Show this matrix is isomorphic to complex number

[Confusedalways]

So the question is show that
$$S=\left\{ \begin{pmatrix} a & b\\ -b & a \end{pmatrix} :a,b \in \Bbb{R} ,\text{ not both zero}\right\}$$ is isomorphic to $\Bbb{C}^*$, which is a non-zero complex number considered as a group under multiplication

So I've shown that it is a group homomorphism by showing how if
$$θ: x+iy → \begin{pmatrix} x & y\\ -y & x \end{pmatrix}$$ then I proved that
θ((x₁+iy₁ )(x₂ +iy₂))=θ(x₁+iy₁)θ(x₂+iy₂)

But I'm stuck on how to show it is both injective and surjective to prove it's an isomorphism?

Thanks!

 

[I like Serena]

Hi Confusedalways,

Doesn't every matrix in S have a unique complex number associated with it through θ?
In other words, every matrix in S has at least one (surjective) and at most one (injective) original in $\mathbb C^*$.

 

[HallsofIvy]

Your title, "Show this matrix is isomorphic to complex number" puzzled me! Individual matrices are not "isomorphic" to anything. What you mean is "Show that the group of all non-zero matrices of this form is isomorphic to the group of all non-zero complex numbers" with multiplication as operation.

That your function is "injective" and "surjective" follows immediately from the definitions.
A function, f, is "injective" if and only if f(x)= f(y) implies x= y. Here, x and y are the given type of matrices. Let $$x= \begin{pmatrix} a & -b \\ b & a \end{pmatrix}$$ and $$y= \begin{pmatrix}a' & -b' \\ b' & a'\end{pmatrix}$$. Then f(x)= a+ bi= f(y)= a'+ b'i. It immediately follows that a= a' and b= b' so that x= y.

A function, f, is "surjective" if and only if, for any y in its range space, there exist x in its domain space, such that f(x)= y. Take y to be the generic a+ ib. It is immediately follows that $$x= \begin{pmatrix}a & -b \\ b & a\end{pmatrix}$$ is the correct x.