Let $M$ be an embedded submanifold of dimension $k$ in $\Bbb R^n$, and let $u^1,\dots,u^k$ be coordinates for a region of $M\subset \Bbb R^n$. Then the inclusion map $M\hookrightarrow \Bbb R^n$ determines $n$ smooth functions $$x_1(u^1,\dots,u^k),\dots, x_n(u^1,\dots,u^k).$$ Let us write $\vec{x}=(x_1,\dots,x_n)$ and $\frac{\partial \vec{x}}{\partial u^i}=(\frac{\partial x_1}{\partial u^i}, \dots, \frac{\partial x_n}{\partial u^i})$. Can we always choose coordinates $(u^1,\dots,u^k)$ so that the symmetric matrix $$(g_{ij}):= \left( \dfrac{\partial \vec{x}}{\partial u^i} \cdot \dfrac{\partial \vec{x}}{\partial u^j}\right)$$ becomes the identity matrix at a fixed point $q\in M$?
(I was reading section 6 of Milnor's Morse Theory, and Milnor assumes that the coordinates are chosen so that $(g_{ij})$ is the identity at $q\in M$. But I am having a hard time seeing that whether this is always possible.)
Edit: Is the matrix $(g_{ij})$ just the matrix of the Riemannian metric induced from the stadard metric on $\Bbb R^n$? If this is true, then I may conclude by choosing normal coordinates at $q$
Given an invertible $k\times k$ matrix $A$, we can define a new set of coordinates $v^i$ by $A^i_jv^j=u^i$. The corresponding basis elements are given by $$ \frac{\partial\vec{x}}{\partial v^i}(p)=A^j_i\frac{\partial\vec{x}}{\partial u^j}(p) $$ Put another way, applying a linear transformation to the coordinates applies the inverse transpose of that linear transformation to the corresponding basis. In particular, for any linear transformation which takes $\frac{\partial\vec{x}}{\partial u^i}(p)$ to an orthonormal basis, in the corresponding change of coordinates $g_{ij}(p)$ is the identity matrix.