An interesting fact in algebra

[Updated: 20-11-2010]
Zorn Leung just discovered an interesting fact in algebra. We define the algebra {(M, +, \times)} by

\displaystyle M:=\left\{\begin{pmatrix} a &b\\ 0 & c \end{pmatrix}\in M_2(\mathbb{R}) \right\}

with ordinary addition and multiplication. For simplicity, we denote the above algebra simply by {M}. Now we define another algebra {(M, + , *)} on the same underlying set and with the same addition, except we define

\displaystyle A*B:=BA \quad (\text{ordinary multiplication})

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

Theorem 1 (Zorn Leung’s theorem, 2010) {M} is isomorphic to {M^*} via the map

\displaystyle  \begin{pmatrix} a &b\\ 0 &c \end{pmatrix} \mapsto \begin{pmatrix} c &b\\ 0 &a \end{pmatrix}.

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

\displaystyle A\mapsto QA^t Q^t

where {Q=\begin{pmatrix} 0 &1\\ 1 &0 \end{pmatrix}\in O(2)} is an orthogonal matrix. It is then easy to see that the above map is a homomorphism. \Box

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 {Q\in O(n)}, the (bijective!) map {A\mapsto Q A^t Q^t} is an algebra isomorphism {M_n(\mathbb{R})\rightarrow M_n(\mathbb{R})^*}. I still have some questions:

  1. Is any algebra isomorphism {M_n(\mathbb{R})\rightarrow M_n(\mathbb{R})^*} necessarily of the above form? (I don’t know the answer)
  2. Is any algebra isomorphism {M\rightarrow M^*} (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. )

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3 Responses to An interesting fact in algebra

  1. kingleunglee says:

    1. Note that M_n(R) = L({\mathbb{R}}^n,{\mathbb{R}}^n). Claim: M_n(R)^{*} \cong L(({\mathbb{R}}^n)^{*},{(\mathbb{R}}^n)^{*}), where isomorphism is given by \phi(A)(x^{t})=x^t A.
    Pf: Let x \in {\mathbb{R}}^n. \phi((A+cB)x)=x^t  (A+cB). So, for any y, (\phi(A+cB)x)(y)=x^t(A+cB)y=x^tAy+c(x^tBy). Therefore, \phi(A+cB)(x^t)=(\phi(A)+c\phi(B))x^t for all x which implies \phi(A+cB)=\phi(A)+c\phi(B).
    Also, (\phi(0_{n \times n})x)y=x^t 0_{n \times n} y=0. Thus, \phi(0_{n \times n})=0_{n \times n}. Therefore, it is a linear isomorphism.
    Finally, \phi(A*B)(x^{t})=\phi(BA) (x^{t})=x^t (BA). Also, \phi(A)(\phi(B) (x^{t}))=\phi(A)(x^t B)=x^t(BA). Thus it is an isomorphism.
    So, the question become what is the isomorphism between L({\mathbb{R}}^n,{\mathbb{R}}^n) and L(({\mathbb{R}}^n)^{*},{(\mathbb{R}}^n)^{*}), which is A \mapsto A^t in some basis(I need to think here). And by change of basis, I think the isomorphism can be given by A \mapsto Q A^t Q^{-1}, where Q is invertible.

  2. KKK says:

    Dear Zorn Leung,

    What does “an isomorporphism…, which is A\mapsto A^t in some basis” mean? How can an (algebra) isomorphism depends on a choice of basis (of their underlying vector spaces)?

  3. kingleunglee says:

    What I mean is that there is an automorphism in L({\mathbb{R}}^n,{\mathbb{R}}^n) and L(({\mathbb{R}}^n)^{*},{(\mathbb{R}}^n)^{*}), and thus all the isomorphism have to composite with those automorphism. So, what I wondering is up to automorphism of L({\mathbb{R}}^n,{\mathbb{R}}^n) and L(({\mathbb{R}}^n)^{*},{(\mathbb{R}}^n)^{*}), does there any other algebra isomorphism?

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