All matrices in $GL(2,\mathbb{R})$ can be written as the product of a orthogonal matrices with a triangular matrices.

Question

I would like to compute the fundamental group of connected components of $GL(2,\mathbb{R})$. A hint for compute this fundamental group, is show that $GL(2,\mathbb{R})$ I homeomorphic to $O(2)\times T^+$, where $O(2)$ is a group of all orthogonal matrices, and $T^+$ is the group of all tringular matrices with positive diagonal.

My attempt is show that $\phi: O(2)\times T^+ \to GL^+(2,\mathbb{R})$ given by $\phi(A,B) = AB $ is a homeomorphism. But, I cannot show that $\phi$ is surjective. I try to use the Gram-Schmidt orthogonalization process , but I cannot show that every matrix whose determinant is non-zero can be written as the product of an orthogonal matrix and a triangular matrix. I don't know how to use the orthonormalization process to accomplish this. I appreciate any kind of help.

Does this answer your question? The Iwasawa decomposition of $\text{GL}(2,\mathbf R)$ — Dietrich Burde, May 26 '23 at 14:36
@DietrichBurde I don't think this is a duplicate, the answer to the linked question just says "use Gram-Schmidt", which the OP seems to already know. This question is about how Gram-Schmidt gives such a factorization. — ronno, May 26 '23 at 19:26
@ronno I thought the answer to the duplicate is more or less the same as your answer, is it not? The QR decomposition obtained by Gram-Schmidt. — Dietrich Burde, May 26 '23 at 19:34
If it is then I don't understand that answer. My reading of it is: "here is how the decomposition changes if you replace $SL(2)$ by $GL(2)$, but in either case you get this decomposition from Gram-Schmidt". — ronno, May 26 '23 at 19:38

ronno · Accepted Answer · 2023-05-26T19:41:44.393

Start with the basis of columns of $A \in \operatorname{GL}(2, \mathbb{R})$. If you perform Gram-Schmidt orthogonalization on this, you get a orthonormal basis, ie one which forms the columns of an orthogonal matrix $Q$. Verify that each operation you perform in the orthogonalization process can be done by multiplying $A$ on the right by an (invertible) upper triangular matrix with positive diagonal. Since $T^+$ is a group, this shows that $AR^{-1} = Q$ for some $R \in T^+$, which is equivalent to $A = QR$.

All of this works equally well for $\operatorname{GL}(n, \mathbb{R})$, although writing down $R$ explicitly becomes more tedious as the number of steps in the Gram-Schmidt process increases.

All matrices in $GL(2,\mathbb{R})$ can be written as the product of a orthogonal matrices with a triangular matrices.

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