# Gram Schmidt process is a homotopy equivalence

How can one see that the Gram Schmidt process is a homotopy equivalence $$SL(m,\mathbb{R})\rightarrow SO(m)$$? Intuitively it seems plausibel to me, but I can't find an explicit homotopy between the Gram Schmidt map and the identity, since this homotopy always has to "stay inside $$SL(m,\mathbb{R}))$$". I am thankful for any help.

Do it step by step. Pick the first vector $$v_1$$ of your frame. Rescaling it to unit is obviously a homotopy equivalence. Rotating the span of $$v_2, \dotsc, v_n$$ to the orthogonal complement of $$v_1$$ is also. Lather, rinse, repeat.
• But it is not a homotopy equivalence which stays inside $SL(m,\mathbb{R})$. I think one would have to modify the Gram Schmidt process quite a bit. Jul 13 at 11:43