Linear (in)dependency of a set of polynomials

What I am unable to understand is why we can "we can associate each polynomial to a vector by taking it's leading coefficients" and then place them in a matrix to compute the determinant. This is in reference to: Answer

When we are talking about plain old vectors in $$R^3$$ for instance, it seems natural but what is it about the coefficients of a polynomial? Would we also be able to reduce this matrix to Echelon form to find a set of linearly independent polynomials from the given set? That is to say, the nonzero rows would be linearly independent.

• If you look at the space $\mathbb{P}_2$ of polynomials with degree less than or equal to 2 as one $\mathbb{R}-vector$ space, then this space is isomorphic to $\mathbb{R}^3$. I can associate each polynomial to a vector by taking it's coefficients. Oct 23 '19 at 18:28
• You can try to verify that $T(a_0,a_1,a_2)=a_0+a_1x+a_2x^2$ is a isomorphism between $\mathbb{R}^3$ and $\mathbb{P}_2$. Oct 23 '19 at 18:32
• But that would be with respect to the standard basis, right? So by taking coefficients, you are using {1,$x$,$x^2$}. Is that what is happening? Oct 23 '19 at 18:38
• @someone111 That is exactly what's happening. Oct 23 '19 at 18:40
• A isomorphism carries a basis to a basis. Is you took the standart basis of $R^3$ then {$T(1,0,0), T(0,1,0), T(0,0,1)$} is a basis of $\mathbb{P}_2$. But {$T(1,0,0), T(0,1,0), T(0,0,1)$} is exactly {$1,x,x^2$}. Oct 23 '19 at 18:43