# Thread: Prove that W is isomorphic to the complex numbers C as a real vector space

1. ## Prove that W is isomorphic to the complex numbers C as a real vector space

The field $W=\left\{ \left[ \begin{matrix} a & -b \\ b & a \end{matrix} \right] : a,b\in R \right\}$ is also a vector space over $R$. Do not prove this fact. Prove that $W$ is isomorphic to the complex numbers $C$ as a real vector space.

So far, this is what I have. It's not much, but it's a start.

Let $\varphi :W\rightarrow C$
$\varphi \left( \left[ \begin{matrix} a & -b \\ b & a \end{matrix} \right] \right) =a+bi\in C$
I've been told that I'm to prove that this is one-to-one, onto, and linear.

That's about as much as I know. My brain's a little muddled at the moment due to a short bout of sickness, so I could use a hand.

2. I've not made that much progress on this, but I'll show right now what else I've done so far. Some of it may not be correct in any case. My mind's not in the best shape right now due to sickness, so I've probably got a few things muddled.

My Prof. told me that for this, I had to show that $\varphi \left( \left[ \begin{matrix} a & -b \\ b & a \end{matrix} \right] \right) =a+bi\in C$ was one-to-one, onto, and linear. Proving these is probably easier than I think it is, but I can't help but feel as if I could easily miss something if I'm not careful.

I think I have the one-to-one part (injective) down, though the onto part (surjective) is much trickier due to there being two variables. I don't remember any similar situation, so I'm kinda stuck.

Primarily, I'm having difficulty showing my proofs. Either I'm over-analyzing the proofs and I'm trying to be too complex in showing them, or I don't know what to do entirely.

3. This a truly simple proof.
In the complex number $a+bi$ both $a~\&~b$ are real numbers. Thus you have onto.

Note that $a+bi=c+di$ if and only if $a=c~\&~b=d$: one-to-one.

4. Originally Posted by Plato
This a truly simple proof.
In the complex number $a+bi$ both $a~\&~b$ are real numbers. Thus you have onto.

Note that $a+bi=c+di$ if and only if $a=c~\&~b=d$: one-to-one.
Argh, I was over-analyzing it again. I do that too much because I'm misled by question wording easily.

I think I got the linearity part down, though I could use a second opinion.

5. Originally Posted by Runty
I think I got the linearity part down, though I could use a second opinion.
The mapping preserves both addition (that is easy to see) and multiplication.

Recall $(a+bi)(c+di)=[(ac-bd)+(bc+ad)i]$.

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