# Homomorphism from the complex to 2*2 matrices

#### ulysses123

is there a homomorphism
Φ:C×C→M₂(R)
Φ must be onto
I have tried variations on the type of entries in the 2 by 2 matrices that will preserve the additive and multiplicative properties of a homomorphism but this seems quite long. Is there a better way for me to find if a homomorphism exists that is onto?

#### HallsofIvy

MHF Helper
Are you sure the question is whether there is a homomorphism between $$\displaystyle C\times C$$ and $$\displaystyle M_2(R)$$ and not just between C and $$\displaystyle M_2(R)$$?

There is an obvious homomorphism $$\displaystyle a+ bi\rightarrow \begin{pmatrix}a & -b \\ b & a}\end{pmatrix}$$.

#### ulysses123

yes i know that homomorphism it is in our text book.The question specifically asks for two input values and this homomorphism doesnt seem to preserve the multiplicative structure.I have tried a similar matrix.

#### Bruno J.

MHF Hall of Honor
Are you sure the question is whether there is a homomorphism between $$\displaystyle C\times C$$ and $$\displaystyle M_2(R)$$ and not just between C and $$\displaystyle M_2(R)$$?

There is an obvious homomorphism $$\displaystyle a+ bi\rightarrow \begin{pmatrix}a & -b \\ b & a}\end{pmatrix}$$.
This is not a surjective homomorphism, however!
If the question was just to find a homomorphism, there would always be the zero homomorphism.

Think of $$\displaystyle \mathbb{C}^2$$ as a $$\displaystyle 4$$-dimensional vector space over $$\displaystyle \mathbb{R}$$. As a vector space, $$\displaystyle M_2(\mathbb{R})$$ is just the same thing!

HallsofIvy

#### ulysses123

i still cant figure this out. I see that there are 4 entries of real numbers but how can i use this fact to get a surgective homomorphism?
the only thing i can think of is that 4 linearly independant 2*2 matrices will generate "all" real 2*2 matrices. But i dont think the the multiplication property of an isomorphism holds in the case where i take the matrix entries to be a,b,c,d.

#### Bruno J.

MHF Hall of Honor
i still cant figure this out. I see that there are 4 entries of real numbers but how can i use this fact to get a surgective homomorphism?
the only thing i can think of is that 4 linearly independant 2*2 matrices will generate "all" real 2*2 matrices. But i dont think the the multiplication property of an isomorphism holds in the case where i take the matrix entries to be a,b,c,d.
I think you are confused about the group operation! The set of square 2x2 matrices is not a group under matrix multiplication, because not all matrices are invertible. It is a group under addition though (just like $$\displaystyle \mathbb{C}^2$$).

#### Jose27

MHF Hall of Honor
Think of $$\displaystyle \mathbb{C}^2$$ as a $$\displaystyle 4$$-dimensional vector space over $$\displaystyle \mathbb{R}$$. As a vector space, $$\displaystyle M_2(\mathbb{R})$$ is just the same thing!
The problem with this is that they're isomorphic as vector spaces but what is asked is an onto ring homomorphism which actually doesn't exist since $$\displaystyle \mathbb{C}^2$$ is a commutative ring and $$\displaystyle M_2(\mathbb{R})$$ is not.

#### ulysses123

but does the homomorphism have to be comutative?
When i chech the multiplication property F{(x,y)(u,v)}=F(x,y)F(u,v)
Do i multiply components only or everything?

eg
F{(a+bi,c+di)(e+fi,g+hi)}
doi have
F{(ae-fi+afi+bei,cg-dh+chi+dgi)}
or do i multiply everything?
Because in this case a homorphism does exist even though it is not surjective, otherwise no homomorphism exists.

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#### Also sprach Zarathustra

General result I think is:

There is homomorphism between $$\displaystyle \mathbb{R}^n$$ and the field of matrices in $$\displaystyle M_n(\mathbb{R})$$

(I think we proved this theorem on linear algebra 1)

#### Jose27

MHF Hall of Honor
but does the homomorphism have to be comutative?
When i chech the multiplication property F{(x,y)(u,v)}=F(x,y)F(u,v)
Do i multiply components only or everything?

eg
F{(a+bi,c+di)(e+fi,g+hi)}
doi have
F{(ae-fi+afi+bei,cg-dh+chi+dgi)}
or do i multiply everything?
Because in this case a homorphism does exist even though it is not surjective, otherwise no homomorphism exists.
In general the homomorphic image of a commutative ring (obviously a ring hom.) is commutative: Take $$\displaystyle f:A\rightarrow B$$ where $$\displaystyle A$$ is commutative and $$\displaystyle B$$ is simply a ring, then without loss of generality $$\displaystyle f$$ is onto (because $$\displaystyle f(A)$$ is a subring) then take $$\displaystyle x,y\in B$$ then $$\displaystyle x=f(a), \ y=f(b)$$ for some $$\displaystyle a,b\in A$$ then $$\displaystyle xy=f(a)f(b)=f(ab)=f(ba)=f(b)f(a)=yx$$.

I'm assuming of course that you're giving $$\displaystyle \mathbb{C}^2$$ the usual structure of a product ring (or some commutative structure at least)