Let $V$ be an open neighborhood of $m$ with $V \subseteq U \subseteq M$.  Let $C^\infty \left({V, \R}\right)$ be the set of smooth mappings $f: V \to \R$.  Let $X_m, Y_m \in T_m M$. Let $\lambda \in \R$. Then, by definition of tangent vector and Equivalence of Definitions of Tangent Vector: :$X_m, Y_m$ are linear transformations on $C^\infty \left({V, \R}\right)$.   Hence $\left({X_m + \lambda Y_m}\right)$ are also linear transformations.  Therefore, it is enough to show that $X_m + \lambda Y_m$ satisfies the Leibniz law.  Let $f, g \in C^\infty \left({V, \R}\right)$.  Then: {{begin-eqn}} {{eqn | l = \left({X_m + \lambda Y_m}\right) \left({f g}\right)        | r = X_m \left({f g}\right) + \lambda Y_m \left({f g}\right)       | c = {{Defof|Linear Transformation}} }} {{eqn | r = X_m \left({f}\right) g \left({m}\right) + f \left({m}\right) X_m \left({g}\right) + \lambda \left({Y_m \left({f}\right) g \left({m}\right) + f \left({m}\right) Y_m \left({g}\right)}\right)       | c = Leibniz law for $X_m, Y_m$  }} {{eqn | r = \left({X_m + \lambda Y_m} \right) \left({f}\right) g \left({m}\right) + f \left({m}\right) \left({X_m + \lambda Y_m}\right) \left({g}\right)       | c = reordering summands }} {{end-eqn}} It follows that: :$X_m + \lambda Y_m \in T_m M$ Hence $T_m M$ is a real vector space.  Again, by definition of tangent vector and Equivalence of Definitions of Tangent Vector: :for all $X_m \in T_m M$ there exists a smooth curve: ::$\gamma : I \subseteq \R \to M$ :with $\gamma \left({0}\right) = m$ such that: {{begin-eqn}} {{eqn | l = X_m \left({f}\right)       | r = \left. {\frac {\mathrm d \left({f \circ \gamma} \right)} {\mathrm d \tau} \left({\tau}\right) }\right\rvert_{\tau \mathop = 0} }} {{eqn | r = \left. {\frac {\mathrm d \left({f \circ \kappa^{-1} \circ \kappa \circ \gamma} \right)} {\mathrm d \tau} \left({\tau}\right)} \right\rvert_{\tau \mathop = 0}       | c = $f \circ \kappa^{-1} \circ \kappa = f$, as $\kappa$ is a homeomorphism, in particular a bijection. }} {{eqn | r = \sum_{i \mathop = 1}^n \left. {\frac {\partial \left({f \circ \kappa^{-1} }\right)} {\partial \kappa^i} \left({\kappa \circ \gamma \left({\tau}\right)}\right) \frac {\mathrm d \left({\kappa^i \circ \gamma}\right)} {\mathrm d \tau}  \left({\tau}\right)} \right\rvert_{\tau \mathop = 0}       | c = Chain Rule for Real-Valued Functions }} {{eqn | r = \sum_{i \mathop = 1}^n \frac {\mathrm d \left({\kappa^i \circ \gamma} \right)} {\mathrm d \tau} \left({0}\right) \frac {\partial \left({f \circ \kappa^{-1} }\right)} {\partial \kappa^i} \left({\kappa \circ \gamma \left({0}\right)}\right)       | c = rearranging }} {{eqn | r = \sum_{i \mathop = 1}^n \frac {\mathrm d \left({\kappa^i \circ \gamma}\right)} {\mathrm d \tau} \left({0}\right) \frac {\partial \left({f \circ \kappa^{-1} }\right)} {\partial \kappa^i} \left({\kappa \left({m}\right)}\right)        | c = as $m = \gamma \left({0}\right)$ }} {{end-eqn}} We define: :$X^i_m := \dfrac {\mathrm d \left({\kappa^i \circ \gamma}\right)} {\mathrm d \tau} \left({0}\right)$ and as above:  :$\left.{\dfrac \partial {\partial \kappa^i} }\right\vert_m \left({f}\right) := \dfrac {\partial \left({f \circ \kappa^{-1}  }\right)} {\partial \kappa^i} \left({\kappa \left({m}\right)}\right)$ Therefore:  :$\displaystyle X_m \left({f}\right) = \left({\sum_{i \mathop = 1}^n X^i_m \left.{\dfrac \partial {\partial \kappa^i} }\right\vert_m}\right) \left({f}\right)$ {{iff}}: :$\displaystyle X_m = \sum_{i \mathop = 1}^n X^i_m \left.{\frac \partial {\partial \kappa^i} }\right\vert_m$ Hence: :$\left\{ {\left.{\dfrac \partial {\partial \kappa^i} }\right\vert_m: i \in \left\{ {1, \dotsc, n}\right\} }\right\}$ forms a basis.  Hence, by Definition:Dimension of Vector Space: :$\dim T_m M = n = \dim M$ This completes the proof.  {{qed}} o9914y9rjsr1fylrsa9ucf849sw6a91