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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

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1 Answer 1

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The business about passing through the origin is not relevant. Your definition is perfectly fine. The issue is whether there exist constants $a$ and $b$, not both $0$, that (simultaneously) work for all $t$. For the purpose of finding such constants, or showing that they do not exist, $t=0$ is completely useless. But you could for example take the two values $t=1$ and $t=-1$ (or $-17$). You obtain the two linear equations $a+b=0$ and $-a+b=0$, which have only the solution $a=b=0$.

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@user6312 @FredrikMeyer Hi, thanks for all your replies. So my definition should include the statement "for all non-trivial $t$ in $I$", is that what all of you are saying? That's the bit that worries me, what happens when $t=0$. But I can't include it otherwise this would render my proof redundant, is that right? Thanks. –  user38268 Apr 2 '11 at 0:08
    
For linear dependence, you want to show that there exist $a$ and $b$, not both $0$, such that $af(t)+bg(t)=0$ for all $t$ (that is, identically). For linear independence, you want to show that there is no $a$ and $b$ (not both $0$) which works for all $t$. The point is that for any functions $f$ and $g$, and any $k$ such that $f(k)=g(k)=0$, the value of the functions at $t=k$ cannot help you to prove linear independence. –  André Nicolas Apr 2 '11 at 0:43
    
I think I can point to part of the source of confusion. You are using matrices, which is perhaps overly fancy. In terms of matrices, however, to prove linear independence it is enough to show that the determinant is non-zero for some choice of $x$ and $y$. That the determinant is $0$ for some other choices does not matter. –  André Nicolas Apr 2 '11 at 0:59

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