Basis for a Subspace of Polynomials of Degree 5 I have the following question: "Consider the subspace $W$ of $P_5(\mathbb{R})$ (the set of polynomials of at most degree five) given by $$W=\{p(x)\in P_5(\mathbb{R})|p(1)=p(-1)=0\}.$$ Find a basis for $W$, and compute its dimension." I have been working on this for a bit, and I know that the standard basis for $P_5(\mathbb{R})$ won't work, so I came up with $$\{x^5-x^3,x^4-2x^2+1,x^3-x,x^2-1\}.$$ However, I'm  afraid that the set doesn't span $P_5(\mathbb{R})$, but I think that it's linearly independent.  Once I have all this, I can easily find the dimension.  Thank you in advance.
 A: The general polynomial of degree $5$ is
$$p(x)=ax^5+bx_4+cx_3+dx^2+ex+f$$
and from the condition $p(1)=p(-1)=0$ we obtain


*

*$a+b+c+d+e+f=0$

*$-a+b-c+d-e+f=0$
from which, by adding and subtracting the two equations, we obtain


*

*$2b+2d+2f=0 \implies f=-b-d$

*$2a+2c+2e=0 \implies e=-a-c$
and finally
$$p(x)=ax^5+bx_4+cx_3+dx^2+(-a-c)x+(-b-d)\\=a(x^5-x)+b(x^4-1)+c(x^3-x)+d(x^2-1)$$
therefore a basis is given by:
$$\{x^5-x,x^4-1,x^3-x,x^2-1\}$$
A: Your answer is correct and span $W$.
One way to see it is the following:


*

*It is linearly independent as the degree are distincts.

*Consider:
\begin{align}
\phi:&P_5(\mathbb{R}) \to \mathbb{R}^2\\
&p \mapsto (p(-1),p(1))
\end{align}
then $W=\ker(\phi)$. But as $\phi((x+1)/2)=(1,0)$ and $\phi((x-1)/2)=(0,1)$ you have $\dim(\phi(P_5(\mathbb{R})))=2$ so:
$$\dim(W)=\dim(\ker(\phi))=\dim(P_5(\mathbb{R}))-\dim(\phi(P_5(\mathbb{R}))=6-2=4$$
and $4$ linearly independent vectors of a subspace of dimension $4$ is indeed a basis.

A: Written in linear algebra terms, given the basis $e_i = x^i$ and a polynomial $p$ expressed as
$$\begin{bmatrix}
p_0  \\
p_1  \\
p_2 \\
p_3\\
p_4\\
p_5
\end{bmatrix}$$
evaluating $p$ at 1 is the same as $[1,1,1,1,1,1]p$, and evaluating it at -1 is  $[1,-1,1,-1,1,-1]p$
We have that both are equal to zero, so 
$$\begin{bmatrix}
1&1&1&1&1&1  \\
1&-1&1&-1&1&-1
\end{bmatrix}p =\begin{bmatrix}
0  \\
0
\end{bmatrix}$$
So we can take the augmented matrix 
$$\begin{bmatrix}
1&1&1&1&1&1 &|0 \\
1&-1&1&-1&1&-1&|0
\end{bmatrix}$$
And reduce it to row-echelon form
$$\begin{bmatrix}
1&0&1&0&1&0 &|0 \\
0&1&0&1&0&1&|0
\end{bmatrix}$$
This tells you that among even $i$, you can pick two $p_i$ to be free and the third will be fixed, and similarly for odd $i$. For instance, we can have $p_i$ free for $i<4$, and then $p_5=-(p_3+p_1)$ and $p_4=-(p_2+p_0)$. If we successively set one free variable to 1 and the other free variables to zero, this gives
$$1-x^4$$
$$x-x^5$$
$$x^2-x^4$$
$$x^3-x^5$$
Another basis, that could be written more succinctly, would be $b_i = e_{i+2}-e_{imod2}$, which gives
$$b_0 = x^2-1$$
$$b_1=x^3-x$$
$$b_2=x^4-1$$
$$b_3 = x^5-x$$
A: $p(1) = p(-1) = 0$ is equivalent to $p(x)$ being divisible by $x^2-1$. So we can start by writing out a couple of simple multiples of $x^2-1$:
$$
x^2-1, x(x^2-1), x^2(x^2-1), x^3(x^2-1)
$$
These are linearly independent: write $0$ as a linear combination of the above polynomials. Only one of the polynomials has a degree 5 term, so its coefficient must be $0$. Once we know that, there is only one of the polynomials with a degree 4 term, so its coefficient must be $0$, and so on.
Also, together they can build any polynomial which is divisible by $x^2-1$ as a linear combination: if $p(x) = q(x)(x^2-1)$, then express $q(x)$ as a linear combination of $1, x, x^2$ and $x^3$ ($q(x)$ can't have degree higher than $3$, cause then $p(x)$ would have degree more than $5$), and the same coefficients will give you $p(x)$ as a linear combination of the polynomials above.
A: The polynomials in $W$ are all divisible by $x^2-1$, and you can consider the set of cubic polynomials
$$\frac{w(x)}{x^2-1}:w\in W,$$ which is in fact $P_3$. It is easy to show that the $p_3(x)(x^2-1)$ are linearly independent and span $W$.
