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 $1\leq j\leq n$.

  • $\begingroup$ 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? $\endgroup$ – Theo Bendit Oct 23 '18 at 12:06
  • $\begingroup$ @TheoBendit Yes, I'll edit to make that clear. $\endgroup$ – 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.

  • $\begingroup$ 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). $\endgroup$ – Matt Oct 24 '18 at 13:45

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