In considering what the dimension of a Grassmanian, $G(k,V)$, is, I came across an interesting property of the tangent space to a Grassmanian. Specifically, given $\mathbb{P}E \subset G(k,V)$, where $E$ is a vector space, we have that:
$$T_E \, G(k,V) \cong E^\ast \otimes V / E$$
Where: $T_E \, G(k,V)$ is the tangent space to the Grassmannian containing all derivatives of curves that start in $E$ and stay in the Grassmannian, $E^\ast$ is the dual space to $E$, and $V / E$ is the quotient space.
I don't know why this is true... It seems to me that if we consider a general curve in $G(k,V)$, it's given by $[c(t)] := [v_1(t) \wedge \cdots \wedge v_k(t)]$. Then, if $c(0) = v_1 \wedge \cdots \wedge v_k \in E$, where the $v_i$ are general (linearly independent) vectors in $E$, we can compute the tangent plane. Specifically:
$$c^\prime (0) = V \wedge v_2 \wedge \cdots \wedge v_k + \cdots + v_1 \wedge \cdots \wedge v_{k-1} \wedge V \in T_E \, G(k,V)$$
Where, above, the sum is not direct, and the expression gives a general vector in the tangent plane. Then, how can we construct an isomorphism to $E^\ast \otimes V /E$?