GL_n is homeomorphic to O(n) × upper triangular

Idea. Existence is Gram-Schmidt; the same recipe is a continuous formula; uniqueness makes the maps inverse to each other.

Step 1: existence by Gram-Schmidt. Let $a_1,\dots,a_n$ be the columns of $A$ (a basis). Define
$$u_j=a_j-\sum_{i<j}\langle a_j,q_i\rangle q_i,\qquad q_j=\frac{u_j}{|u_j|},$$
where $u_j\ne0$ because $a_j\notin\operatorname{span}(a_1,\dots,a_{j-1})$. Then $(q_j)$ is orthonormal and $a_j=\sum_{i\le j}R_{ij}q_i$ with $R_{ij}=\langle a_j,q_i\rangle$ for $i<j$ and $R_{jj}=|u_j|>0$. So $A=QR$ with $Q=(q_1|\dots|q_n)$ orthogonal and $R\in T_+$.

Step 2: uniqueness. If $QR=Q'R'$ then $M:=R'R^{-1}=Q'^TQ$ is orthogonal and upper triangular with positive diagonal. An orthogonal matrix satisfies $M^{-1}=M^T$; but $M^{-1}$ is upper triangular, and $M^T$ is lower triangular, so $M$ is diagonal. A diagonal orthogonal matrix has entries $\pm1$, and positive ones force $M=I$. Hence $R'=R$, $Q'=Q$.

Step 3: continuity of $A\mapsto(Q,R)$. Each step of Gram-Schmidt uses sums, products, inner products, and $u\mapsto u/|u|$, which is continuous on $u\neq0$. By induction on $j$, the vectors $q_j$ and the numbers $R_{ij}$ are continuous functions of $A$ on $GL_n$, as $u_j\ne0$ throughout.

Step 4: the inverse. $(Q,R)\mapsto QR$ is continuous (Problem 67) with values in $GL_n$, and by uniqueness and existence it is the inverse of $A\mapsto(Q,R)$.

Remark.
- $T_+\cong(0,\infty)^n\times\mathbb{R}^{n(n-1)/2}$ is contractible, so this decomposition again shows that $GL_n$ deformation retracts onto $O(n)$ (compare Problem 90).
- It is the $GL_n(\mathbb{R})=K\cdot A\cdot N$ Iwasawa decomposition, with $K=O(n)$, $A$ the positive diagonal matrices, and $N$ the unipotent upper triangular matrices.
- Numerically, `numpy.linalg.qr` does not enforce $R_{jj}>0$; one must multiply by the signs of the diagonal (checked).