Consider a (not necessarily connected) matrix group $G$, and define the identity component $G^0$ of $G$ as the set of all the elements $g$ of $G$ for which there exists a path $\gamma\colon [0,1]\to G$ with $\gamma(0)=I$ and $\gamma(1)=g$.
I am trying to show that $G^0$ is a matrix group.
I have already shown it is a subgroup of $G$, but I'm struggling to show that it is closed in $GL_n(\mathbb{K})$.
My idea was to show that $G^0$ is path connected (which follows from it being the union of path connected sets that have a point in common) and that if $D\subseteq G$ is another path connected set in $G$, then $D\subseteq G^0$, so the set $G^0$ is closed as it is the intersection of two closed sets, $G^0=GL_n^+(\mathbb{R})\cap G$ or $G^0=GL_n(\mathbb{K})\cap G$. But now I have to show that $GL_n^+(\mathbb{R})$ is path connected and that $GL_n(\mathbb{K})$ is path connected for $\mathbb{K}\in \{\mathbb{C},\mathbb{H}\}$.
Is there another way to show this?
How can I show the last part of my idea?
P.S.: A matrix group is a closed (with respect to the subspace topology) subgroup of $GL_n(\mathbb{K})$, where $\mathbb{K}\in \{\mathbb{R},\mathbb{C},\mathbb{H}\}$.