Prove that the monoid $M_n(\mathbb{R})$, where the operation is matrix multiplication, is isomorphic to $M_n(\mathbb{R})^{op}$

Question

I just need a hint as to what to pick for a function. I know that I have to prove the there exists a bijective mapping that preserves the operations and maps the identity from $M_n(\mathbb{R})$ to the identity in $M_n(\mathbb{R})^{op}$. I am just having difficulty in choosing a mapping.

Answer 1

Take$$\begin{array}{ccc}M_n(\mathbb{R})&\longrightarrow&M_n(\mathbb{R})\\M&\mapsto&M^t.\end{array}$$

Answer 2

We want to show that there is a bijective mapping $f: M_n(\mathbb{R}) \rightarrow M_n(\mathbb{R})^{op}$ such that $f(I_M)=\hat{I}_M$, $f(x*y)=f(x)\ \hat{*}\ f(y)$, where $I_M$ is the identity in $M_n(\mathbb{R})$, $\hat{I}_M$ is the identity in $M_n(\mathbb{R})^{op}$, and $x,y \in M.$

First we show that $f$ is a homomorphism. Let $f(M) = M^t$, where $M^t$ is the transposition of matrix $M.$ The identity is immediately available: $f(I_M)=(I_M)^t=\hat{I}_M$. To show that the operation is preserved, for some $A,B \in M_n(\mathbb{R})$ take $f(A*B)=(A*B)^t=B^t * A^t = A^t\ \hat{*}\ B^t = f(A)\ \hat{*}\ f(B).$

To show $f$ is an isomorphism, we want to show that $f^{-1}: M_n(\mathbb{R})^{op} \rightarrow M_n(\mathbb{R})$ exists such that $f^{-1}f= 1_{M_n(\mathbb{R})}$ and $ff^{-1}=1_{M_n(\mathbb{R})^{op}}$. If $f(M)=M^t$, then let $f^{-1}(M^t)=(M^t)^t=M$. Then for any $X^t \in M_n(\mathbb{R})^{op}$, we have $f^{-1}(X^t)=X$ and $ff^{-1}(X^t)=f(f^{-1}(X^t))=f(X)=X^t$, which implies $ff^{-1}=1_{M_n(\mathbb{R})^{op}}$. And for any $X \in M_n(\mathbb{R})$, we have $f(X)=X^t$ and $f^{-1}f(X)=f^{-1}(f(X))=f^{-1}(X^t)=X$, which implies $f^{-1}f= 1_{M_n(\mathbb{R})}$.

Therefore the monoid $M_n(\mathbb{R})$ is isomorphic to $M_n(\mathbb{R})^{op}$.