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Let $f_1(x)=x^2, f_2(x)=x \cdot |x|$ , then they are linearly dependent on $[0,\infty)$, and also linearly dependent on $(-\infty,0]$. But the question is, are they linearly dependent on $(-\infty,\infty)$?

They should be linearly dependent as on both the interval $[0,\infty)$ and $(-\infty,0]$, the functions are linearly dependent. But they are linearly independent as given in my reference book.

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  • $\begingroup$ Why don't you set up the definition of linear dependence? $\endgroup$ – zipirovich Feb 6 at 4:03
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    $\begingroup$ Which theorem are you using when you said "it should be linearly dependent as ... " ? $\endgroup$ – Eclipse Sun Feb 6 at 4:06
  • $\begingroup$ as it is linearly dependent on both the interval so it should be .... $\endgroup$ – mSourav Feb 6 at 4:07
  • $\begingroup$ Are there non-zero solutions for $(a,b)$ such that $af_1 + bf_2 = 0, \forall x$? $\endgroup$ – Dylan Feb 6 at 4:15
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    $\begingroup$ @mSourav That is not a theorem I have heard of. You should at least try to prove the statement. And it is easier to show they are linearly independent by definition. $\endgroup$ – Eclipse Sun Feb 6 at 4:16
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Consider the vector space $\mathscr{C}(\Bbb R)$ of functions $\Bbb R\to\Bbb R$. Our two functions $f_1, f_2\in\mathscr{C}(\Bbb R)$ are defined by $f_1(x)=x^2$ and $f_2(x)=x\cdot\lvert x\rvert$. To see that $\{f_1, f_2\}$ is linearly independent, suppose that $$ c_1\cdot f_1+c_2\cdot f_2=0\tag{$\ast$} $$ where $c_1, c_2\in\Bbb R$. This equation means that $$ c_1\cdot x^2+c_2\cdot x\cdot\lvert x\rvert=0 $$ for all $x\in\Bbb R$. Plugging in $x=-1$ and $x=1$ gives the system of linear equations $$ \begin{array}{rcrcrc} c_1 &-& c_2 &=& 0 \\ c_1 &+& c_2 &=& 0 \end{array} $$ How many solutions are there to this system of equations?

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If two functions, each of which is identically not zero, are linearly dependent, then the first must be the linear combination of the second: $$f_1=cf_2 \\ x^2=c\cdot x|x| \Rightarrow \begin{cases}c> 0, if \ \ x> 0\\ c< 0, if \ \ x< 0\\ c\in \mathbb R, if \ \ x=0\\ \end{cases}$$ However, there is no unique $c$, for which $x^2=c\cdot x|x|$ holds true for all $x\in \mathbb R$. Hence, the two functions are independent for $x\in \mathbb R$.

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    $\begingroup$ A set of two vectors is linearly dependent if and only if one is a scalar multiple of the other; in general, you cannot assert a priori that the first is a multiple of the second, though you can do that after you observe that the second is not the zero vector. Note that $v$ and $\mathbf{0}$ are always linearly dependent, but $v$ is not a a multiple of $\mathbf{0}$ unless they are both $\mathbf{0}$. $\endgroup$ – Arturo Magidin Feb 6 at 5:18
  • $\begingroup$ If one vector is equal to the sum of scalar multiples of other vectors, it is said to be a linear combination of the other vectors... It is not assertion, but a necessary and sufficient condition, which fails in this situation, no? $\endgroup$ – farruhota Feb 6 at 5:27
  • $\begingroup$ @ArturoMagidin, please, see if the update is suitable now? $\endgroup$ – farruhota Feb 6 at 5:48
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    $\begingroup$ Yes, the statement is now correct. The problem you had is that you went from “if two vectors are linearly dependent” to “the first is a multiple of the second”. While the latter statement implies the first, the first may hold while the second is false, because of the example I gave. $\endgroup$ – Arturo Magidin Feb 6 at 6:29
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Hint for a short way : just draw graphs of both functions and look whether they are scalar multiple of each other .

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