[Math] The Gram-Schmidt process is a deformation retraction

algebraic-topologylie-groupslinear algebra

Consider the Gram-Schmidt process $r : GL(n) \rightarrow O(n)$ that sends invertible matrices to orthogonal matrices. I need to show this is a deformation retraction and, by restrictions of $r$, establish each space is a deformation retract of the following: $SO(n) \subset SL(n) \subset GL^{+}(n)$ where $GL^{+}(n)$ is the set of matrices with positive determinant.

I'm not very experienced with linear algebra so I'm having some difficulty in this. The idea I had was to define, for vectors $u, v \in R^{n}$ and real $t \in [0, 1]$, the altered projection:
$P(u, v, t) = \displaystyle\frac{<u, (1 – t)u + tv> u}{<u, u>}$.

Then we'd have the homotopy $H((a_{1}, …, a_{n}), t) = (w_{1}, …, w_{n})$ where

$w_{1} = a_{1}$

$w_{2} = a_{2} – P(w_{1}, a_{2}, t)$

.
.

$w_{n} = a_{n} – \displaystyle\sum_{i = 1}^{n – 1} P(w_{i}, a_{n}, t)$.

But actually now I think this doesn't work, because I can't really guarantee it's invertible for all $t$. I have no other ideas.

Best Answer

The Gram-Schmidt process without normalization of the vectors is determinant preserving, can be done in a continuous motion, and commutes with scalar multiplication. So first rescale continuously until $|\det |=1$ and then continuously perform the non-normalized orthogonalization to get to $O(n)$.