0
$\begingroup$

Studying the Lyapunov equation (given $X,Y$ with disjoint spectrum, $XA+AY=Z$; find $A$) I came to this more general question to which I did not find any convenient answer so far. Let $V$ be a real (fin.dim) vector space. Consider the vectorspace (not the algebra!) $\text{End}(V)$. Suppose there is a isomorphism $\phi:\text{End}(V) \to \text{End}(V)$. The natural questions are then

1.: When is the preimage of an iso again an iso?

2.: When is the image of an iso again an iso?

The very first idea I had, was that this is 'iff', but this is clearly wrong. One could understand (maybe even should? picking a basis in $V$) the elements of $\text{End}(V)$ as vector in $\mathbb{R}^{n^2}$ ($n=\dim V$), so that $\phi$ can be understood as $\in \text{End}(\mathbb{R}^{n^2})$. The problem then is, that I don't see any way to adress 'iso' for $\mathbb{R}^{n^2}$ that is somehow connected to the concept of iso in $\text{End}(\mathbb{R}^{n^2})$. Personally I would be more interessted in the first question.

Any ideas? It seems that there are some answers around but they seem to tackle different problems, as far, as I understood them. Thanks in advance!

$\endgroup$
  • $\begingroup$ Why would you consider vector space automorphisms of $\operatorname{End}(V)$ that are incompatible with the algebra structure to begin with? $\endgroup$ – Christoph Feb 28 '18 at 12:57
  • 1
    $\begingroup$ Because $\phi$ would not be iso anymore. In the more specific picture of the Lyapuno eq., given $X,Y$ with disjoint spectrum, there is a unique solution to any $Z$ and $\phi:A \mapsto XA+AY$ is linear and its inverse as well (of course). However $\phi$ does not respect multiplication. $\endgroup$ – Caroline Feb 28 '18 at 13:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.