# Disrupt a linear one to one map $L_0$ gives a linear one to one map $L_t$

I have a continuous function $f$ from $[0,1]\to\mathcal{L}(\Bbb{R}^n,\Bbb{R}^p)$ such that $f(t)=L_t.$

Assume that $L_0$ is one ton one, I would like to prove that $L_t$ is one to one for $t$ sufficiently small.

Set $\varepsilon>0$ there exists $\gamma>0$ such that for any $t\in[0,\gamma)$ we have $$\Vert L_t-L_0\Vert<\varepsilon$$

Now for any $x\in \Bbb{R}^n$ I have $\Vert (L_t-L_0)(x)\Vert\le \Vert L_t-L_0\Vert \Vert x\Vert$

In fact, I am not sure how can I use the linearity here.

If $x\ne0$ I can choose $\varepsilon=\frac{1}{\Vert x\Vert}$, I will have $\Vert L_t(x)-L_0(x) \Vert<1$ for $t\in [0,\gamma)$, not sure that help.

Geometrically I cannot see why $L_t$ will be one to one, I can imagine that $L_t(x)$ is equal to $L_0(t)+\vec{v}$ and the norm of linear map can be visualize as the taking the biggest vector on the unit sphere.

Question: Can someone give a nice view of why is going to be one to one ?

Notice first that that there exists $c$ such that $\|L_0(x)\| > c > 0$ for all vectors $x$ such that $\|x\| =1$. This holds since $L_0$ is linear (which implies $L_0(0) = 0$), $L_0$ is one-to-one, and the set $\{ x : \|x\| = 0\}$ is closed.
Thus linearity implies that $\|L_0(x)\| \geq c \|x\|$ for any vector $x$.
Continuity says that there exists $\delta>0$ such that if $t < \delta$, then $\|L_t - L_0 \| \leq c/2$.
Then for $t < \delta$, we have \begin{align*} \| L_t(x) - L_t(y) \| &= \|L_t(x-y)\| \\ &\geq \| L_0(x-y)\| - \|[L_t -L_0](x-y) \| \\ &\geq c\|x-y\| - c/2(\|x - y\|) \\ &= c/2\|x-y\|\\ & > 0 \qquad \qquad \qquad \text{ if } x \neq y, \end{align*} so $L_t$ is one-to-one.
• Your solution is a bit confusing, my problem is that I don't see where you are using the fact that $L_t$ is linear, it's seems that your proof will works for any continuous maps. – Alex Sep 23 '16 at 16:56