Showing the polynomials are a basis of $P_2$ I'm trying to show $A = \{x^2 + 1, x^2 − 1, 2x − 1\}$ is a basis of $P_2$ and I know that I have to show that it's both linearly independent and that the polynomials in $A$ span $P_2$.
I have shown that the polynomials are linearly independent by showing the Wronskian is non-zero: $-8$. However I'm unsure how to show that it spans $P_2$?
 A: If you are comfortable with the idea that $\{1,x,x^2\}$ is a basis for the polynomials of degree at most 2, then write 
\begin{eqnarray*}
e_1 & =& x^2 +1\\
e_2 & = & x^2 - 1\\
e_3 & = & 2x-1.
\end{eqnarray*}
You can then find
\begin{eqnarray*}
1 & = & \frac{1}{2}(e_1 - e_2) \\
x & = & \frac{1}{2} \left [e_3 + \frac{1}{2}(e_1 - e_2) \right ] \\
x^2 & = & \frac{1}{2}(e_1 + e_2).
\end{eqnarray*}
A: You can explicitly show that $1,x,x^2$ are in the span of $A$, but you can also note that if you know that the elements of $A$ are linearly independent, then the span of $A$ is a 3-dimensional subspace of  the 3-dimensional vector space $P_2$, so $A$ spans $P_2$ automatically.
A: We have a set of vectors $A=\{1+x^2, -1+x^2,-1+2x\}$. We can see that if we have a basis for $P_2$, wherein $P_2$ is the set of all polynomials of degree at most 2, then it is of the form 
$$\{1, x, x^2\}.$$
Hence, we can write our vectors in terms of the basis as $A=\{(1,0,1),(-1,0,1),(-1,2,0)\}$. If the set is a basis, then it spans $P_2$, which is a way of saying that every $p\in P_2$ can be expressed as a linear combination of these three vectors, and that the three vectors are linearly independent. Firstly, to show that the vectors are linearly independent, we can form the matrix
$$\left(\begin{array}{ccc}
1 & -1 & -1\\
0 & 0 & 2\\
1 & 1 & 0\\
\end{array}\right)\sim\left(\begin{array}{ccc}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1\\
\end{array}\right).$$
Hence we can see that the vectors are linearly independent. Now, do they span $P_2?$ Or, in other words, can we show that every vector in $P_2$ can be expressed in the form
$$\mathbf{p}=a(1,0,1)+b(-1,0,1)+c(-1,2,0).$$
That is, does the system
$$\left(\begin{array}{ccc}
1 & -1 & -1\\
0 & 0 & 2\\
1 & 1 & 0\\
\end{array}\right)\left(\begin{array}{c}
a\\
b\\
c\\
\end{array}\right)=\left(\begin{array}{c}
p_1\\
p_2\\
p_3\\
\end{array}\right),$$
have a solution, wherein our vector $\mathbf{p}$ is defined as $\mathbf{p}=(p_1,p_2,p_3)$. Since the matrix on the left has a non-zero determinant, then a non-trivial solution exists, which means we can express every polynomial in $P_2$ as a linear combination of our three vectors, and hence the vectors span $P_2$.
Because of this, the vectors form a basis for $P_2$. 
