# 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 $$1\leq j\leq n$$.

• When you write $V^n$, do you mean a linear space $V$ of dimension $n$? Also, what do you mean by the curves being linearly independent? Are you referring to pointwise linear independence? Oct 23 '18 at 12:06
• @TheoBendit Yes, I'll edit to make that clear.
– Matt
Oct 23 '18 at 12:07

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.

• Thanks for the counterexample. I've apparently simplified the problem a bit too far. Your first approach is similar to something I had sketched out as well (also to no avail).
– Matt
Oct 24 '18 at 13:45