Let $F:\mathbb{R}^n\to\mathbb{R}^m$ have an inverse function ${F^{-1}}:\mathbb{R}^m\to\mathbb{R}^n$ .If $F$ is differentiable at $a\in R^{n}$ and $F^{-1}$ is differentiable at $b=F(a)\in R^{m}$, then show that $m=n$.
Hint: $tr(AB)=tr(BA)$, $A$ is a $m \times n$ matrix and $B$ is a $n \times m$ matrix.
I have considered $DF(a)$, inverse function theorem, etc. I just don't know how to use the hint to solve it. Can somebody teach me how to use that hint?
On
The hint does not help me in any way (at least as long as $\rm tr$ denotes the trace of a matrix.
You can use the chain rule on
$$ (f \circ f^{-1}) (x) = x $$
to conclude
$$ (Df)(f^{-1}(x)) \cdot (D f^{-1})(x) = I_m, $$
where $I_m$ is the $m$-dimensional identity matrix.
Do the same for the converse "order" of $f$, $f^{-1}$. Why does this help you?
$$tr(D(F^{-1} \cdot F)(a)) = tr(DF^{-1}(b) \cdot DF(a)) = tr(DF(a) \cdot DF^{-1}(b)) = tr(D(F\cdot F^{-1} )(b))$$ $$tr(D(F^{-1} \cdot F)(a)) = tr (I_n) = n$$ $$tr(D(F\cdot F^{-1} )(b)) = tr (I_m) = m$$ So $ n = m $