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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.

Yiorgos S. Smyrlis
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user126279
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    http://math.stackexchange.com/questions/23139/proof-of-linear-independence-of-eat –  Feb 05 '14 at 01:09
  • The closers of this question, M. Turgeon, T. Bongers, user127.0.0.1, Nicholas R. Peterson, mathematics2x2life, have erred in their assertion that this is a duplicate of the question they site, since that question assumes the coefficients of $t$ in the exponentials are real, whereas this one doesn't. This is a difference of consequence, since the proof techniques are different in each case (See the comment threads below for more on this.) How can a group of persons so manifestly competent at mathematics misread such a thing? – Robert Lewis Feb 05 '14 at 02:14
  • It does bug me. I'm trying to be civil, or as they say on MO, keep my pants on in the seminar. But I would like someone to address my points with the rigor which I have come to believe mathematicians are capable. My name's Bob; thanks for letting me share (vent?)! – Robert Lewis Feb 05 '14 at 02:29
  • Trying to maintain my sense of humor and good will, without which all is lost! – Robert Lewis Feb 05 '14 at 03:08

2 Answers2

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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 $(\dagger)$.

Clement C.
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  • This works for real $r_1,r_2$. What if they are purely imaginary? Then $\lim_{t \to \infty} \exp((r_1−r_2)t) \ne \infty$! I think the demonstration can be patched up in this case, however. – Robert Lewis Feb 05 '14 at 01:49
  • I assumed they were real — the OP didn't specify their domain. If they are purely imaginary (or, similarly, have equal real part, in which case we can reduce it to the purely imaginary case), we can fix it like this (roughly): multiply each side by the right value to end up with something of the form $0=\gamma_1 e^{i\alpha t} + \gamma_2$ where $\gamma_1,\gamma_2,\alpha$ are real (and $\alpha \neq 0$); taking $t=\frac{\pi}{2\alpha}$ ends in $0=i\gamma_1+\gamma_2$, and allows to conclude. – Clement C. Feb 05 '14 at 01:57
  • How do you get $\gamma_1$ and $\gamma_2$ both real? I think your demonstration is close, however. Regards. – Robert Lewis Feb 05 '14 at 02:02
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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!!!

Robert Lewis
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