

Question: Verify that $M_n(\mathbb{H})$ is complete with respect to the norm $\lVert\ \rVert$. Use this to define an exponential function $\text{exp}: \text{M}_n(\mathbb{H}) \rightarrow \text{GL}_n(\mathbb{H})$ with properties analogous to those for the exponential functions on $\text{M}_n(\mathbb{R})$ and $\text{M}_n(\mathbb{C})$.

My problem is that I do not really know how to put a bound on $\lVert A\rVert = sup\{\frac{|Ax|}{|x|}:\ 0 \neq x \in \mathbb{H}^n\}$ where $A \in \text{M}_n(\mathbb{H})$, since I am not sure what is the set of basis in $\mathbb{H}^n$. Also, for the exponential function, I don't understand what I need to do to "define" the function... is it defined by the power series in this case just like $\text{M}_n(\mathbb{R})$ and $\text{M}_n(\mathbb{C})$?

  • 1
    $\begingroup$ what norm? There are lots of norms on matrix rings and (while they should all work) we whould know the one you have in mind to use. $\endgroup$
    – rschwieb
    May 8, 2017 at 19:02
  • $\begingroup$ Yes I should have mentioned it. It is $\Lvert A\Rvert = sup\{\frac{|Ax|}{|x|}:\ 0 \neq x \in \mathbb{H}^n\} $\endgroup$
    – Noob4398
    May 9, 2017 at 1:44
  • $\begingroup$ Presumably the analogous properties are just $\exp(A+B)=\exp(A)\exp(B)$ when $A,B$ commute, or that $X(t)=\exp(tA)$ satisfies $X'=AX$ (as a function of real $t$)? And then, proving the properties requires justifying convergence with respect to the norm as $\exp$ is defined by the power series. Also, presumably the operator norm matches the Frobenius norm, as for $\Bbb R$ or $\Bbb C$. Oh, and be aware you can edit/delete comments until you figure out how to display them properly, so you don't have to leave a comment with unrendered LaTeX hanging like you did. $\endgroup$
    – anon
    May 9, 2017 at 2:21


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