Subspace Topology of $GL_n(\Bbb R)$

Question

I'm trying to understand $GL_n(\mathbb R)$. I've read that we can think its topology with Euclidean metric. I know that $GL_n(\mathbb R)$ is a subspace of $Mat_n(\mathbb R)$ but I cannot understand well the subspace topology of $GL_n(\mathbb R)$ with respect to $Mat_n(\mathbb R)$. I'm in Euclidean metric that is $d(x,y)=\|x-y\|$ is not a metric on $GL_n(\mathbb R)$ because we cannot define norm since it is not a vector space (it does not have $0$)

I appreciate for any help. Please pardon me for this basic question.

Answer 1

On $\operatorname{Mat}_n(\Bbb R)$ you can consider, say, the distance $$d\bigl((a_{ij})_{1\leqslant i,j\leqslant n},(b_{ij})_{1\leqslant i,j\leqslant n}\bigr)=\sum_{i,j=1}^n|a_{ij}-b_{ij}|.$$And now, since $GL_n(\Bbb R)$ is a subset of $\operatorname{Mat}_n(\Bbb R)$, this induces a distance on $GL_n(\Bbb R)$, which is simply the restriction of $d$ to $GL_n(\Bbb R)\times GL_n(\Bbb R)$.

Answer 2

A norm-based metric (which is defined for all matrices) can be restricted to any subset and it's still a metric: the axioms for a metric don't mention a linear structure and $\|x-y\|$ is still well-defined.

Any norm will do, as they are all equivalent on the matrix space which is esentially $\Bbb R^{n^2}$ anyway.