Homomorphism from the complex to 2*2 matrices

Mar 2010
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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

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Apr 2005
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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 \(\displaystyle a+ bi\rightarrow \begin{pmatrix}a & -b \\ b & a}\end{pmatrix}\).
 
Mar 2010
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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.

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Jun 2009
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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!
 
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Mar 2010
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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.

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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

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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.
 
Mar 2010
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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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Dec 2009
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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

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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)