Gram Schmidt process is a homotopy equivalence

Question

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.

Answer 1

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.