<s>[INST] Spaces of polynomials 
Let $p_{1}(x)=-2x+2$, $p_{2}(x)=x+2$, $p_{3}(x)=x^{2}+2x+3$, $p_{4}(x)=x^{2}-x+3$.
a) From the above four polynomials, determine a linearly independent subset that spans the polynomials.
b) What is the dimension of the (sub)space that this basis spans?
c) Does this basis span the space of the second degree polynomials? If not, describe the  subspace it spans in terms of the coefficients of a polynomial that belongs to this subspace?

I said since the Wronskian is 0, the functions are linearly dependent. We see three non-zero rows there. This means three of the functions form a linearly independent set of functions. So, {$p_{1}(x),p_{2}(x),p_{3}(x)$} is subset that spans the space of second degree polynomials and is a basis for this space. The dimension is three since there are three functions that form the basis. But parts a and c yelling at me: "Wrong!". Can somebody help me?
 [/INST] (a) Let $\vec p_i$ be the coefficient vector of $p_i(x)$. Then, 
$$M= 
\begin{bmatrix}
 \vec p_1 & \vec p_2 & \vec p_3 & \vec p_4
\end{bmatrix}
= 
\begin{bmatrix}
0 & 0 & 1 & 1 \\
-2 & 1 & 2 & -1 \\
2 & 2 & 3 & 3
\end{bmatrix}
\sim
\begin{bmatrix}
1 & 0 & 0 & 1 \\
0 & 1 & 0 & -1 \\
0 & 0 & 1 & 1
\end{bmatrix}
.$$
Columns $1$, $2,$ and $3$ are pivot columns $\Rightarrow T =\{p_1, p_2, p_3 \}$ is a linearly independent subset that spans $\{p_1, p_2, p_3, p_4 \} \Rightarrow T$ forms a basis for $\operatorname{Span} \{p_1, p_2, p_3, p_4 \}$.
(b) $\dim (\operatorname{Span} T) = \operatorname{Rank} M = 3.$
(c) Let $p \in \mathbb{P}^2$. Then, $p$ is of the form $ax^2+bx+c$. The corresponding system to solve is 
$$
\begin{bmatrix}
\vec p_1 & \vec p_2 & \vec p_3 & \vec p
\end{bmatrix}
=
\begin{bmatrix}
0 & 0 & 1 & a \\
-2 & 1 & 2 & b \\
2 & 2 & 3 & c
\end{bmatrix}
\sim
\begin{bmatrix}
1 & 0 & 0 & \frac{1}{6} a + \frac{1}{6}c - \frac{1}{3} b \\
0 & 1 & 0 & - \frac{5}{3} a + \frac{1}{3}c + \frac{1}{3} b  \\
0 & 0 & 1 & a
\end{bmatrix}.
$$
We can see that the system is consistent, and $$p = \left( \frac{1}{6} a + \frac{1}{6}c - \frac{1}{3} b \right)p_1 +
\left( -\frac{5}{3} a + \frac{1}{3}c + \frac{1}{3} b \right) p_2 + a p_3. $$
Hence, $\operatorname{Span} T = \mathbb{P}^2$. But, we showed that $T$ was linearly independent $\Rightarrow T$ forms a basis for $\mathbb{P}^2$.
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