Is $(p, q, r)$ linearly independent?

Question

I have an exercise where I have to figure out if the set $(p, q, r)$ is linearly independent where $p=x, q=x^2, r=2$ and $p,q,r\in P_2$.

I know that they are independent if $c_1p+c_2q+c_3r=0$ where $c_1,c_2,c_3\in\mathbb{R}$.

The next step I should do is write this into a matrix but I'm not sure how to do this since there is only one polynomial and not a set of them.

Answer 1

If (p,q,r) were linearly independent then they would form a basis for P_2 but there are many elements in P_2 (such as 1 and x-1) which cannot be written as a linear combination of p,q,r

Apologies for the incorrect answer. I mistakenly thought $P_2$ is over the integers (been working with algebraic integers all day!). Anyway, the idea is same. Can we show that $p,q,r$ form a basis for $P_2$? The answer is yes, and we can show this by observing that $p,q,r$ generate the natural basis of $P_2$: $(x^2,x,1)$. So since $p,q,r$ form a basis of $P_2$, they are linearly independent.

Answer 2

You can directly prove either that $p, q, r$ are linearly independent by the definition: $$c_1x+c_2x^2+2c_3=0+0x+0x^2$$ as these are polynomials, and $c_i$ scalars, the coefficients must be equal: each $c_i=0$.

Or, prove that each polynomial $a+bx+cx^2$ of $P_2$ can be uniquely written as a linear combination of $p, q, r$.
(By the same argument as above, we must have $c_1=b,\ c_2=c,\ c_3=\frac a2$.)

Or, if you insist on the matrix, consider the standard basis $1,x,x^2$ of $P_2$, and write the coordinates of $p, q, r$ according to this basis in the columns: $$\pmatrix{0&0&2\\1&0&0\\0&1&0}$$ (As, for e.g. $p=x$ we have $p=0\cdot 1+1\cdot x+0\cdot x^2$.)

Then row-eliminate it, or calculate its rank, determinant, or inverse, whatever..