# Subspace of Polynomials of degree 2

• Tomblue

#### Tomblue

I'm trying to show that a set W of polynomials in P2 such that p(1)=0 is a subspace of P2. Then find a basis for W and dim(W).

I have already found that the set W is a subspace of P2 because it is closed under addition and scalar multiplication and have showed that. The thing I'm stuck on is finding a basis of W and the dim(W), but i think if i figure out how to find the basis i would be able to see what the dim(W) is. So my question is can anyone give me a hint on how to find the basis for W? Thanks

Welcome to PF!

Hi Tomblue! Welcome to PF!
I have already found that the set W is a subspace of P2 because it is closed under addition and scalar multiplication and have showed that. The thing I'm stuck on is finding a basis of W and the dim(W) …

Hint: what is a basis for the whole of P2? How large is it?

So how large do you expect a basis for W to be?

Now adjust the basis you got for P2.

W=p(x)={a0 + a1x + a2x2}
so I'm not sure if this is right but I'm trying to work through it, so since the set W is a set of polynomials in P2 then p(x)={a0+a1 +a2} and a0 would be zero since p(1)=0. Then therefore the basis for W would be = (0,-1,1) but i think the dim(W) is 2-dimensional but if that was correct then i should of had 2 free variables when finding my basis and it would be different. So I'm still not quite sure i did it right.

I can find three linear independent solutions for $$\{a_0, a_1, a_2\}$$; it looks like you stopped after you found one.

You are jumping to conclusions when you say $$a_0$$ would be zero since p(1)=0.

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i found that a0=0, then a2=t, and a1=-t those are the three linearly independent solutions, so what i thought the basis would look like in vector form would bex € P2 such that x=t(0,-1,1)

What if you let $$a_1=0$$ or $$a_2 = 0$$, first of all is it possible to let these variables take that value, and if so, what are then the corresponding values of the other two variables?

If i let a2=0, then a0=-a1 and a1=-a0 but i don't see how that would help me

It helps you because it gives you two more linear independent sets of coefficients $$\{a_0,a_1,a_2\}$$ that help to form a basis for W.

Maybe it would help if you see it like this: (I'm no good with latex, sorry).

$$\left[ \begin{array}{ccc} a_{11} & a_{12} & 0 \\ a_{21} & a_{22} & -1 \\ a_{31} & a_{32} & 1\end{array} \right] [ \begin{array}{ccc} a_{0} \\ a_{1} \\ a_{2}\end{array} ] = 0$$

Hi Tomblue!

(just go up :zzz: …)
W=p(x)={a0 + a1x + a2x2}
so I'm not sure if this is right but I'm trying to work through it, so since the set W is a set of polynomials in P2 then p(x)={a0+a1 +a2} and a0 would be zero since p(1)=0. Then therefore the basis for W would be = (0,-1,1) but i think the dim(W) is 2-dimensional but if that was correct then i should of had 2 free variables when finding my basis and it would be different. So I'm still not quite sure i did it right.

I'm not convinced you understand what a basis is.

(it would have helped if you'd actually answered my question)

One basis for P2 would be the three elements 1 x and x2, which we can write as {1, x, x2}.

Another basis would be {1, 1 + x, 2 + x2}.

Any set of three independent polynomials will do.

ok, now fiddle about with {1, x, x2} to get a (smaller) basis for W.

Here's a different approach.

W is the space of polynomials a0+a1x+a2x2
this is really R3, or {a0 a1 a2}.

Requiring that a0+a1+a2=0, is the same as requiring that the euclidean inner product of any element of W with {1 1 1} be zero.

Hence the subspace W forms a plane in R3, namely the plane that is orthogonal to the vector {1 1 1}, and thus should have dimension 2.

To find a basis simply take the cross product of {1 1 1} with any two other linearly independent vectors, for instance using {1 0 0} and {0 1 0} gives the basis {0 1 -1} and {-1 0 1}, which correspond to the polynomials x-x2 and -1+x2.

Not so general, but gives a good mental picture.

Any polynomial in P2 can be written uniquely as $a_0+ a_1x+ a_2x^2$ (which is the same as saying that {1, x, x2} is a basis).

Any polynomial, p, in P2 such that p(1)= 0 must satisfy $a_0+ a_1(1)+ a_2(1^2)= a_0+ a_1+ a_2= 0$. We can solve for anyone of those coefficients in terms of the other two. For example, we can say that $a_0= -a_1- a_2$. That means that $a_0+ a_1x+ a_2x^2= -a_1- a_2+ a_1x+ a_2x^2$$= (-1+x)a_1+ (-1+ x^2)a_2$. Now do you see what you can use as a basis? You should also be able to find two other bases for the same subspace by solving for $a_1$ in terms of $a_2$ and $a_0$ and by solving for $a_2$ in terms of $a_0$ and $a_1$.