Let $W= \text{span}(p_1,p_2,p_3), W \subseteq R_2[x]$
$p_1(x)=1+2x+x^2$
$p_2(x)=3-9x^2$
$p_3(x)=1+4x+5x^2$
From $p_1,p_2,p_3$ choose a basis B for $W$.
I got a problem because they are asking for a basis but my solution says that they are linearly dependent... How can I get a basis anyway?
Form these polynomials to vectors, the form is $a+bx+cx^2$, so we have:
$$\vec{p_1}=\begin{pmatrix} 1\\ 2\\ 1 \end{pmatrix}, \vec{p_2}=\begin{pmatrix} 3\\ 0\\ -9 \end{pmatrix}, \vec{p_3}=\begin{pmatrix} 1\\ 4\\ 5 \end{pmatrix}$$
Now we need to check if they are linearly independent. I used determinant trick:
$$\begin{vmatrix} 1 & 3 & 1\\ 2 & 0 & 4\\ 1 & -9 & 5 \end{vmatrix}\begin{matrix} 1 & 3\\ 2 & 0\\ 1 & -9 \end{matrix}$$
If we use Saruss, indeed, the determinant is zero and thus the vectors are linearly dependent. So we cannot choose any of them as a basis..
Is there a way to get a basis anyway? Maybe I can choose one of the polynomials $p_1,p_2,p_3$ as a basis because a single one of them should be linearly independent.
Use row-reduction to find the redundant vector/column/polynomial(s); the remaining one(s) form a basis of $W$.
More details: $$\begin{bmatrix} 1 & 3 & 1 \\ 2 & 0 & 4 \\ 1 & -9 & 5 \end{bmatrix} \Longrightarrow \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & -1/3 \\ 0 & 0 & 0 \end{bmatrix} $$ tells you that $p_3(x) = 2p_1(x) - 1/3 p_2(x)$, hence $p_3$ is redundant of $p_1 $ and $p_2$.