derivative after composition with linear map Let $f: \mathbb{R}^3 \to \mathbb{R}$ be a polynomial function and let $T: \mathbb{R}^3 \to \mathbb{R}^3$ be an invertible linear map. If $\nabla f(P) \neq 0$ for all $P \in \mathbb{R}^3 - \{0\}$ does it follow that $\nabla (f \circ T)(P) \neq 0$ for all $P \in \mathbb{R}^3 - \{0\}$?
By using the chain rule I get that $\nabla (f \circ T)(P) = \nabla f (T(P)) \cdot \nabla T(P) = \nabla f (T(P)) \cdot T(P)$, but just the fact that both $\nabla f(T(P))$ and $T(P)$ are nonzero.
 A: Let $a \in \mathbb{R}^{3}$. By definition, $\nabla f(a)$ is the vector in $\mathbb{R}^{3}$ such that, for all $h \in \mathbb{R}^{3}$ :
$$ \mathrm{D}_{a}f \cdot h = h^{\top} \nabla f(a) $$
where $\mathrm{D}_{a}f$ is the differential of $f$ at $a$ and $\mathrm{D}_{a}f \cdot h$, its value at $h$. By the chain rule :
$$ 
\begin{align*}
\mathrm{D}_{a} \big( f \circ T \big) \cdot h &= {} \mathrm{D}_{T(a)} f \cdot \Big( \mathrm{D}_{a} T \cdot h \Big) \\
 &= \big[ \mathrm{D}_{a} T \cdot h \big]^{\top} \nabla f \big( T(a) \big) \\
\end{align*}
$$
But, since $T$ is linear, for all $a$, $\displaystyle \mathrm{D}_{a}T \equiv T$. As a consequence, 
$$
\begin{align*}
\mathrm{D}_{a} \big( f \circ T \big) \cdot h &= {} \big[ Th \big]^{\top} \nabla f \big( T(a) \big) \\
 &= h^{\top} T^{\top} \nabla f \big(T(a)\big) \\
\end{align*}
$$
So,
$$ \nabla \big( f \circ T \big)(a) = T^{\top} \nabla f \big( T(a) \big) $$
Now, as $T$ is invertible, $\displaystyle \nabla \big( f \circ T \big)(a) = 0$ if and only if $\displaystyle \nabla f \big( T(a) \big) = 0$. With your assumption on $\nabla f$, this is equivalent to $T(a) = 0$, which is also equivalent to $a=0$.
As a consequence, 
$$ \nabla \big( f \circ T \big)(a) = 0 \quad \text{if and only if} \quad a = 0. $$
