Show that a polynomial family is a base to a subspace QUESTION
Taking $P4 = ax^4+bx^3+cx^2+dx+e$ the vector space of polynomials of degree $4$ or lower and $H$ the subspace of $P_4$ such that its elements satisfy the conditions $p(1)= 0$, $p(-1)=0$
Show that $B = \{q_1,q_2,q_3\}$ is a base of $H$ if the polynomials are the following:
$q_1(x) = x^2-1$
$q_2(x)=x^3-x$
$q_3(x) = (x^2-1)^2$
ANSWER ?
So I know that to be a base you need to be linearly independent and you have to generate $P_4$ (in this specific case). 
To prove so, I tried to calculate the matrix of the canonical coordinates of $B$, which gave me
$$\begin{pmatrix}
0&0&1\\
0&1&0\\
1&0&-2\\
0&-1&0\\
-1&0&1
\end{pmatrix}$$
Then I used Gaussian elimination and got to the following point (I had to transform $(x^2-1)^2$ into $x^4-2x^2+1$:
$$\begin{pmatrix}
-1&0&1\\
0&1&0\\
0&0&1\\
0&0&0\\
0&0&0
\end{pmatrix}$$
But I am stuck because how can I show that $B$ is a base of $H$ if $\dim(B) = 3$.
It should be $5$? I recall reading that to be a base of something, you need to have the same dimension.
 A: Note that given the conditions $p(1)=p(-1)=0$, it follows that $H$ is the subspace of $P_4$ consisting of all fourth degree polynomials divisible by both $x-1$ and $x+1$ and therefore by $x^2-1$. Using the division algorithm we know that for all $p(x)\in P_4$
\begin{eqnarray}
p(x)&=&ax^4+bx^3+cx^2+dx+e\\
    &=&[ax^2+bx+(a+c)](x^2-1)+(b+d)x+(a+c+e)
\end{eqnarray}
so we know that for $p(x)\in H$ the remainder must be zero.
Thus $d=-b$ and $e=-(a+c)$ and every element of $H$ can therefore be expressed in the form
\begin{eqnarray}
p(x)&=&ax^4+bx^3+cx^2-bx-(a+c)\\
    &=&a(x^4-1)+b(x^3-x)+c(x^2-1) \tag{1}
\end{eqnarray}
So we know that $B^\prime=\{x^4-1,x^3-x,x^2-1\}\,$ forms a basis for $H$. But we want $(x^2-1)^2=x^4-2x^2+1$, not $x^4-1$. But if in equation $(1)$ we let $a=1,\,b=0,\,c=-2\,$ we get
\begin{equation}
 1\cdot(x^4-1)+0\cdot(x^3-x)-2\cdot(x^2-1)=(x^2-1)^2
\end{equation}
Since $(x^2-1)^2$ is a linear combination of the elements of $B^\prime\,$ we may substitute it in place of $x^4-1$ in $B^\prime$ to get the desired basis $B$.
A: You are correct in that the number of elements in the basis must equal the dimension of the space. You are asked to show that $\{q_1,q_2,q_3\}$ is a basis for $H$, not for $P_4$, so you should verify that


*

*The space $H$ is $3$-dimensional.

*The basis elements $q_1$, $q_2$ and $q_3$ are contained in $H$.

*The basis elements $q_1$, $q_2$ and $q_3$ span $H$.


For the latter, it suffices to show that $q_1$, $q_2$ and $q_3$ are linearly independent. 


*

*The space $P_4$ is $5$-dimensional, and $H\subset P_4$ is defined as
the subspace of polynomials $p\in P_4$ satisfying $p(1)=p(-1)=0$, or
equivalently the $ax^4+bx^3+cx^2+dx+e\in P_4$ satisfying
$$a+b+c+d+e=0\qquad\text{ and}\qquad a-b+c-d+e=0.$$ These are two
(inequivalent) linear equations, so $H$ is indeed $3$ dimensional,
not $5$-dimensional!

*It is easily verified that $q_1,q_2,q_3\in H$ by plugging in $1$ and
$-1$.

*Your Gaussian elimination yields a matrix of rank $3$, which means that its column vectors $q_1$, $q_2$ and $q_3$ span a $3$-dimensional space. So they span all of $H$.
Alternatively, you could argue that $q_1$, $q_2$ and $q_3$ are linearly independent along the following lines: Because $q_1$ and $q_2$ have different degrees they are certainly not scalar multiples of eachother, so they are linearly independent. Any linear combination of $q_1$ and $q_2$ has degree at most $3$, whereas $q_3$ has degree $4$, so $q_3$ is linearly independent of $q_1$ and $q_2$, and we are done.
