Why is a vector space equal to its tangent space for any point?

I'm self-studying Guillemin and Pollack, but I'm stuck on Problem 3 of section 2. It says that if $V$ is a vector subspace of $\mathbb{R}^N$, then $T_x(V)=V$ if $x\in V$.

If $x\in V$, then since $V$ is a manifold, there is a local parametrization $\phi: U\to V$ where $U$ is open in $\mathbb{R}^k$. Without loss of generality, we can require $\phi(0)=x$. Then $T_x(V)$ is defined to be the image of $d\phi_0$ on $\mathbb{R}^k$.

An arbitrary element of $T_x(V)$ looks like $$d\phi_0(v)=\lim_{t\to 0}\frac{\phi(0+tv)-\phi(0)}{t}=\lim_{t\to 0}\frac{\phi(tv)-x}{t}$$

but this doesn't seem very useful to show $T_x(V)=V$. What is the right approach?

• You can parameterize $V$ by a linear function $\phi$ whose image under $\mathbb{R}^k$ is just $V$. Since $d\phi=\phi$... – Alex Youcis Sep 30 '13 at 1:18
• thanks for the hint @AlexYoucis – Geovanna Anthony Sep 30 '13 at 1:36

Hint: Fix a basis $\{v_1,...,v_k\}$ of $V$ and consider the following parametrization: $\Phi:\mathbb R^k\to V$ given by $(a_1,...,a_k)\mapsto \sum_{i=1}^ka_iv_i$