I am trying to show that the functions $t^3$ and $|t|^3$ are independent on the whole real line. To do this, I try and prove it by contradiction. So assume that they are dependent. So then there must exists constants $a$ and $b$ such that $at^3+b|t|^3=0$ for all $t \in (-\infty,\infty)$. Now pick two points $x$ and $y$ in this interval and assume without loss of generality that $x<0$, $y\geq0$. Now form the simultaneous linear equations
$ax^3 + b|x|^3 = 0$, $ay^3 + b|y|^3 = 0$, viz.
$\left[\begin{array}{cc} x^3 & |x|^{3}\\ y^3 & |y|^{3}\end{array}\right]\left[\begin{array}{c} a\\ b\end{array}\right]=\left[\begin{array}{c} 0\\ 0\end{array}\right]$
Now here's my problem. If I look at the determinant of the coefficient matrix of this system of linear equations, namely $x^3 |y|^3 - y^3 |x|^3$ and noting that $x<0$ and $y>0$, I have that the determinant is non-zero which implies that the only solution is $a=b=0$, i.e. the functions $t^3$ and $|t|^3$ are linearly independent. However what happens if indeed $y=0$? Then the determinant of the matrix is $0$ and I have got a problem.
Is there something that I am not getting from the definition of linear independence?
The definition (I hope I state this correctly) is: If $f$ and $g$ are two functions such that the only solution to $af+bg = 0$ $\forall t$ in an interval $I$ is $a=b=0$, then the two functions are linearly independent.
But what happens if my functions pass through the origin, like the above? Then I've just shown that there exists a $t$ in an interval containing zero such that the two functions are zero, viz. I can plug in any $a$ and $b$ such that $af+bg = 0$.
Please help, I am confused with the logic and definitions.
Thanks, Ben