Tangent space of Grassmannian $Gr_k (\mathbb{R}^n)$ I am trying to show that the tangent space of the Grassmannian $Gr_k (\mathbb{R}^n)$ at $L,$ is naturally/canonically isomorphic to $Hom(L,\mathbb{R}^n/L).$ However, I cannot see even intuitively what this isomorphism is.
Any help would be appreciated!
 A: First of all, you can understand the manifold structure of the Grassmannian by representing $k$-dimensional subspaces near a fixed $\Lambda_0\in Gr_k(\Bbb R^n)$ as graphs of linear maps $\Lambda_0 \to \Lambda_0^\perp$. This chart then gives you tangent vectors with the same sort of representation.
A more intuitive geometric interpretation comes this way. Think of a $k$-plane as an equivalence class of $k$-vectors $v_1\wedge\dots\wedge v_k$, where $v_1,\dots,v_k$ give a basis; two such $k$-vectors are equivalent if they differ by a nonzero scalar multiple. Consider a smooth curve $\gamma(t)$ in $Gr_k(\Bbb R^n)$ with $\gamma(0) = \Lambda_0$. If we choose a smoothly varying orthonormal basis $e_1(t),\dots,e_k(t)$ for $\gamma(t)$, then we can  identify $\gamma(t)$ with $\Gamma(t)=e_1(t)\wedge\dots\wedge e_k(t)$, and the tangent vector $\gamma'(0)$ can be visualized as $\Gamma'(0) \pmod{\Lambda_0}$. That is, we want to know how the basis vectors are twisting out of the given $k$-plane $\Lambda_0$. 
$$\Gamma'(0) = e_1'(0)\wedge e_2(0)\wedge\dots\wedge e_k(0) + \dots + e_1(0)\wedge\dots\wedge e_{k-1}(0)\wedge e_k'(0)\,. \tag{$\star$}$$
To see a change in the $k$-plane (i.e., something nonzero mod $\Gamma(0)$), we need some $e_j'(0)\ne 0 \in \Bbb R^n/\Lambda_0$. So what one needs to check is this: Given a linear map $A\colon \Lambda_0\to\Bbb R^n/\Lambda_0$, setting $e_j'(0) = Ae_j(0)$ in ($\star$) gives us a bijection between tangent vectors to such curves $\gamma(t)$ at $t=0$ and linear maps.
