# Thread: General Linear Group GLn _ Problem from Artin's Chapter on Linear Groups

1. ## General Linear Group GLn _ Problem from Artin's Chapter on Linear Groups

Can anyone help with the following problem from Artin - Algebra Ch9 on Linear Groups.

Is $\displaystyle GL_n$( $\displaystyle \mathbb{C}$) isomorphic to a subgroup of $\displaystyle GL_ {2n}$($\displaystyle \mathbb{R}$)?

How do I approach proving this one way or the other?

Peter

2. ## Re: General Linear Group GLn _ Problem from Artin's Chapter on Linear Groups

Originally Posted by Bernhard
Can anyone help with the following problem from Artin - Algebra Ch9 on Linear Groups.

Is $\displaystyle GL_n$( $\displaystyle \mathbb{C}$) isomorphic to a subgroup of $\displaystyle GL_ {2n}$($\displaystyle \mathbb{R}$)?

How do I approach proving this one way or the other?

Peter
Start with $\displaystyle n=1$, and see where that takes you. Can you find a 2x2 real matrix which squares to -I, where I is the 2x2 real identity matrix.

One you have found this matrix, you want to "expand" $\displaystyle GL_n(\mathbb{R})$ by this matrix. Try and work out what I mean by "expand"...

3. ## Re: General Linear Group GLn _ Problem from Artin's Chapter on Linear Groups

Taking your advice, it looks like something like the following may work for n = 1

$\displaystyle \phi$: (a + ib) $\displaystyle \rightarrow$ $\displaystyle \left(\begin{array}{cc}a&b\\-b&a\end{array}\right)$:

should be OK for n = 1

But how to 'expand' $\displaystyle GL_{2n}$($\displaystyle \mathbb{R}$)???

Can you help?

Peter

4. ## Re: General Linear Group GLn _ Problem from Artin's Chapter on Linear Groups

Originally Posted by Bernhard
Taking your advice, it looks like something like the following may work for n = 1

$\displaystyle \phi$: (a + ib) $\displaystyle \rightarrow$ $\displaystyle \left(\begin{array}{cc}a&b\\-b&a\end{array}\right)$:

should be OK for n = 1

But how to 'expand' $\displaystyle GL_{2n}$($\displaystyle \mathbb{R}$)???

Can you help?

Peter
Let A and B be nxn matrices. Then let $\displaystyle \phi: (A+iB) \mapsto \left( \begin{array}{cc} A & B\\-B & A\end{array} \right)$...

5. ## Re: General Linear Group GLn _ Problem from Artin's Chapter on Linear Groups

Will check this out!

Thanks so much for the help. Appreciate your assistance!

Will now try to go further with Artin's chapter on the Linear Groups!

Peter

6. ## Re: General Linear Group GLn _ Problem from Artin's Chapter on Linear Groups

Originally Posted by Bernhard
Can anyone help with the following problem from Artin - Algebra Ch9 on Linear Groups.

Is $\displaystyle GL_n$( $\displaystyle \mathbb{C}$) isomorphic to a subgroup of $\displaystyle GL_ {2n}$($\displaystyle \mathbb{R}$)?

How do I approach proving this one way or the other?

Peter
Perhaps a more conceptual way of looking at it, if that kind of thing makes you happy, is that if $\displaystyle V,W$ are isomorphic vector spaces then $\displaystyle \text{GL}(V),\text{GL}(W)$ are isomorphic groups. Now, evidently $\displaystyle \dim_\mathbb{R}\mathbb{C}^n=2n$ so that $\displaystyle \mathbb{C}^n\cong\mathbb{R}^{2n}$ as real vector spaces, and so $\displaystyle \text{GL}_n(\mathbb{C})\cong \text{GL}(\mathbb{C}^n)\cong\text{GL}(\mathbb{R}^{ 2n})\cong\text{GL}_{2n}(\mathbb{R})$.

7. ## Re: General Linear Group GLn _ Problem from Artin's Chapter on Linear Groups

Originally Posted by Drexel28
Perhaps a more conceptual way of looking at it, if that kind of thing makes you happy, is that if $\displaystyle V,W$ are isomorphic vector spaces then $\displaystyle \text{GL}(V),\text{GL}(W)$ are isomorphic groups. Now, evidently $\displaystyle \dim_\mathbb{R}\mathbb{C}^n=2n$ so that $\displaystyle \mathbb{C}^n\cong\mathbb{R}^{2n}$ as real vector spaces, and so $\displaystyle \text{GL}_n(\mathbb{C})\cong \text{GL}(\mathbb{C}^n)\cong\text{GL}(\mathbb{R}^{ 2n})\cong\text{GL}_{2n}(\mathbb{R})$.
do you mean GL:Vect-->Grp is a functor?

8. ## Re: General Linear Group GLn _ Problem from Artin's Chapter on Linear Groups

Originally Posted by Deveno
do you mean GL:Vect-->Grp is a functor?
You bet I do.