## 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?