How can i show that two polynomials are linearly independent How can I show that two polynomials, specifically : $x^3 - x$ , $x^2 - x$ are linearly independent?
Intuitively I know they are since one is cubed and the other is squared. But how do I prove it?
 A: Suppose $a(X^3 - X) + b(X^2 -X) = 0$. We want to show that $a = b = 0$.I 
Suppose we are working in $F[X]$ where $F$ is some field. Then equality is defined as equality of coefficients ($\sum_i a_i x^i = \sum_i b_i x^i$ iff for all $i$ we have $a_i = b_i$. Then we have $aX^3 + bX^2 + (-a + -b)X = 0 = 0X + 0X^2 + 0X^3$, so $a = 0, b = 0$ by comparing the coefficients of $X^3$ and $X^2$ directly.
If we are working in the space of polynomial functions on the reals, say, the equality $aX^3 + bX^2 + (-a -b)X = 0$ should hold as functions, so the identity should hold for all values of $X$ (in the reals say). So pick e.g. $X= 2$ and $X=3$ to get two equations in $a$ and $b$ and solve to show that $a=b=0$.   
A: Just identify them with their coordinates in the canonical base $\{1,X,X^2,X^3\}$. They are $(0,1,0,1)$ and $(0,1,1,0)$. You can clearly see that they are linearly independent. If not, try solving $a_1(X^3-X)+a_2(X^2-X)=0$ and check accordingly to the definition of linearly independent. You obtain $a_1X^3 +a_2X^2-(a_1+a_2)X=0$ where $0$ is the polynomial $0$. It is true iff $a_1=a_2=0$ hence the two vectors are linearly independent.
A: Assume $\alpha(x^3-x)+\beta(x^2-x)=0$ so that should be correct for any $x$. Thus for $x=2$ you get $3\alpha+\beta=0$ and for $x=3$ you get $4\alpha+\beta=0$ those both equations imply $\alpha=\beta=0$ which means that your polynomials are linearly independent
