Show that ${\boldsymbol v}_1$ and ${\boldsymbol v}_2$ are linearly independent Show that if $r_1 \neq r_2$, the vectors (functions) $${\boldsymbol v}_1 = \exp(r_1t),\,\,\,\,\,\,{\boldsymbol v}_2 = \exp(r_2t)$$ are linearly independent in the space of continuous functions $-\infty<t<\infty$.
I know I need to show that $$c_1{\boldsymbol v}_1+c_2{\boldsymbol v}_2 = 0$$ is linearly independent if $c_1, c_2 = 0$ but I am not sure how to go about it.  
 A: Assume wlog $r_1 > r_2$; multiplying both sides by $\frac{1}{\mathbf{v_2}}$ (which never cancels), you get that for every $t\in\mathbb{R}$,
$$ 0 = c_1 e^{(r_1-r_2)t}+ c_2 \tag{$\dagger$}$$
but since $e^{(r_1-r_2)t}\xrightarrow[t\to\infty]{} +\infty$, this implies $c_1=0$. And in turn $c_2=0$ by plugging back in $(\dagger)$.
A: Here's one quick way to see this:
Suppose that $\exp(r_1 t)$ and $\exp(r_2 t)$ are not linearly independent.  Then there exist scalars $a, b$ such that
$a \exp(r_1 t) + b \exp(r_2 t) = 0, \tag{1}$
and if we set $t = 0$ we obtain
$a + b = 0. \tag{2}$
Now if we differentiate (1) with respect to $t$ we obtain
$ar_1 \exp(r_1 t) + br_2 \exp(r_2 t) = 0, \tag{3}$
and again taking $t = 0$ we find
$ar_1 + br_2 = 0. \tag{4}$
But if $r_1 \ne r_2$ the only solution to the linear system in $a, b$, given by (2), (4) is
$a = b = 0$.  You can see this by writing the equations as a matrix-vector system
$\begin{bmatrix} 1 & 1 \\ r_1 & r_2 \end{bmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = 0, \tag{5}$
and noting that 
$\det(\begin{bmatrix} 1 & 1 \\ r_1 & r_2 \end{bmatrix}) = r_2 - r_1 \ne 0 \tag{6}$
since $r_1 \ne r_2$.  Since $a = b = 0$, $\exp(r_1 t)$ and $\exp(r_2 t)$ are linerly independent.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
