I can not answer the question but maybe the following perspectives helps clarify the obstruction. let $k \leq n$ and consider $$ GL(n) \overset{span}{\longrightarrow} Gr(k, n) \overset{\approx}{\longleftrightarrow} O(n) / (O(k) \times O(n - k)) \overset{\pi_{n,k}}{\longleftarrow} O(n) $$ $GL(n)$ is the group of invertible matrices on $\mathbb{R}^n$, i.e. this represents your iniital bases. $Gr(k, n)$ is the Grassmannian, the differentiable manifold of k-dimensional subspaces of $\mathbb{R}^n$. One way to construct the Grassmannian is to start with the orthogonal group $O(n)$, i.e. the orthogonal matrices on $\mathbb{R}^n$, and take the quotient by the embedding $(O(k) \times O(n - k)) \longrightarrow O(n)$ given by the matrix which applies the $O(k)$ part to the first $k$ dimensions and the $O(n - k)$ part to the remaining $n - k$ dimensions. Basically, an element of $O(n) / (O(k) \times O(n - k))$ is an equivalence classes of orthogonal matrices on $\mathbb{R}^n$ which leave invariant some $k$-dimensional subspace. What would be natural to do is to invert the projection $\pi_{n,k}$, so that $\pi_{n,k}^{-1} \circ \approx \circ span$ gives you your desired map. Of course $\pi_{n,k}$ is not injective, so you can not invert it. But maybe one can study this map and find an atlas of natural subsets and a natural choice of $\pi_{n,k}^{-1}$ on each, somehow.

Here I don't understand well what happens so I'm guessing, but I think $O(n) / (O(k) \times O(n - k)) \overset{\pi_{n,k}}{\longleftarrow} O(n)$ is indeed a covering and therefore the universal covering space of the Grassmannian $Gr(k, n)$ should give you information about the obstruction to inverting $\pi_{n,k}$. I read that the universal covering space is the "oriented Grassmannian" and it is a double cover. This would suggest that perhaps by splitting $O(n)$ into two parts (orientation). Then for each half you should be able to choose some inverse of $\pi_{n,k}$ and obtain two orthogonalization procedures, one which produces positive orientation and one which produces negative orientation bases.

I can not answer the question but maybe the following perspectives helps clarify the obstruction. Let $k \leq n$ and consider $$ \operatorname{GL}(n) \overset{\operatorname{span}}{\longrightarrow} Gr(k, n) \overset{\approx}{\longleftrightarrow} O(n) / (O(k) \times O(n - k)) \overset{\pi_{n,k}}{\longleftarrow} O(n). $$ $\operatorname{GL}(n)$ is the group of invertible matrices on $\mathbb{R}^n$, i.e. this represents your iniital bases. $\operatorname{Gr}(k, n)$ is the Grassmannian, the differentiable manifold of $k$-dimensional subspaces of $\mathbb{R}^n$. One way to construct the Grassmannian is to start with the orthogonal group $O(n)$, i.e. the orthogonal matrices on $\mathbb{R}^n$, and take the quotient by the embedding $(O(k) \times O(n - k)) \longrightarrow O(n)$ given by the matrix which applies the $O(k)$ part to the first $k$ dimensions and the $O(n - k)$ part to the remaining $n - k$ dimensions. Basically, an element of $O(n) / (O(k) \times O(n - k))$ is an equivalence classes of orthogonal matrices on $\mathbb{R}^n$ which leave invariant some $k$-dimensional subspace. What would be natural to do is to invert the projection $\pi_{n,k}$, so that $\pi_{n,k}^{-1} \circ {\approx} \circ \operatorname{span}$ gives you your desired map. Of course $\pi_{n,k}$ is not injective, so you can not invert it. But maybe one can study this map and find an atlas of natural subsets and a natural choice of $\pi_{n,k}^{-1}$ on each, somehow.

