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. 
 A: 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!
