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.Q:

Proving that the partial derivatives are linearly independent in the set of all polynomials of degree $\leq n$

Let $k$ be a field. We have $n+1$ functions $\alpha_{0},\alpha_{1},…,\alpha_{n}:\mathbb{R}^{n}\to k$ and we define $\alpha:=\alpha_{0}-\alpha_{1}+…+\alpha_{n}$ (where the minus sign indicates the opposite way of adding the terms). The (germs at the origin) of the functions $\alpha_{i}$’s are linearly independent if their partial derivatives $\frac{\partial \alpha_{0}}{\partial x_{0}},\frac{\partial \alpha_{1}}{\partial x_{1}},…,\frac{\partial \alpha_{n}}{\partial x_{n}}$ are not all zero at the same time.
We want to show that the germs of the $\alpha_{i}$’s are linearly independent iff the derivative of $\alpha$ with respect to every of its variables is not zero.
$\Rightarrow:$ Let us suppose that the germs of the $\alpha_{i}$’s are linearly independent. Then, $\frac{\partial \alpha_{0}}{\partial x_{0}},\frac{\partial \alpha_{1}}{\partial x_{1}},…,\frac{\partial \alpha_{n}}{\partial x_{n}}$ are linearly independent.
Then, for $i=0,1,…,n$ and $j=0,1,…,n$ we have

\left\{\begin{matrix}
\frac{\partial \alpha_{0}}{\partial x_{0}}\cdot x_{0}+\frac{\partial

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