rank of a matrix with two columns s.t. their dot product is zero I have function $\sigma(u,v)=(f(u,v),g(u,v),h(u,v))$ s.t. $\sigma_u$ x $\sigma_v\neq(0,0,0)$   (cross-product) 
also, there is the $3\times 2$ matrix :              $$ 
                                       \begin{bmatrix}
                                       f_u & f_v \\
                                        g_u & g_v \\
                                       h_u & h_v \\
                                       \end{bmatrix}$$
also $\sigma_u(0,v)$ and $\sigma_v(0,v)$ are penpendicular.
the matrix has rank 2(when $u=v=0$). why? could you please help.
($f_u$ means derivative of $f$ wrt. $u$ )
 A: If the matrix had rank $1$, we would then have
$$f_v = a f_u; \,\,g_v = a g_u; \,\,h_v = a h_u$$
But since $\sigma_u(0,v)$ and $\sigma_v(0,v)$ are perpendicular, this would mean $a=0$, i.e., $f_v=g_v = h_v =0$ at $(0,v)$ or $f_u^2+g_u^2+h_u^2 = 0$ at $(0,v)$, i.e., $f_u=g_u = h_u =0$ at $(0,v)$. Both of which, I assume, doesn't make sense in the context of your problem.
A: The matrix has rank $2$ at $(0, v)$ if $\sigma_v \ne 0 \ne \sigma_u$ there.  For we have $\sigma_u \bot \sigma_v$ there, or, algebraically, $\sigma_u \cdot \sigma_v = 0$.  If these two vectors were linearly dependent, then we would have
$a\sigma_u + b\sigma_v = 0 \tag{1}$
with $a \ne 0 \ne b$.  Taking the dot product with $\sigma_u$ yields
$0 = a \sigma_u \cdot \sigma_u + \sigma_v \cdot \sigma_u = a\sigma_u \cdot \sigma_u; \tag{2}$
since $\sigma_u \ne 0$ this forces $a = 0$; a similar argument  shows $b = 0$.  But this conclusion contradicts the hypothesis placed on $a, b$.  Thus $\sigma_u, \sigma_v$ are linearly independent, and the matrix has rank $2$ at $(0, v)$.  Away from $(0, v)$ what happens is anybody's guess, though by the continuity is $\sigma_u$ and $\sigma_v$ we can affirm linear independence in some neighborhood of the set of points $(0, v)$.
Hope this helps.  Cheers,
and as always,
Fiat Lux!!!
