Prove the function $f(a+bi)=\begin{bmatrix}a & b\\-b & a\end{bmatrix}$ is an isomorphism from $\mathbb{C}$ to $L$.

Question

Let $L$ be the following subset of $M_2 (\mathbb{R})$ $$L=\left\{\begin{bmatrix}a & b\\-b & a\end{bmatrix}:a,b \in \mathbb{R}\right\}.$$

Prove the function $f(a+bi)=\begin{bmatrix}a & b\\-b & a\end{bmatrix}$ is an isomorphism from $\mathbb{C}$ to $L$.

I know that I will have to prove that $f(a+bi*c+di)=f(a+bi)f(c+di)$ for any $a+bi, c+di \in \mathbb{C}$, which I have done. I am struggling to show the 1-1 correspondence.

Answer 1

For the inverse of $f$, consider $\varphi: L \to \mathbb{C}$ defined by $\varphi\left( \begin{bmatrix} a & b \\ -b & a \\ \end{bmatrix}\right)= a + bi.$

I also wanted to point out that $$(a+bi) \begin{bmatrix} 1 \\ i \end{bmatrix} = \begin{bmatrix} a + bi \\ -b + ai \end{bmatrix} = \begin{bmatrix} a & b \\ -b & a \\ \end{bmatrix}\begin{bmatrix} 1 \\ i \end{bmatrix}$$

Answer 2

If the function is a one-to-one correspondence, it is both injective and surjective. To prove injectivity, we must show that $f(a + bi) = f(c + di) \implies a = c \space \wedge \space b =d $. Suppose $f(a+bi)=f(c+di)$. Then, $$\begin{bmatrix}a&b\\-b&a\end{bmatrix}=\begin{bmatrix}c&d\\-d&c\end{bmatrix}$$ Since the matrixes are the same size, they are component-wise equivalent. So, $a=c$ and $b=d$ implying injectivity.

To prove surjectivity, we must show that $\forall f(x) \in L, \exists x \in \mathbb{C}$. This is easy. If our matrix is $\begin{bmatrix}a&b\\-b&a\end{bmatrix}$, take $a+bi$ as the element of the domain that corresponds to it. Therefore, the function is surjective.

Since the function is both surjective and injective, it is bijective, or a one-to-one correspondence. Lastly, we need to show there is a homomorphism. That is,$f(a+bi*c+di)=f(a+bi)f(c+di)$, which is what you've done.