Showing a basis for a polynomial I am having trouble with this basic basis problem. Need to show that $\{z^4,z^4-z^3,z^4-z^3+z^2,z^4-z^3+z^2-z,z^3-1\}$ is basis for $P_4$. I figured out that it is linearly independent but having problem show that it spans $P_4$
 A: From $z^4$ and $z^4-z^3$, you get $z^3$ by a linear combination.
From $z^4-z^3+z^2$, you get $z^2$ since you have $z^4$ and $z^3$.
From $z^4-z^3+z^2-z$, you get $z$ since you have $z^4$, $z^3$, $z^2$.
From $z^3-1$, you get $1$ since you have $z^3$.
So, $z^4$, $z^3$, $z^2$, $z$, $1$ are all in the subspace generated by $z^4$, $z^4-z^3$, $z^4-z^3+z^2$, $z^4-z^3+z^2-z$, $z^3-1$. Therefore, this subspace must be the whole space $P_4$.

More systematically, write the given polynomials with respect to the basis $z^4$, $z^3$, $z^2$, $z$, $1$; you'll get a triangular matrix that is invertible because it has $\pm1$ in the diagonal.
A: Hint: Consider the following line of thinking.
Theorem: If the vectors $v_1, \ldots, v_m$ span a vector space $E$ then any other set with more than $m$ vectors in $E$ is L.D.
Proof: Exercise.
Claim: Let $E$ be a vector space of dimension $n$. A set with $n$ vectors spans $E$, if, and only if, this set is L.I..
Proof: In fact, if $X=\{v_1 , \ldots, v_n\}$ spans $E$ and is not L.I., then one of the elements of $X$ must be a linear combination of the $n-1$ vectors left. These $n-1$ vectors would still be a spanning set of $E$ which contradicts the Theorem above (why?). 
Conversely, suppose $X$ is L.I.. If $X$ doesn't span $E$ then there would be $v \in E$ such that $v$ isn't a linear combination of the elements in $X$. Then we may take a new set $\{v_1 ,\ldots, v_n, v\}$ that is L.I. and has more elements than $X$, which contradicts the Theorem above.   
Can you conclude from the Claim the answer to your question?
