Path-connectedness of Grassmannian Let $\mathrm{Gr}(k,X^n)$ be Grassmannian. How to prove that $\mathrm{Gr}(k,X^n)$ is path-connected (without using homology theorem) ?
 A: Given two $k$-planes through the origin, one may rotate one plane into the other.  This rotation defines a path in the Grassmanian from the point representing the first plane to the point representing the second plane.
A: I am assuming you are working with an $\mathbb{R}$ or $\mathbb{C}$ vector space (if not, it makes sense to talk about the Grassmannian as an algebraic variety, but "path connected" is a little funny, although in fact I think it's true that the Grassmannian is algebraically $\mathbb{P}^1$-"path connected" as well). I'll write this up over $\mathbb{R}$ but the proof is the same over $\mathbb{C}$.
Pick a basis for the ambient vector space which we regard as fixed forever. 
Now given any $k$-plane in $n$-space, suppose we have a basis of $k$ vectors for that plane. The matrix expressing the basis for the $k$-plane in terms of the given basis for the whole space is a $k \times n$ matrix. Moreover, it has at least one non-vanishing $k \times k$ minor. We would like to say that this matrix "is" the point in the Grassmannian, but that's not well-defined. We could always change the basis for the small space without changing the point of the Grassmannian. So we introduce an equivalence relation; namely we say that two $k \times n$ matrices (with at least one non-vanishing $k \times k$ minor) are the same if and only if they differ by left-multiplication by a matrix in $GL_k(\mathbb{R})$.
$$
G_k(n) = \{ (k \times n)-\text{matrices with at least one non-vanishing } (k \times k) \text{ minor}\} /\sim
$$
Let $\Sigma$ be the set of subsets of $\{ 1, 2, \ldots, n \}$ with exactly $k$ elements. For each $\sigma \in \Sigma$, let 
\begin{align}
G_\sigma(n) &= \{ (k \times n)-\text{matrices }  M | \text{ the minor of } M \\ &\text{ whose columns correspond to } \sigma \text{ doesn't vanish.}\}.
\end{align}
Then the sets $G_\sigma(n)$ cover $G_k(n)$. Moreover, for $\sigma \neq \sigma'$, we have $G_\sigma(n) \cap G_{\sigma'}(n) \neq \emptyset$. Finally, $G_\sigma(n)$ is homeomorphic to $\mathbb{R}^{nk - k^2}$, because for each matrix in $G_\sigma(n)$, we may take the representative of the equivalence class whose matrix at the columns indexed by $\sigma$ is the identity matrix. This implies path connectedness, among other things.
