Showing independence of a set of functions $\{1,x,x^2,x^3,(x-3)^3_+,(x-5)^3_+\}$ I want to show that the set of functions $\{1,x,x^2,x^3,(x-3)^3_+,(x-5)^3_+\}$ are linearly independent.

I.e., I want to show that if $p(x)=c_1+c_2x+c_3x^2+c_4x^3+c_5(x-3)^3_+ + c_6(x-5)^3_+=0$ then $c_1=c_2=c_3=c_4=c_5=c_6=0$.

The last two functions is what is confusing me. 

Since $p(x)=c_1+c_2x+c_3x^2+c_4x^3+c_5(x-3)^3_+ + c_6(x-5)^3_+=0$ means $p(x)=0$ for all $x$. Could I consider three different cases: $x&lt;3, 3 \leq x &lt;5, x \geq 5$, and for each case show how if $p(x)=0$ then all coefficients are $0$ using the argument posted in the answer to this question Prove that $1, x, x^2, \dots , x^n$ are linearly independent in $C[-1,1]$ ?

I was thinking along these lines, because for those three cases of $x$, you have a cubic polynomial.

If this is not a correct approach, how could I solve this question?

EDIT:

$(x-3)^3_+$ is $(x-3)^3$ if $x \geq 3$ and $0$ otherwise



If I understand your argument correctly you are saying the following: On each of the three regions, the spline is a cubic polynomial.  Since cubic polynomials can have only three roots, unless the polynomial is zero, the only way we can linearly combine the basis functions to be the zero function is if we set all the $ci=0$.  Is this right?

But then consider applying the argument to the following linear combination of basis functions

$c_1 + c_2x + c_3x^2 + c_4x^3 + c_5(x + 2x^2 + 3x^3)$

Then we can claim that the above is a cubic polynomial, hence it has at most three roots, and hence the assumption that the linear combination is the zero function forces c$1 = c2 = .. = c5 = 0.$   But in this case that is not true because we can get the zero function with $c1=0, c2=-1, c3=-2, c4=-4$ and $c5=1.$    The problem with the argument is the assumption that the zero function forces all ci=0.
Your idea is correct. To add more to it, consider eliminating coefficients as you go. Start with $x&lt;3$. Then if $p(x)=0$ for all $x\in \mathbb{R}$, then when $x&lt;3$,
$$
c_1+c_2x+c_3x^2+c_4x^3=0
$$
If $c_j \neq 0$ for some $j$, then a nonzero polynomial has a root at every element in $(-\infty,3)$, a contradiction. Hence, $c_1=c_2=c_3=c_4=0$.

Then move on to $3\leq x &lt;5$ and show $c_5=0$. Finally, finish with $x\geq 5.$
To show that a set of functions $\{f_{1}, f_{2}, \cdots, f_{n}\}$ is linearly independent you must proove that
$$P(x) = c_{1}f_{1} + c_{2}f_{2} + \cdots c_{n}f_{n} = 0$$
Only for 
$$c_{1} = c_{2} = \cdots = c_{n} = 0 $$  

Which can also be proven by using the Wronskian determinant. This is if $W\{f_{1},f_{2},\cdots,f_{n}\} \neq 0 \implies \{f_{1},f_{2},\cdots,f_{n}\} \text{ is LI} $

This particular set is actually linearly dependant for any $x$:
$$ (x-5)^{3} =  x^{3} - 15x^{2} + 75x - 125$$
$$ (x-3)^{3} =  x^{3} - 9x^{2} + 27x -27$$
$$ (x-5)^{3} - \frac{125}{27}(x-3)^3 = \frac{-98x^{3}}{27} + \frac{80x^2}{3} - 50x$$
Then you can see easily that:
$$ (x-5)^{3} - \frac{125}{27}((x-3)^3) + \frac{98}{27}(x^{3}) - \frac{80}{3}(x^{2}) + 50(x) + 0(1) = 0$$

For the cases when the $(x-n)^3$ terms nullify it is trivial to prove that if there is a zero in the set then such set is linearly dependant.