Trying to understand some basic facts about tangent space of Grassmannian. I am reading Harris's 'Algebraic Geometry: A first course'. I am trying to understand its identification of the tangent space to a Grassmannian. Let $G(k,n)$ be the Grassmannian of $k$-planes in $K^{n}$. If $\Gamma$ in a $(n-k)$-plane in $K^{n}$, set
$$
U_{\Gamma}=\{\Lambda\in G(k,n):\Lambda\cap\Gamma=\{0\}\}.
$$ 
It is said that, if we fix $\Lambda_{0}\in U_{\Gamma}$, then any $\Lambda\in U_{\Gamma}$ is the graph of a homomorphism $\Lambda_{0}\rightarrow \Gamma$, so that
$$
U_{\Gamma}\simeq \mathrm{Hom}(\Lambda_{0},\Gamma).
$$
I don't get to understand why any $\Lambda\in U_{\Gamma}$ is the graph of a homomorphism $\Lambda_{0}\rightarrow \Gamma$. 
Any hint would be appreciated.
 A: Fixed the $(n-d)$-plane $\Gamma$ in $\mathbb{K}^n$; for simplicity, let $\{e_1,\dots,e_n\}$ be a basis of $\mathbb{K}^n$ such that $\Gamma=\langle e_{d+1},\dots,e_n\rangle$. The set $U_{\Gamma}$ is in bijection with the set $\{u_1,\dots,u_d\in\mathbb{K}^n\}$ such that $\{u_1,\dots,u_d,e_{d+1},\dots,e_n\}$ is a basis; that is, $U_{\Gamma}$ is in bijection with the set of complementary subspace of $\Gamma$ in $\mathbb{K}^n$.
Let
$$
\forall i\in\{1,\dots,d\},\,u_i=\sum_{j=1}^na_i^je_j,\\
A=\begin{pmatrix}
a_1^1 & a_1^2 & \dots & a_1^n\\
a_2^1 & a_2^2 & \dots & a_2^n\\
\vdots & \ddots & \ddots & \vdots\\
a_d^1 & a_d^2 & \dots & a_d^n
\end{pmatrix}\in M_{d\times n}(\mathbb{K});
$$
by the previous reasoning: $U_{\Gamma}$ is in bijection with the set of the matrices $A$ with maximal rank, that is $U_{\Gamma}$ is in bijection with the set
$$
\{A\in M_{d\times n}(\mathbb{K})\mid\operatorname{rank}A=d\}\simeq\mathbb{K}^{d(n-d)}.
$$
$\mathbb{K}^{d(n-d)}$ is in bijection with $\hom_{\mathbb{K}}(\Lambda_0,\Gamma)$ where $\Lambda_0$ is a $\mathbb{K}$-vector space of dimension $d$, because the set of $\mathbb{K}$-linear function from $\Lambda_0$ to $\Gamma$ is in bijection with $\mathbb{K}^{d(n-d)}$; then $U_{\Gamma}$ is in bijection with $\hom_{\mathbb{K}}(\Lambda_0,\Gamma)$.
