In what sense is the differential of a linear map is itself?

Question

Given $R^n$ and $R^k$ and $L$ is a linear map from $R^n$ to $R^k$. I was told that the differential of $L$ at $p$, $dL_p:T_pR^n \to T_{L(p)}R^k$ is $L$ itself. Here $dL_p(v)(g) = v(g\circ L)\ \forall v \in T_pR^n$. This is really confusing since $L$ and $dL_p$ does not even have the same source space. In what sense should they be "equal"?

Answer 1

The point is that if $V$ is any finite-dimensional vector space, there is a canonical (i.e., basis-independent) isomorphism between each tangent space $T_pV$ and $V$ itself. Using the definition of the tangent space $T_pV$ as the space of derivations of $C^\infty(V)$ at $p$, this isomorphism is given by sending $v\in V$ to the derivation $D_v\colon C^\infty(V)\to \mathbb R$, defined by $$ D_v f = \left.\frac{d}{dt}\right|_{t=0} f(p+tv). $$

Once you make the identifications $T_p\mathbb R^n\cong \mathbb R^n$ and $T_{L(p)}\mathbb R^k\cong \mathbb R^k$, the equation $DL_p = L$ falls out quickly from the definition of the differential.