An $n$-dimensional subgroup of $GL_{n+1}(\mathbb R)$

Could somebody please tell me if my answer to the following exercise is correct?:

Describe a subgroup of $GL_{n+1}(\mathbb R)$ that is isomorphic to $\mathbb R^n$ under vector addition.

It's not clear to me why the exercise considers $GL_{n+1}(\mathbb R)$ instead of $GL_{n}(\mathbb R)$, I believe $GL_{n}(\mathbb R)$ should also contain a subgroup isomorphic to $\mathbb R^n$. My answer is to define $D_i$ to be the diagonal matrix consisting of $1$ on the diagonal except at position $i$ where it is $-1$. Then $D_1, \dots, D_n$ span an $n$-dimensional subspace.

• What do you mean "It's not clear to me why the exercise considers $GL_{n+1}(\mathbb R)$ instead of $GL_{n+1}(\mathbb R)$", these are the same objects? Jul 14 '14 at 19:22
• But the multiplication of such diagonal matrices (=the group operation of $GL_{n+1}$ won't correspond to addition of vectors (=the group operation of $\Bbb{R}^n$). Jul 14 '14 at 19:23
• Hint: The matrix multiplication of the group of $2\times2$ real matrices of the form $$A(x)=\pmatrix{1&x\cr0&1\cr}$$ corresponds to the addition of the numbers: $$A(x_1)A(x_2)=A(x_1+x_2)$$ for all $x_1,x_2\in\Bbb{R}^1$. What modifications do you need to make to this for it to work and give a copy of $(\Bbb{R}^n,+)$ inside $(GL_{n+1}(\Bbb{R}),\cdot)$? Jul 14 '14 at 19:26
• @JyrkiLahtonen I will need some time to think about this, thank you for the hint. Jul 14 '14 at 19:55

I will show how we get the additive group of $\Bbb{R}^3$ as a subgroup of $GL_4(\Bbb{R})$. You can generalize this to arbitrary $n$ easily.

Consider matrices of the form $$A(x_1,x_2,x_3)=\left(\begin{array}{cccc} 1&x_1&x_2&x_3\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{array}\right).$$ It is easy to verify that products of matrices of this form follow the rule $$A(x_1,x_2,x_3)A(y_1,y_2,y_3)=A(x_1+y_1,x_2+y_2,x_3+y_3).$$ Therefore the mapping $(x_1,x_2,x_3)\mapsto A(x_1,x_2,x_3)$ is an injective homomorphism of groups from $(\Bbb{R}^3,+)\to (GL_4(\Bbb{R}),\cdot).$

Mind you it is possible to also a copy of $(\Bbb{R}^4,+)$ inside $GL_4$. To achieve that we need to be a bit more clever, and use matrices of the form $$B(x_1,x_2,x_3,x_4)=\left(\begin{array}{cccc} 1&0&x_1&x_2\\ 0&1&x_3&x_4\\ 0&0&1&0\\ 0&0&0&1 \end{array}\right)$$ instead of stuffing all the coordinates on the first row.

Another Exercise for the OP: There are other ways of realizing $\Bbb{R}^3$ as a subgroup of $GL_4$. Find at least two more!

• I never saw that there was an answer! I'm sorry. Nov 15 '14 at 1:33
• But if I take the subsets of diagonal matrices with last diagonal entry equal to $0$ and define $$(D + D')_{ii} = (\lambda_i + \lambda_i')$$ why does it not give an $n$-dimensional additive subgroup? Nov 15 '14 at 1:37
• Oh, the group operation of multiplication has to map to addition? Nov 15 '14 at 1:38
• Regarding your exercise: I guess I can put the three variables into the last row (instead of the first) but I'm not sure this counts as different from what you did. Nov 15 '14 at 1:41
• Yes. using the last row is ok. Or the first column (or the last). A matter of taste whether they count as different, but that's what I had in mind :-) Nov 15 '14 at 7:02