# Thread: Homomorphism from the complex to 2*2 matrices

1. ## Homomorphism from the complex to 2*2 matrices

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?

2. 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 $\displaystyle a+ bi\rightarrow \begin{pmatrix}a & -b \\ b & a}\end{pmatrix}$.

3. 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.

4. Originally Posted by HallsofIvy
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!

5. 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.

6. Originally Posted by 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.
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 $\displaystyle \mathbb{C}^2$).

7. Originally Posted by Bruno J.
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.

8. 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.

9. 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)

10. Originally Posted by 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.
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)

11. Originally Posted by Jose27
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.
Ulysses, are you looking for a group homomorphism, or a ring homomorphism?

12. Originally Posted by 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)
The matrices $\displaystyle M_n(\mathbb{R})$ are not a field, no more than $\displaystyle \mathbb{R}^n$!

In any case, even as vector spaces over $\displaystyle \mathbb{R}$, these two objects are non-isomorphic (not the same dimension).

13. the question specifcally says:is there ANY homomrphism, so i'm presuming he wants us to think about this question.
I have basically showed 1 example that is a ring homomorphism but is not surjective, and a map that is surjective but does not preserve the ring homomorphism properties.And additionally stated that C*C is commutative but matrices aren't.
Does a homomrphism exist though that is NOT necassarily a ring homomorphism?Because i have answered this as no, because i cant find any homomorphism that is surjective.