Problem. Let $U \subset \mathbb{R}^{m}$ be an open set, $f: U \to \mathbb{R}$ differentiable in $a \in U$ and $M = \lbrace (x,y) \in \mathbb{R}^{m+1}:x \in U,y=f(x) \rbrace$ the graphic of $f$. The set $E$ of all vectors $v = (\alpha_{1},...,\alpha_{m+1}) \in \mathbb{R}^{m+1}$ such that $\alpha_{m+1} = \sum_{1}^{m}\alpha_{i}\cdot \frac{\partial f}{\partial x_{i}}(a)$ is a subspace $m$-dimensional of $\mathbb{R}^{m+1}$. Show that $E$ is equal to the set $V$ of all velocity vectors $\lambda'(0)$ of the paths $\lambda: (-\epsilon,\epsilon) \to \mathbb{R}^{m+1}$, differentiables in $0$, with $\lambda(0) = (a,f(a))$ and such that $\lambda(t) \in M$ for all $t$. Determine $\beta_{1},...,\beta_{m+1}$ for that $v = (\beta_{1},...,\beta_{m+1})$ be non-zero and orthogonal to $E$.
I solved almost the whole problem, but I have some questions. I showed that:
$E$ is a subspace of $\mathbb{R}^{m+1}$
$E = V$ (Here, I need show that $\dim E = m$, since I use that $V = T_{p}\mathbb{R}^{m+1}$ that implies $\dim V = m$, then $V = E$ because I showed that $V \subset E$)
There is a $v \perp E$, since $E \oplus \mathrm{span}(v) = \mathbb{R}^{m+1}$
I couldn't show that:
$\dim E = m$
To make explicit $\beta_{1},...,\beta_{m+1}$
I appreciate any hint.
Hints:
For dimension, consider the vectors $(e_i,df/dx_i(a))$ where $e_i$, $i=1,...,m$ are the canonical vectors of $\Bbb{R}^m$, and prove that are l.i. (This prove that $\text{dim} E\geq m$ but is less that $m+1$ because you have a non trivial combination of coordinates)
For betas, think about the normal of tangent plane