ring homomorphism mapping c to 2x2 matrices

[renegade05]

Hello there, trying to show that the map:

\(\displaystyle \phi:a+ib \to \begin{bmatrix} a&-b \\
b&a
\end{bmatrix}\)

is a ring homomorphism.

I have got the $\displaystyle \phi(z1)\pm\phi(z2)=\phi(z1\pm z2)$ and
$\displaystyle \phi(z1z2)=\phi(z1)\phi(z2)$

however proving \(\displaystyle \phi(0)=\begin{bmatrix} 0&0 \\
0&0
\end{bmatrix}\) and \(\displaystyle \phi(1)=\begin{bmatrix} 1&0 \\
0&1
\end{bmatrix}= I\)

is proving difficult for me for some reason.

Should i start with letting $\displaystyle z=a+ib=0$ then $\displaystyle -a=ib$ and $\displaystyle \phi(0)= ?$ ... I dont know.. how do i work with that? Should i convert to polar? $\displaystyle z=a+ib=0=rcis{\theta}$ But im not sure where that leads either. I think I'm over thinking this. Please some guidance.

Also (SIDENOTE), is there no latex command for the shorthand 'cis' ? \cis doesn't work?

 

[pka]

\(\displaystyle \phi:a+ib \to \begin{bmatrix} a&-b \\
b&a
\end{bmatrix}\)

is a ring homomorphism.

What are the requirements to show a ring homomorphism?

[ tex]\text{cis}(\theta)[/tex] gives $\displaystyle \text{cis}(\theta)$

 

[renegade05]

$\displaystyle \phi(z1)\pm\phi(z2)=\phi(z1\pm z2)$
$\displaystyle \phi(z1z2)=\phi(z1)\phi(z2)$
$\displaystyle \phi(1)=1$
$\displaystyle \phi(0)=0$

I can show the first two, but the last two I dont know how to do it applied to this problem.

 

[pka]

$\displaystyle \phi(z1)\pm\phi(z2)=\phi(z1\pm z2)$
$\displaystyle \phi(z1z2)=\phi(z1)\phi(z2)$
$\displaystyle \phi(1)=1$
$\displaystyle \phi(0)=0$

I can show the first two, but the last two I dont know how to do it applied to this problem.

Surely \(\displaystyle \phi(1)=\phi(1+0i)=\left[ {\begin{array}{*{20}{c}} 1&{ - 0} \\ 0&1\end{array}} \right]\)

What do you not understand?

 

[renegade05]

Surely \(\displaystyle \phi(1)=\phi(1+0i)=\left[ {\begin{array}{*{20}{c}} 1&{ - 0} \\ 0&1\end{array}} \right]\)

What do you not understand?

why surely?

if $\displaystyle \phi(1)$ then $\displaystyle z=1=a+bi$ does this make it self evident that $\displaystyle 1-a=bi$ and $\displaystyle a=1, b=0$ is the only solution? Is there anyway to show this algebraically? is that the only solution? I'm new to complex analysis: is there any other possible solutions to $\displaystyle 1-a=bi$ ?

 

[pka]

why surely?

if $\displaystyle \phi(1)$ then $\displaystyle z=1=a+bi$ does this make it self evident that $\displaystyle 1-a=bi$ and $\displaystyle a=1, b=0$ is the only solution? Is there anyway to show this algebraically? is that the only solution? I'm new to complex analysis: is there any other possible solutions to $\displaystyle 1-a=bi$ ?

I am sorry to tell you that it appears that you have no idea what any of this is about.
I suggest that you start over with a course in pre-algebra.
You are wasting your time unless you know basic algebra.

 

[daon2]

why surely?

if $\displaystyle \phi(1)$ then $\displaystyle z=1=a+bi$ does this make it self evident that $\displaystyle 1-a=bi$ and $\displaystyle a=1, b=0$ is the only solution? Is there anyway to show this algebraically? is that the only solution? I'm new to complex analysis: is there any other possible solutions to $\displaystyle 1-a=bi$ ?

I think you are missing a key, very basic, fact. Perhaps you are thinking too hard. The fact that $\displaystyle a+bi$ is real if and only if $\displaystyle b=0$ tells you that a real number $\displaystyle r$ is the same as $\displaystyle r+0i$. There is no algebra that needs to be done.

 

[renegade05]

I am sorry to tell you that it appears that you have no idea what any of this is about.
I suggest that you start over with a course in pre-algebra.
You are wasting your time unless you know basic algebra.

I am a 3rd year honors math major who has aced every math course I have been in... I don't need a course in pre-algebra. Thanks for that remark though?

Thanks daon2 thats the answer I was looking for. Just a case of over thinking it.