# Implicit Function Theorem and linear dependence

Suppose that $x\in B_1$, the unit ball in $\mathbb{R}^{3}$ and suppose that $A(x)\in C^\infty(B_1,\mathbb{R}^{3\times 3})$ is a smooth matrix valued function on $B_1$.

If I know that $\det A(x)=0$ for all $x\in B_1$, what does the Implicit Function Theorem tell me?

Pointwise, without the implicit function theorem, I can tell that there exists a unit vector $U(x)$ such $$A(x)U(x)=0.$$ Now does the Implicit Function Theorem help in any way w.r.t to the type of $U$ I can choose? Clearly, $U$ is a function of two variables, as it lives on the sphere.

As pointed out by copper.hat, if $A\equiv0$, $U$ could be anything. So I am allowed to choose for example $U\equiv(1,0,0)$, that is, a function independent of everything.

My more precise question (which could be misdirected) is : does the implicit function theorem tell me anything about $U$, namely that if I am given $U(x_0,y_0,z_0)$, I can choose a smooth parameterization of $U$ near that point, for example?

• The implicit function theorem usually requires invertibility of sorts... – copper.hat Jul 31 '16 at 21:20
• @copper.hat Yes, but surely not of A in this case. – username Jul 31 '16 at 21:21
• Are you looking for something in particular? – copper.hat Jul 31 '16 at 21:22
• Well it should imply that I am on an 2-d manifold, shouldn't it? – username Jul 31 '16 at 21:23
• Take A to be zero. – copper.hat Jul 31 '16 at 21:24