tangent vector in $\text{End}(V)$ Consider the $\text{End}_{\Bbb{R}}(V)$ for $\Bbb{R}-$ vector space $V$, and a smooth curve $\gamma(t)\subset \text{End}(V)$，it's known there is a identification $$\Phi:T_p\text{End}(V) \to \text{End}(V)\\ D_v|_p \mapsto v$$
Where $D_v|_p f = \frac{d}{dt}|_{t=0}(f(tv+p))$ which can be shown is a derivation.
Prove under this identification $$\Phi(\dot\gamma(0))(X) = \frac{d}{dt}|_{t=0}(\gamma(t)(X)) \tag{*}$$
where $\Phi(\dot\gamma) \in \text{End}(V)$, $X \in V$, $\gamma(t) \in \text{End}(V)$, and $\frac{d}{dt}|_{t=0}(\gamma(t)(X)) \in T_ pV \cong V$ and $\Phi(\dot{\gamma})(X) \in V$ also.
Can someone explain it clearly, there are so many identification here.
 A: The equation means that the identification $\Phi$ commutes with evaluation at $\mathbf x$, i.e., we can first differentiate at $0$ and then evaluate at $\mathbf x$ or the other way around.
To prove it, fix a basis $B=\{\mathbf e_1,\dots,\mathbf e_n\}$ of $\mathbb V$. Let $E_{ij}\colon\mathbf x\mapsto x_i\mathbf e_j\in\textrm{End}(\mathbb V)$, where $x_i$ is the $i$th coordinate of $\mathbf x$ in $B$. Then $(E_{ij})_{1\le i,j\le n}$ is a basis of $\textrm{End}(\mathbb V)$. Therefore, given $A\in\textrm{End}(\mathbb V)$, consider the curve $\gamma_{ij}\colon t\mapsto A+tE_{ij}$. We have
$$
  \Phi(\dot\gamma_{ij}) = \dot\gamma_{ij}(0) = E_{ij}.
$$
On the other hand,
$$
  \gamma_{ij}(t)(\mathbf x)= A+tE_{ij}(\mathbf x),
$$
hence
$$
  \frac{d(\gamma_{ij}(t)(\mathbf x))}{dt}\Big|_{t=0}=E_{ij}(\mathbf x).
$$
Thus,
$$
  \Phi(\dot\gamma_{ij})(\mathbf x)
    = E_{ij}(\mathbf x)
    = \frac{d(\gamma_{ij}(t)(\mathbf x))}{dt}\Big|_{t=0}.
$$
Therefore, the equation
$$
  \Phi(\dot\gamma)(\mathbf x) = \frac{d(\gamma(t)(\mathbf x))}{dt}\Big|_{t=0}
$$
holds for $\gamma=\gamma_{ij}$. And since $(\dot\gamma_{ij}(0))_{1\le i,j\le n}$ is a basis of $T_{\mathbf c}(\textrm{End}(\mathbb V))$, the equation holds for every smooth curve $\gamma$ passing through $\mathbf c\in\textrm{End}(\mathbb V)$.

Addendum
What follows comes from the first comment below.
Let's first add precision to our expressions. If $\gamma$ is a smooth curve on $X$ with $\gamma(0)=p$, denote by $D_\gamma\in T_p(X)$ the derivation associated to it, namely $D_\gamma(g)=(g\circ\gamma)'(0)$ for germs $g$. Note that if $\phi\colon X\to Y$, then
\begin{align*}
   d\phi(p)\colon T_p(X)&\to T_{\phi(p)}(Y)\\
   D_\gamma&\mapsto D_{\phi\circ\gamma}.
\end{align*}
Let $\mathbb V$ be a finite dimensional vector space. For $v\in\mathbb V$ let $\Phi_{\mathbb V,v}\colon T_v(\mathbb V)\to\mathbb V$ be the identification $\Phi_{\mathbb V,v}(D_\gamma)=\gamma'(0)$. Then, if/ $v\in\mathbb V$ and $L\colon\mathbb V\to\mathbb W$ is linear, the following diagram commutes

In other words, the identification $\Phi$ is a natural transformation from the tangent functor to the identity when restricted to vector spaces. To see this, walk along the diagram

The right-bottom equality follows form the linearity of $L$.
Now the equation of the question can be obtained in the particular case where $\textrm{End}(\mathbb V)$ plays the role of $\mathbb V$, $\mathbb V$ the role of $\mathbb W$ and the evaluation at $v\in\mathbb V$ the role of $L$.
