Prove GL(R) is not isomorphic to GL(C)

[ArcanaNoir]

Homework Statement

Prove GL(R) is not isomorphic to GL(C)

Homework Equations

The Attempt at a Solution

Well I don't really have a good grasp of the issue at hand as this is another appendix problem, but my thoughts are I think GL(R) is a subgroup of GL(C) so maybe I could show they don't have the same cardinality? Would that be enough? Been a while since I did algebra... Not sure how I would would show that either.

 

[micromass]

Do you know why $\mathbb{R}\setminus \{0\}$ is not isomorphic to $\mathbb{C}\setminus \{0\}$? Try to generalize.

 

[ArcanaNoir]

Do you know why $\mathbb{R}\setminus \{0\}$ is not isomorphic to $\mathbb{C}\setminus \{0\}$? Try to generalize.

Umm f(I)=f(I^2)=f(I)^2 implies f(I) is the identity in GL(C). [STRIKE]But f(-I)^2=f(I) so f(-I) is also the identity in GL(C). [/STRIKE]

Is that how to do it?

Errr. Flaw in that logic. meh.

 

[ArcanaNoir]

Okay here's what I got. Comments are appreciated.

Suppose there exists an isomorphism \phi : \text{GL}(\mathbb{C}) \rightarrow \text{GL}(\mathbb{R}). Then \phi(\text{I})=\text{I}. We also have \phi (-\text{I})^2=\phi((-\text{I})^2)=\phi(\text{I})=\text{I}. This implies \phi(-\text{I})=\pm \text{I}. Since \phi(i\text{I})^2=\phi(-\text{I}), and \phi(i\text{I}) is a matrix of real numbers, it cannot be that \phi(-\text{I})=-\text{I}. Hence, \phi(-\text{I})=\text{I}, which shows that \\phi is not injective.

 

[micromass]

Why does $A^2 = I$ imply that $A=\pm I$? And why can $A^2 = -I$ not occur?

 

[ArcanaNoir]

Why does $A^2 = I$ imply that $A=\pm I$?

Because the square root of I is ±I right?

And why can $A^2 = -I$ not occur?

Because real numbers squared cannot be negative 1.

Right?

 

[micromass]

Because the square root of I is ±I right?





Because real numbers squared cannot be negative 1.

Right?

You seem to assume that these things hold for matrices because they hold for real numbers. You need to prove whether they also hold for arbitrary matrices.

 

[Office_Shredder]

Because the square root of I is ±I right?

Haha you wish

http://en.wikipedia.org/wiki/Square_root_of_a_matrix#Properties

 

[ArcanaNoir]

Haha you wish

http://en.wikipedia.org/wiki/Square_root_of_a_matrix#Properties

Alas!

You seem to assume that these things hold for matrices because they hold for real numbers. You need to prove whether they also hold for arbitrary matrices.

Well... Do they hold? Seems like they don't based on the link Office_Shredder provided.

Your original direction was to show it holds for numbers, so if you pretend I wrote 1's instead of I's, that's my logic. Now I just need to get this stuff to generalize. Did you have any hint on how to do that?

 

[micromass]

Maybe you can find the centers of both groups?

 

[ArcanaNoir]

Maybe you can find the centers of both groups?

That would be any scalar (except for 0) times the identity matrix?

 

[micromass]

That would be any scalar (except for 0) times the identity matrix?

Yes. Are the centers isomorphic?

 

[ArcanaNoir]

Yes. Are the centers isomorphic?

I would guess not. I would like to say because of my proof, but I know I'm missing something.

 

[I like Serena]

I would guess not. I would like to say because of my proof, but I know I'm missing something.

Following up on mm's suggestion, can you think of a group that is isomorphic with a real non-zero scalar times the identity matrix?
How about a group that is isomorphic with a complex non-zero scalar times the identity matrix?I also have an alternative solution method: what you can say about the eigenvalues of the matrix A if $A^2=I$?Btw, I'm somewhat confused about the meaning of $GL(\mathbb R)$.
What does it mean?
I suspect you do not mean what wiki says about the General linear group.

 

[ArcanaNoir]

Following up on mm's suggestion, can you think of a group that is isomorphic with a real non-zero scalar times the identity matrix?
How about a group that is isomorphic with a complex non-zero scalar times the identity matrix?

Reals minus 0 with multiplication?

Btw, I'm somewhat confused about the meaning of $GL(\mathbb R)$.
What does it mean?
I suspect you do not mean what wiki says about the General linear group.

Either nxn invertible matrices or infinite ones, not sure what my professor intended. If I can get it figured out for nxn I'm confident it will generalize.

 

[I like Serena]

Reals minus 0 with multiplication?

Yep.

Do you know why $\mathbb{R}\setminus \{0\}$ is not isomorphic to $\mathbb{C}\setminus \{0\}$? Try to generalize.

Now consider this previous comment of mm...