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

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.

• Try to define a set of continuous functions on $\operatorname{GL}(n)$ such that $H$ is the set of common zeros of these functions. Nov 14, 2021 at 21:26

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.