The set of linear combinations is unchanged Show that, if $\sigma (u, v)$ is a surface patch, the set of linear combinations
of $\sigma_u$ and $\sigma_v$ is unchanged when $\sigma$ is reparametrized. 
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I have done the following: 
Let $\tilde{\sigma}(\tilde{u}, \tilde{v})$ be a reparametrization of $\sigma$, i.e., $\sigma (u, v)=\tilde{\sigma}(\tilde{u}(u,v), \tilde{v}(u,v))$. 
Then we have $$\sigma_u=\frac{\partial{\overline{u}}}{\partial{u}}\tilde{\sigma}_{\tilde{u}}+\frac{\partial{\overline{v}}}{\partial{u}}\tilde{\sigma}_{\tilde{v}} \\ \sigma_v=\frac{\partial{\overline{u}}}{\partial{v}}\tilde{\sigma}_{\tilde{u}}+\frac{\partial{\overline{v}}}{\partial{v}}\tilde{\sigma}_{\tilde{v}}$$ 
Is this correct? 
How could we continue? 
 A: You note that the Jacobian matrix of a change of variables  is an isomorphism( a change of variables  is a diffeomorphism).
In particular the tangent plane spanned by $\sigma_u , \sigma_v$ is equal to the tangent plane spanned by $\sigma_\tilde{u} , \sigma_\tilde{v}$ .
A: You need to show that $\left \{ \vec r_u(u,v), \vec r_v(u,v)\right \}$ and $\left \{ \vec r_\tilde u(\tilde u,\tilde v), \vec r_\tilde v(\tilde u,\tilde v)\right \}$ span the same vector space. The easiest way to do this is to compute the Jacobian $\left | \frac{\partial (u,v)}{\partial (\tilde u,\tilde v)} \right |$. The definition of reparameterization requires that this determinant $\neq 0$ which in turn says that the two pairs of basis vectors span the same space.
If you are being asked to do the calculation from scratch, then proceed as follows:
$\vec r(\tilde u,\tilde v)=x(u(\tilde u,\tilde v),v(\tilde u,\tilde v))\vec i+y(u(\tilde u,\tilde v),v(\tilde u,\tilde v))\vec j$
$\vec r_\tilde u=\left ( \frac{\partial x}{\partial u}\frac{\partial u}{\partial \tilde u}+\frac{\partial x}{\partial v}\frac{\partial v}{\partial \tilde u} \right )\vec i+\left ( \frac{\partial y}{\partial u}\frac{\partial u}{\partial \tilde u}+\frac{\partial y}{\partial v}\frac{\partial v}{\partial \tilde u} \right )\vec j$
$\vec r_\tilde v=\left ( \frac{\partial x}{\partial u}\frac{\partial u}{\partial \tilde v}+\frac{\partial x}{\partial v}\frac{\partial v}{\partial \tilde v} \right )\vec i+\left ( \frac{\partial y}{\partial u}\frac{\partial u}{\partial \tilde v}+\frac{\partial y}{\partial v}\frac{\partial v}{\partial \tilde v} \right )\vec j$ 
which is 
$\vec r_\tilde u=\frac{\partial u}{\partial \tilde u}\vec r_u+\frac{\partial v}{\partial \tilde u}\vec r_v$
and 
$\vec r_\tilde v=\frac{\partial u}{\partial \tilde v}\vec r_u+\frac{\partial v}{\partial \tilde v}\vec r_v$
and it is precisely because the Jacobian is non-zero that we can conclude that 
$\left \{ \vec r_\tilde u, \vec r_\tilde v \right \}$ are independent and therefore span the same space as $\left \{ \vec r_u, \vec r_v \right \}$.
