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This is perhaps something standard in linear algebra(a subject I am somewhat weak in). So I apologize in advance. I would be grateful if someone can guide me.

The statement of implicit function theorem from wikipedia is the following:

Let $f: \mathbb{R}^{n + m} \rightarrow \mathbb{R}^m$ be a continuously differentiable function, and let $\mathbb{R}^{n+m} $ have coordinates $z = ( x, y)$, where $x \in \mathbb{R}^n$ and $y\in \mathbb{R}^m$. Fix a point $( a , b) = (a_1 , \ldots , a_n , b_1 , \ldots, b_m )$ with $ f( a, b) = c$, where $c \in \mathbb{R}^m$. If the matrix $( \partial f_i/\partial y_j)(a,b)$ is invertible, then there exists an open set $U$ in $\mathbb R^n$ containing $a$, and an open set $V$ in $\mathbb R^m$ contntaining $b$, and a unique continuously differentiable function $g: U \rightarrow V$ such that $$ \{ (\mathbf{x}, g(\mathbf{x}))|\mathbf x \in U \} = \{ (\mathbf{x}, \mathbf{y}) \in U \times V| f(\mathbf{x}, \mathbf{y}) = \mathbf{c} \}.$$

My question is, whether the condition that the matrix $( \partial f_i/\partial y_j)(a,b)$ where $1 \leq i, j \leq m$ is invertible can be weakened to that the matrix $$\frac{\partial f_i}{\partial z_j}(a,b)$$ where $1 \leq i \leq m$ and $1 \leq j \leq n+m$, has rank $m$.

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As a side: I think you should use a new variable to denote $y_i$ for $i=1\cdot\cdot\cdot n+m$, as $y_i$ only goes from $1$ to $m$. –  awllower Feb 22 '13 at 5:28
    
@awllower : Indeed. Fixed. –  Rolanda Hooch Feb 22 '13 at 5:32

2 Answers 2

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Let $n=m=1$. And let $f(x,y)= x$. Then indeed your matrix has rank=$1$. Now choose any $x$, so that $f(x,y)=x$ for any $y$. It is clear then that no $g(y)$ can exist to fulfill the conditions in the theorem. Hence it is not possible to weaken that invertibiliity criterion.
Inform me of any errors that occur, thanks.

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I think the main issue is just notational. You want to change the indices of $x=\{x_{i},y_{j}\}$ by some recombination so that the matrix is now invertible with respect to the new $y_{j}$s.

Just in case of confusion, you should note that $f(x,y)\rightarrow \mathbb{R}^{m}$ has only $m$ factors $f_{1},...,f_{m}$ via the projection map. So there is no $n\times m$ matrix involved in $[\frac{\partial f_{i}}{\partial y_{j}}]$. It must be $m\times m$.

I do remember seeing a statement of implicit function theorem use the rank of the matrix 4-5 years ago, and it should be in standard analysis books. But once you can prove all the statements yourself, this kind of notional difficulty should not be a source of trouble.

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$f_1, \ldots , f_n$ has only $n$-factors. But they can be differentiated wrt any of the $n + m$ variables in $\mathbb{R}^{n+m}$. So the derivatives fill in an $m \times (n+m)$ matrix. I intended that this matrix has rank $m$. –  Rolanda Hooch Feb 22 '13 at 5:05
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I believe you want to differentiate $f_{j}$ with all the variables $x_{i},y_{j}$. I encourage you to find an easy counter-example yourself that your claim does not hold. –  Bombyx mori Feb 22 '13 at 5:10

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