How to show that the set of functions $x_1,x_2,x_1x_2,x_1^2,x_2^2$ are linearly independent? I know how to show $x_1,x_1^2$ and $x_2,x_2^2$ are linearly independent using the methods given here. But not sure how to how to extend it to $x_1,x_2,x_1x_2,x_1^2,x_2^2$. The $x_1x_2$ term is confusing me. Any suggestions on how to proceed?
 A: First, you set up the generic linear combination from the definition of linear dependence/independence (you said that you know that part). There will be five coefficients in it, and your goal is to show that all of then are equal to $0$.
General idea: Plug in five different pairs of values for $x_1,x_2$ — that will give you five linear equations for the coefficients, and solving this system of five equations with five unknowns you should find that it has only the trivial solution (all $c_i=0$).
For a more efficient solution, you can start with what was suggested on the comments: plug in $x_1=0$ and separately $x_2=0$. That will give you some additional simple relations that will allow you to eliminate two of the unknown coefficients. Then use the same general strategy to find the remaining three coefficients.
A: If $x$, $y$, $x^2$, $xy$ and $y^2$ were linearly independent then, for constant $a,b,c,d,e$, we have $ax+by+cx^2+dxy+ey^2 = 0$ for all $x$ and $y$ if and only if $a=b=c=d=e=0$.
If this is true for all $x$ and $y$ then it's true for particular choice of $x$ and $y$.
For example, $(x,y)=(1,0)$, $(x,y)=(0,1)$, $(x,y)=(-1,0)$, $(x,y)=(0,-1)$ and $(x,y)=(1,1)$.
These give us the equations
$$a+c=0$$
$$b+e=0$$
$$-a+c=0$$
$$-b+e=0$$
$$a+b+c+d+e=0$$
From $a+c=0$ and $-a+c=0$ we get $a=c=0$.
From $b+e=0$ and $-b+e=0$ we get $b=e=0$.
If $a=b=c=e=0$ then $d=0$.
$ax+by+cx^2+dxy+ey^2 = 0$ for all $x$ and $y$ if and only if $a=b=c=d=e=0$.
