I feel very lost. Can someone help me find out what I am doing wrong here.
$S = \{ 2-x^2, 2x^3-3x+1, 2x+3, 4\}$
Standard basis for $P^4$ is $\{1,x,x^2,x^3,x^4\}$
$p_1 = 2-x^2\\
p_2 = 2x^3-3x+1\\
p_3 = 2x+3\\
p_4 = 4\\
p_5 = 0\\
$
$0 = a_1p_1 + a_2p_2 + a_3p_3 + a_4p_4 + a_5p_5 = (2a_1 + a_2 +3a_3 +4a_4) + (-3a_2 + 2a_3)x + (-a_1)x^2 + (2a_2)x^3 + (0)x^4$
$2a_1 + a_2 +3a_3 + 4a_4$
$-3a_2 + 2a_3$
$-a_1$
$2a_2$
$0$
$$A=\begin{bmatrix}
2 &1 &3 &4\\
0&-3& 2 &0\\
-1& 0& 0& 0\\
0& 2& 0& 0\end{bmatrix}
\begin{bmatrix}
a_1\\
a_2\\
a_3\\
a_4\end{bmatrix}
=\begin{bmatrix}
0\\
0\\
0\\
0\end{bmatrix}$$
Then I row reduce to get:
$$B = \begin{bmatrix} 1\\&1\\&&1\\&&&1\end{bmatrix}$$
Meaning that all nonzero rows of $B$ are a basis for the row space of $A$?
That is, they form a basis for the subspace spanned by A?
$v_1 = (1,0,0,0) v_2= (0,1,0,0) v_3= (0,0,1,0) v_4= (0,0,0,0,1)$ make the basis.
Best Answer
A basis is any linearly independent subset of vectors that spans the space.
Choose sufficiently many columns from $A$ such that they are linearly independent and span the space.
If you can row reduce $A$ into $B$ then your 4 vectors are linearly independent. Hence, you need all 4 column vectors of $A$ to form your basis.