# Show that the set $W$ of all polynomials in $P_2$ such that $p(1)=0$ is a subspace of $P_2$. Find a basis for $W$.

a.) Show that the set $W$ of all polynomials in $P_2$ such that $p(1)=0$ is a subspace of $P_2$.

b.) Make a conjecture about the dimension of $W$.

c.) Confirm your conjecture by finding a basis for $W$.

I know how to show $W$ is a subspace of $P_2$, by showing closure under addition and multiplication by a scalar. However, I am clueless as to how to find a basis. I can't see how this is sufficient information to answer the question. All I can see is that if $p(1)=0$, then $a_01+a_1x+a_2x^2=0$ implies $$a_0+a_1+a_2=0$$

• Can you guess the dimension of $W$ first, before finding a basis? Apr 18, 2014 at 18:00

A polynomial $p(x)=ax^2+bx+c$ satisfies $p(1)=0$ if and only if $$a+b+c=0$$ Hence every $p\in W$ is of the form $$p(x)=ax^2+bx+(-a-b)=a(x^2-1)+b(x-1)\tag{1}$$ Can you use equation $(1)$ to find a basis for $W$ and thus compute $\dim W$?

Sidenote: There is a more fun way to do this problem. The map $T:P_2\to \Bbb R$ given by $T(p)=p(1)$ is linear and $\ker T=W$. Since $T(1)=1$ it is clear that $T$ has rank $1$. The Rank-Nullity theorem then gives $\dim W$.

• It looks like $(x^2-1),(x-1)$ would be a basis, and $dim=2$? Apr 18, 2014 at 18:08
• @Lanfranco Sounds good to me! Apr 18, 2014 at 18:09
• With that info, we can find 3 solutions. However, the book gives: $(x-1),(-x+x^2)$, which I didn't get in any of my solutions. Apr 18, 2014 at 18:21
• Note that $x^2-x=(x^2-1)-(x-1)$ so those "three" solutions are not linearly independent. Apr 18, 2014 at 18:23
• @Lanfranco That's one way to think about it. The general polynomial $ax^2+bx+c$ has three degrees of freedom $a$, $b$, and $c$. Imposing the condition $p(1)=0$ gives the relation $c=-a-b$ which means we now only have two degrees of freedom. Apr 18, 2014 at 19:08

a.Since $P(1) = 0$, $x - 1$ is a factor of $P$. So any $P$ in $W$ is of the form: $P(x) = (x - 1)\cdot q(x)$. So if $P$ and $S$ are two elements in $W$, then, $(P + S)(1) = 0 + 0 = 0$. So $P + S \in W$, and $rP(1) = r\cdot 0 = 0$. So $rP \in W$ for any real number $r$. So $W$ is a subspace of $P_2$

b. To get a dimension of $W$ we find a spanning set $T$ of $W$ and independent as well. Consider $T = \{x - 1, (x - 1)^2\}$. $T$ is independent. We need to show $T$ spans $W$. Let $P$ be any element of $W$. So $P(x) = (x - 1)\cdot q(x)$. If $q(x)$ is a scalar, say $q(x) = m$, then $P = m(x - 1) + 0(x - 1)^2$, and if $q(x)$ is not a scalar, then $q(x) = ax + b$ for some number $a, b$. So $P(x) = (x - 1)(ax + b) = (b - a)(x - 1) + a(x - 1)^2$. So $T$ spans $W$. Thus: $T$ is a basis with $2$ elements. So $dim(W) = 2$