Linear Independence with Absolute Value Question Are functions $t^3$ and $|t|^3$ linearly independent on $(−∞,∞)$?
I'm fairly certain $t^3$ is linearly independent, as I don't see anything that would cause it to be linearly dependent.  Please do correct me if I'm mistaken.
However, I don't know if $|t|^3$ changes the nature of things.  What difference does the absolute value mean, if it means anything at all?  Thanks.
 A: Two functions $f(x)$ and $g(x)$ are linearly dependent on a set $I$ $$\iff$$$$\left\{\begin{array}{c}\exists a,\,b: af(x)+bg(x)=0\quad \forall x\in I\\|a|+|b|\ne 0\end{array}\right.$$
Suppose your function are linearly dependent, then we can write
$$at^3+b|t|^3=0,\quad t\in \Bbb R.$$
if $t<0$, then it simplifies to $t^3 (b-a)=0$, which gives $a=b$, but when we plug it into the case $t>0$, we obtain $2at^3=0$, which yields $a=0$. Thus we can conclude that $t^3$ and $|t|^3$ are linearly independent on $\Bbb R$.
A: HINT: Prove the following proposition.

Proposition. Let $V$ be any vector space. Then $\{u,v\}\subseteq V\setminus\{0\}$ is linearly independent if and only if $v$ is not a scalar multiple of $u$.

Then apply it to your function space.
A: Can you find non-zero $a$, $b$ such that $at^3 + b|t|^3 = 0$ for all $t$? Or can you prove that if $at^3 + b|t|^3 = 0$ for all $t$, then $a=b=0$?
Consider $t=1$. If equality holds, then $a+b = 0$. And consider $t = -1$, the equation becomes
$$\begin{align}
a(-1)^3 + b|-1|^3 =& 0\\
-a+b =& 0
\end{align}$$
What do you see with these?
A: A single function cannot be linearly independent or linearly dependent.  These are terms applied to a set of functions.  A single function in a set can be linearly independent of the others if it cannot be written as a linear combination of the others in the set.
Consider the linear combination $a f(x) + b g(x)$, with $f$ and $g$ described as in the question.
If $t > 0$, then we end up with $|t| = t$, so the equation becomes ($a+b) t^3 = 0$
This solves as $a = -b$. So, for $a = 1$, $b = -1$, we have a non-trivial linear combination of $f$ and $g$ producing $0$, making $f$ and $g$ linearly dependent when $t > 0$.
If $t < 0$, then we end up with $|t| = -t$, so the equation becomes $(a-b) t^3 = 0$
This solves as $a = b$. So, for $a = 1$, $b = 1$, we again have a non-trivial linear combination of $f$ and $g$ producing $0$, making $f$ and $g$ linearly dependent when $t < 0$.
Hence, $f$ and $g$ are linearly dependent for all $t \neq 0$.
A: Here we can use simply the ratio test. For t<0 the ratio of the functions is -1 where for t> 0, it is 1. Clearly both are not same. Therefore independent on R. Although functions are dependent only in positive R and independent in negative R saperately but for whole R these are independent.
