The matrix $A$ has the interesting property that its square is minus the identity matrix: $A^2 = \begin{bmatrix}-1&0\\0&-1\end{bmatrix}$. That suggests that it should correspond to the complex number $i$.
In fact, $K$ consists of all matrices of the form $\begin{bmatrix}a&-b\\b&a\end{bmatrix}$ where $a$ and $b$ are real numbers. To prove that $K$ is isomorphic to $\Bbb C$ you should show that the map taking $\begin{bmatrix}a&-b\\b&a\end{bmatrix}$ to $a+ib$ is an isomorphism. That is, it is a bijective map that preserves addition and multiplication.
#3
Kansas Boy
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It is sufficient to map $\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$ to 1 and map $\begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}$ to i.
Are there known conditions under which a Markov Chain is also a Martingale? I know only that the only Random Walk that is a Martingale is the symmetric one, i.e., p= 1-p =1/2.