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I came across the following problem that says:

Consider $f(x)=|x|e^{ax},g(x)=xe^{ax}.$ Then the pair $\{f,g\}$ is
$1.$ linearly independent on $\mathbb R$
$2.$ linearly independent on $(0,\infty)$
$3.$ linearly independent on $(-\infty,0)$
$4.$ linearly dependent on $(-\epsilon,\epsilon)$ for some $0<\epsilon<1$.

My Attempt: I know that the pair $\{f,g\}$ is linearly independent iff $\alpha f(x)+\beta g(x)=0 \implies \alpha=0,\beta =0$ where $\alpha ,\beta$ are scalars.So from the given problem I see that $\alpha |x|+ \beta x=0$ as $e^{ax} \neq 0.$Now on $(0,\infty)$ ,$\alpha+\beta=0$ and on $(-\infty,0)$, $-\alpha+\beta=0.$ So on $(0,\infty)$ ,where $\frac {\alpha}{\beta}=k=-1$ and on $(-\infty,0)$ , $\frac {\alpha}{\beta}=k=1$. So options $(2),(3)$ can not be right as the functions will be L.D. in those interval. Option $(4)$ also can not be right as no single value of $k$ is available as in $(-\epsilon,0)$, $k=1$ and on $(0,\epsilon)$,$k=-1.$ So,I can conclude option $(1)$ is the right choice.

Am I going in the right direction? If not, can someone point me in the right direction? Thanks in advance for your time.

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up vote 5 down vote accepted

By the very definition

$$f(x)=|x|e^{ax}=\begin{cases}\;\;\;xe^{ax}&\,\;\;x\ge 0\\{}\\-xe^{ax}&,\;\;x<0\end{cases}$$

Now, two vectors are linearly dependent iff one of them is a multiple scalar of the other one, so if your set is linearly dependent then there exists $\,k\in\Bbb R\,$ s.t.

$$f(x)=kg(x)\Longleftrightarrow |x|e^{ax}=kxe^{ax}\Longleftrightarrow|x|=kx$$

so...what do you think? You may want to check for positive and negative values of the variable separatedly...

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Thank you for the nice explanation. From $|x|=kx,$ I see $k=1$ or $k=-1$ depending on $x>0$ or $x<0$ respectively.So here the given function are L.D. So only the option $(4)$ is true. – user52976 Mar 3 '13 at 4:57
@user33640, I didn't understand you: the only true option is (1)...(4) cannot be as there're negative and positive values in that interval and thus not one single positive or negative $\,k\,$ will do... – DonAntonio Mar 3 '13 at 5:03
thanks a lot sir for the feedback.I have got it now. – user52976 Mar 3 '13 at 16:01

Let me more explain the @DonAntonio's idea. First, you must understand with what you deal: $f$ and $g$ are two vectors (functions) in the vector space of functions $I\rightarrow \mathbb{R}$ and in different case, the interval $I$ is replaced by $\mathbb{R}, (0,\infty),\ldots$

Now, $f$ and $g$ are is linearly independent iff $$\alpha f+\beta g =0 \Rightarrow \alpha=\beta=0.$$ The first $0$ above is the function zero (zero vector) and the second $0$ is the real zero. So otherwise, we can write, $f$ and $g$ are is linearly independent iff $$\forall x\in I,\quad\alpha f(x)+\beta g(x) =0 \Rightarrow \alpha=\beta=0.$$

If you understand this, you can decide which of the above cases is true.

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