I need to find a base from the polynomial space $P_3[x]$
for $[1 − x + 3x^2 − x^3]_D = $$ \begin{pmatrix} 1\\ 0\\ 2\\ 0\\ \end{pmatrix}. $$ $
If it helps, in the first part of the question I was asked to show that the following is true about the given space:
$B=\{1 + x, 1 − x, x^3 + x^2, x^2 − x^3\}$
$[1 − x + 3x^2 − x^3]_B = $$ \begin{pmatrix} 0\\ 1\\ 1\\ 2\\ \end{pmatrix} $$ $
and I found it was.
Can someone show me the full algebraic way to do it?
There are many bases in which $1-x+3x^2-x^3$ has co-ordinates $(1,0,2,0)$. One possibility is $\{1-x+x^2-x^3, x, x^2, x^3\}$.