[Updated: 20-11-2010]

Zorn Leung just discovered an interesting fact in algebra. We define the algebra by

with ordinary addition and multiplication. For simplicity, we denote the above algebra simply by . Now we define another algebra on the same underlying set and with the same addition, except we define

Let us denote the later algebra simply by . If is the whole matrix algebra (instead of the above ), of course taking the transpose map will make an algebra isomorphism from to (similarly defined), but what surprised me is the following result:

Theorem 1 (Zorn Leung’s theorem, 2010)is isomorphic to via the map

Well, of course, it is simple to prove it by direct checking. ~~What’s intriguing to me is that I don’t know the fundamental reason, other than using brute force, why this is true.~~ Zorn Leung had a proof which gives a clearer picture, but my understanding is still not very good. (Perhaps this is a very well-known fact to algebraists, but since I am quite rusty in algebra, I hope somebody could come up and say something. )

Here’s a proof by Zorn Leung:

*Proof:* Note that the homomorphism is given by

where is an orthogonal matrix. It is then easy to see that the above map is a homomorphism.

It’s easy to see that the above proof extends to the higher dimensional case. And, in addition, it is also easy to see that for any orthogonal , the (bijective!) map is an algebra isomorphism . I still have some questions:

- Is any algebra isomorphism necessarily of the above form? (I don’t know the answer)
- Is any algebra isomorphism (as in Zorn Leung’s theorem, or its higher dimensional analogue) necessarily of the form given by the theorem above? (I think the answer is no. )

1. Note that . Claim: , where isomorphism is given by .

Pf: Let . . So, for any , . Therefore, for all which implies .

Also, . Thus, . Therefore, it is a linear isomorphism.

Finally, . Also, . Thus it is an isomorphism.

So, the question become what is the isomorphism between and , which is in some basis(I need to think here). And by change of basis, I think the isomorphism can be given by , where is invertible.

Dear Zorn Leung,

What does “an isomorporphism…, which is

in some basis” mean? How can an (algebra) isomorphism depends on a choice of basis (of their underlying vector spaces)?What I mean is that there is an automorphism in and , and thus all the isomorphism have to composite with those automorphism. So, what I wondering is up to automorphism of and , does there any other algebra isomorphism?