How can prove that a set of matrices is a closed Lie subgroup of $GL(n)$?

Question

I want to prove that the set $H\subset GL(n)$ of invertibles matrices of the form $$\begin{pmatrix} A & 0\\ C&B \end{pmatrix}$$ where $A\in GL(k)$, $B\in GL(n-k)$ and $C\in M_{(n-k)\times k}$ is a Lie subgroup of $GL(n)$. I only need to know why is closed, I don't see this part.

Answer 1

Hint The standard technique is to construct a continuous map $$\Phi : GL(n) \to X$$ for some space $X$ such that $H$ is the preimage $\Phi^{-1}(x)$ of some $x \in X$. (In fact, we can replace $\{x\}$ with any closed subset $Y \subseteq X$.)

Example The special linear group $$SL(n) := \{A \in GL(n) : \det A = 1\}$$ is, by definition, the inverse image $\det^{-1}(1)$ of the determinant map $\det : GL(n) \to \Bbb R$. The determinant of a matrix is a polynomial in its entries and in particular continuous, so $SL(n)$ is closed in $GL(n)$.

Now, by definition the only requirement for an element of $GL(n)$ to be in $H$ is for its upper-right $k \times (n - k)$ submatrix to be zero.