Invariant form for the Implicit Function Theorem

Question

I am trying to show an invariant form (by coordenates) of the Implicit function Theorem and, the point I am having trobule is, exactly, the begining: Is not clear for me that there is a subspace $S\subset \mathbb{R}^{d+k}$ $k$-dimensional where $Df(a_0)|_S$ is bijective. My assumptions are:

Let $f:U\subset \mathbb{R}^{d+k} \rightarrow \mathbb{R}^k$ be a $C^\ell$ map, with $f(a_0)=c$ for some fixed $a_0 \in U$ and $Df(a_0) \in \mathcal{L}(\mathbb{R}^{d+k},\mathbb{R}^k)$ is surjective.

Could someone give me a tip?

Answer 1

Your question actually has nothing to do with analysis, it's purely linear algebra, and as such I think it's beneficial to phrase the situation such that we keep only the essential details:

Let $V,W$ be finite-dimensional vector spaces over a field $\Bbb{F}$, and $T:V\to W$ a surjective linear transformation. Then, there is a (not necessarily unique) subspace $S$ of $V$ such that the restriction $T|_S:S\to W$ is an isomorphism.

The proof is simple. You know that $\ker(T)$ is a subspace of $V$, and thus it has a complementary subspace. It doesn't matter which complement we choose, just take any one of them and call that $S$. So, we now have an (internal) direct sum decomposition $V=\ker(T)\oplus S$. Now, by definition of the kernel, it follows that \begin{align} T[S]= T[\ker(T)+S]=T[V]=W, \end{align} where the last equality is because $T$ is surjective by hypothesis. In other words, the restriction $T|_S:S\to W$ is surjective. It is also injective because \begin{align} \ker(T|_S)&=\ker(T)\cap S = \{0\}, \end{align} where the last equality is by definition of $S$ being a complement to $\ker(T)$. Therefore, $T|_{S}:S\to W$ is indeed bijective (hence an isomorphism).

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Some simple things to reflect on:

• Think about what exactly a choice of complementary subspace to the kernel means, and how that relates to row/column operations once you reduce everything to matrices.
• Also, why does a complementary subspace exist?

Answer 2

One can be a little more explicit:

Since $\operatorname{rk} Df(a_0) = k$, we know that $Df(a_0)$ has $k$ linearly independent columns. Let the indices of these columns be $i_1,...,i_k$ and let $S = \operatorname{sp} \{ e_{1_1},...,e_{i_k} \}$, where $e_i$ is the $i$th unit vector. Then $Df(a_0)\mid_S$ is bijective.