Vector spaces and basis I came across this problem on vector space basis:
Verify that $2-x^2, x^3-x, 2-3x^2$ and $3-x^3$ form a basis for $P^4$ and express each of the polynomial $x^2$ as a linear combination of this basis.
I have tried solving the problem with my knowledge of vector space but still don't know how to go about it.
Any helpful solution to this?
 A: I think that $P^4$ is the space of all polynomials with degree $ \le 3$:
Since $ \dim P^4 =4$, you have to show: if $a,b,c,d \in \mathbb R$ and if 
$$0=a(2-x^2)+b(x^3-x)+c(2-3x^2)+d(3-x^3)$$
then $a=b=c=d=0$.
A: 1) Let's verify that the system of vectors $(2−x^2,x^3−x,2−3x^2, 3−x^3)$(1) is a basis for $P^4$. Write the canonical basis for $P^4$: $(1 ,x ,x^2 ,x^3)$(2) and decompose vectors of the system (1) by basis (2):
\begin{pmatrix}
2 & 0 & 2 & 3\\
0 & -1 & 0 & 0\\
-1 & 0 & -3 & 0\\
0 & 1 & 0 & -1\\
\end{pmatrix} 
Now we'll proof linear independence of (1):
$
A=\begin{pmatrix}
2 & 0 & 2 & 3\\
0 & -1 & 0 & 0\\
-1 & 0 & -3 & 0\\
0 & 1 & 0 & -1\\
\end{pmatrix}
\sim
\begin{pmatrix}
1 & 0 & -1 & 3\\
0 & -1 & 0 & 0\\
-1 & 0 & -3 & 0\\
0 & 0 & 0 & -1\\
\end{pmatrix} 
\sim
\begin{pmatrix}
1 & 0 & -1 & 3\\
0 & -1 & 0 & 0\\
0 & 0 & -4 & 3\\
0 & 0 & 0 & -1\\
\end{pmatrix}
\sim
\begin{pmatrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1\\
\end{pmatrix}$
Proofed,(1) is linear independent, $rankA=rank(1)=4$. But it is maximal linear independent for $P^4$( if we add any vector to (1), it becomes linear dependent (definition of max. linear independent system)($dimP^4=rank(1)=4$)), so we can state that (1) is a basis for $P^4$.
2)Since (1) is a basis for $P^4$, we can decompose any vectors of the $P^4$ by (1)(definition of basis), including each of the polynomial $x^2$. 
