Linear independence of curves in a Vector Space I've come across a statement in a proof, that I'm not sure I follow.  I believe I broken it down into a (potentially) basic linear algebra result which I'm a bit rusty on.
Proposition:  Let $V$ be an $n$-dimensional linear space with basis $\{v_j\}$.  Let $\{u_j(t)\}$ be a collection of curves in $V$ which are (pointwise) linearly independent for all $t\in(0,\delta)$ for some $\delta>0$. Suppose $\{w_j(t)\}$ is another collection of curves in $V$ which satisfy
$$w_j(0)=u_j(0)=0,\qquad w_j'(0)=u_j'(0)\neq0,$$
for $1\leq j\leq n$.  Then there exists $\epsilon>0$ such that $\{w_j(t)\}$ are (pointwise) linearly independent for all $t\in(0,\epsilon)$.
Is this true?  If so, any help on the proof, or idea of the proof would be appreciated.
Note in the above, that all collections range $1\leq j\leq n$.
 A: Counterexample: The statement is not true. Consider in $\mathbb{R}^2$ the curves 
$$
\begin{align*}
 u_1(t) &= (t,0), &  u_2(t) &= (t,t^2), \\
 w_1(t) &= (t,0), &  w_2(t) &= (e^t-1,0).
\end{align*}
$$
Note that $u_j(0)=w_j(0)=(0,0)$ and $u_j'(0)=w_j'(0)=(1,0)$ where $j=1,2$. The collection $\{u_1(t),u_2(t)\}$ is linearly independent for every $t>0$, but the collection $\{w_1(t),w_2(t)\}$ is linearly dependent for every $t>0$.
Some additional considerations: My initial idea for a proof was to consider the function
$$ 
f\colon \mathbb{R}^+ \to \mathbb{R}: t \mapsto \det[u_1(t)\;u_2(t)\;\ldots\;u_n(t)]. 
$$
Note that this function is differentiable, since the curves are differentiable and the determinant a composition of sums and products. One would like to proof that this function is non-zero at some point. Since $f(0)=0$, one would like to have $f'(0)\neq 0$, since this would imply that $f$ becomes non-zero for $t>0$ small enough.
Unfortunately this approach will never work. Write $A=\det[u_1(t)\;u_2(t)\;\ldots\;u_n(t)]$. Then
$$ 
  f'(t) = \frac{d}{dt}\left(\det A\right) 
        = \mathrm{trace}\left(\mathrm{adj}\,A \frac{dA}{dt}\right).
$$ 
At $t=0$ the matrix $A$ is zero by the assumption. Hence $\mathrm{adj}\,A=0$ at $t=0$, so $f'(0)=0$.
Note that it is possible that $f(t)$ becomes non-zero even if $f'(0)=0$; the collection $\{u_1(t),u_2(t)\}$ above is an example.
Another approach would be to use the fact that the function $f$ is continuous. If there is one point where the function $f$ for the collection $\{w_j(t)\}$ is non-zero at some point, then $f(t)$ is non-zero on a neighbourhood of that point. But you are maybe not in this setting.
Edit: If you know that $u_1'(0), \ldots, u_n'(0)$ are linearly independent, then for $t$ small enough, you will have $u_j(t)\approx t u_j'(t)$. This allows you to show that $\{u_j(t)\}$ is an independent collection for $t$ small enough.
I hope these considerations are useful to you.
