How to show that the set of vectors are a basis for $P_2$ How would I show that {$x^2 - x, x - 1, x$} is a basis for the vector space $P_2$? After putting them in matrix form and reducing to row-echelon form, the matrix had rank 2. I'm really confused
 A: Note that it suffices to show that this set spans $P_2$, then it follows directly that it is a basis (since $P_2$ has dimension 3). Observe that $$(x^2-x)+x=x^2$$ $$-(x-1)+x=1$$ And we know that {$1,x,x^2$} is a basis for $P_2$. Hence this shows {$x^2 - x, x - 1, x$} is a basis.
A: You wrote in your post that:

After putting them in matrix form and reducing to row-echelon form, the matrix had rank 2.

Here is the matrix corresponding to the given polynomials. (The matrix is obtained as the coefficients when these polynomials are expressed in using the basis $x^2,x,1$ for the space $P_2$.) And also the computation of rref.
$$\begin{pmatrix}
1 &-1 & 0\\
0 & 1 &-1\\
0 & 1 & 0
\end{pmatrix}\sim
\begin{pmatrix}
1 &-1 & 0\\
0 & 1 & 0\\
0 & 1 &-1
\end{pmatrix}\sim
\begin{pmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 &-1
\end{pmatrix}\sim
\begin{pmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{pmatrix}$$
A: Outline : 
(1) Show that any polynomial $ax^2 + bx + c$ in $P_2$ can be expressed as a linear combination of these.
(2) Show that if $a_1(x^2 - x) + a_2(x-1) + a_3x = 0$, then $a_1 = a_2 = a_3 = 0$.
This became a little to big for comment and hence posting it as an outline answer.
A: Hint: ( I suppose $P^2$ is the space of second degree polynomial on $\mathbb{R}$)
Note that, $\forall a,b,c \in \mathbb{R}$ we have:
$$
a(x^2-x)-c(x-1)+(a+b+c)x= ax^2 +bx+ c
$$
