# Why is this determinant zero?

Let $f:U\rightarrow \mathbb{R}^m$ be a differentiable function over an open set $U$ and let $\phi:\mathbb{R}^m\rightarrow\mathbb{R}$ a $C^1$ function such that $\phi (f(x))=0$, $\forall x\in U$. Let $a\in U$; if $(D\phi)_b\neq 0$, where $b=f(a)$, then $det (Df)_a=0$.

I need to prove this result, and in order to do so, I defined $g=\phi \circ f$. Through the Chain Rule I reached this equation:

$$(\bigtriangledown\phi)_b (Df)_a=0$$

That's true for all $a\in U$. I know I can't just say that this means that $(Df)_a=0$ so $det (Df)_a=0$. Can someone help?

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I said it: "differentiable function over an open set U". – Marra Mar 25 '12 at 20:33

Hint: It's not true in general that $(Df)_a = 0$. But if $u$ is a nonzero row vector and $A$ a matrix with $u A = 0$, what does that tell you about $A$?

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That ker(A) is not trivial? – Marra Mar 25 '12 at 21:49
No. If $Au=0$ then what I said would be correct. – Marra Mar 25 '12 at 21:51
Try $\text{ker}(A^T)$ – Robert Israel Mar 26 '12 at 2:07
I was just thinking about this today. If $uA=0$, then $u$ (or its transpose, which is also a vector) belongs to $Ker(A^T)$ (so it's not trivial), which means $det(A^T)=det(A)= 0$ – Marra Mar 27 '12 at 0:26

$D\phi_b$ is a linear form. Suppose it is not zero. Then its kernel has dimension $m-1$. But since $g=0$, $Dg_a = D\phi_b \circ Df_a = 0$ which means $$Im( Df_a) \subset Ker(D\phi _b)$$

So $Df_a$ is not surjective, and that implies $det(Df_a) = 0$

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