Here I don't understand well what happens so I'm guessing, but I think $O(n) / (O(k) \times O(n - k)) \overset{\pi_{n,k}}{\longleftarrow} O(n)$ is indeed a covering and therefore the universal covering space of the Grassmannian $\operatorname{Gr}(k, n)$ should give you information about the obstruction to inverting $\pi_{n,k}$. I read that the universal covering space is the "oriented Grassmannian" and it is a double cover. This would suggest that perhaps by splitting $O(n)$ into two parts (orientation). Then for each half you should be able to choose some inverse of $\pi_{n,k}$ and obtain two orthogonalization procedures, one which produces positive orientation and one which produces negative orientation bases.

I can not answer the question but maybe the following perspectives helps clarify the obstruction. let $k \leq n$ and consider $$ GL(n) \overset{span}{\longrightarrow} Gr(k, n) \overset{\approx}{\longleftrightarrow} O(n) / (O(k) \times O(n - k)) \overset{\pi_{n,k}}{\longleftarrow} O(n) $$ $GL(n)$ is the group of invertible matrices on $\mathbb{R}^n$, i.e. this represents your iniital bases. $Gr(k, n)$ is the Grassmanian, the differentiable manifold of k-dimensional subspaces of $\mathbb{R}^n$. One way to construct the Grassmanian is to start with the orthogonal group $O(n)$, i.e. the orthogonal matrices on $\mathbb{R}^n$, and take the quotient by the embedding $(O(k) \times O(n - k)) \longrightarrow O(n)$ given by the matrix which applies the $O(k)$ part to the first $k$ dimensions and the $O(n - k)$ part to the remaining $n - k$ dimensions. Basically, an element of $O(n) / (O(k) \times O(n - k))$ is an equivalence classes of orthogonal matrices on $\mathbb{R}^n$ which leave invariant some $k$-dimensional subspace. What would be natural to do is to invert the projection $\pi_{n,k}$, so that $\pi_{n,k}^{-1} \circ \approx \circ span$ gives you your desired map. Of course $\pi_{n,k}$ is not injective, so you can not invert it. But maybe one can study this map and find an atlas of natural subsets and a natural choice of $\pi_{n,k}^{-1}$ somehow.

I can not answer the question but maybe the following perspectives helps clarify the obstruction. let $k \leq n$ and consider $$ GL(n) \overset{span}{\longrightarrow} Gr(k, n) \overset{\approx}{\longleftrightarrow} O(n) / (O(k) \times O(n - k)) \overset{\pi_{n,k}}{\longleftarrow} O(n) $$ $GL(n)$ is the group of invertible matrices on $\mathbb{R}^n$, i.e. this represents your iniital bases. $Gr(k, n)$ is the Grassmannian, the differentiable manifold of k-dimensional subspaces of $\mathbb{R}^n$. One way to construct the Grassmannian is to start with the orthogonal group $O(n)$, i.e. the orthogonal matrices on $\mathbb{R}^n$, and take the quotient by the embedding $(O(k) \times O(n - k)) \longrightarrow O(n)$ given by the matrix which applies the $O(k)$ part to the first $k$ dimensions and the $O(n - k)$ part to the remaining $n - k$ dimensions. Basically, an element of $O(n) / (O(k) \times O(n - k))$ is an equivalence classes of orthogonal matrices on $\mathbb{R}^n$ which leave invariant some $k$-dimensional subspace. What would be natural to do is to invert the projection $\pi_{n,k}$, so that $\pi_{n,k}^{-1} \circ \approx \circ span$ gives you your desired map. Of course $\pi_{n,k}$ is not injective, so you can not invert it. But maybe one can study this map and find an atlas of natural subsets and a natural choice of $\pi_{n,k}^{-1}$ on each, somehow.

Here I don't understand well what happens so I'm guessing, but I think $O(n) / (O(k) \times O(n - k)) \overset{\pi_{n,k}}{\longleftarrow} O(n)$ is indeed a covering and therefore the universal covering space of the Grassmannian $Gr(k, n)$ should give you information about the obstruction to inverting $\pi_{n,k}$. I read that the universal covering space is the "oriented Grassmannian" and it is a double cover. This would suggest that perhaps by splitting $O(n)$ into two parts (orientation). Then for each half you should be able to choose some inverse of $\pi_{n,k}$ and obtain two orthogonalization procedures, one which produces positive orientation and one which produces negative orientation bases.