Basis for a Subspace of Polynomials of Degree 5

Question

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.

Answer 1

$p(1) = p(-1) = 0$ 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.

Answer 2

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.

Answer 3

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:

$$\{x^5-x,x^4-1,x^3-x,x^2-1\}$$

Answer 4

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 $W$.

Answer 5

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

$$b_0 = x^2-1$$ $$b_1=x^3-x$$ $$b_2=x^4-1$$ $$b_3 = x^5-x$$