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.


1 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.

  • $\begingroup$ 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. $\endgroup$ Jul 13 at 11:43

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