Prove that the tangent space of a hyperplane is itself

Question

I know this might sound really stupid: I was trying to show that the tangent space of a hyperplane is itself.

I started by parametrising the hyperplane locally at $x$ with a diffeomorphism $\phi : U \rightarrow X$, where $U\subset R^n$ and $X\subset R^m$ is the hyperplane we're considering, such that $\phi(0)=x$.

Now the tangent space $T_x(X)$ is given by the image of the derivative map $d\phi_0: R^n \rightarrow R^m$. If we can prove that $\phi$ is a linear map, then we are done, because the derivative map of a linear map is itself. But it seems that I can't prove that $\phi$ is linear.

Also I realised there is a loophole in my proof. Since we're proving a global result, not just at $x$, parametrising $X$ locally won't work...

Any ideas?

Answer 1

You can choose $\phi$ to be a global and linear map. Since the hyperplane is a linear subspace of $\mathbb{R}^m$, you may pick a basis for your hyperplane. That defines a linear isomorphism from $\mathbb{R}^n$ to the hyperplane taking the standard basis of $\mathbb{R}^n$ to your chosen basis, where $n$ is the dimension of the hyperplane. Since a linear isomorphism is automatically a diffeomorphism, $\phi$ indeed serves as a chart.