Subspace of Polynomials of degree 2

[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

 

[tiny-tim]

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.

 

[Tomblue]

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.

 

[epkid08]

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.

 

[Tomblue]

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)

 

[epkid08]

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?

 

[Tomblue]

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

 

[epkid08]

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.

 

[Dafe]

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

<br /> <br /> \left[ \begin{array}{ccc} a_{11} &amp; a_{12} &amp; 0 \\ a_{21} &amp; a_{22} &amp; -1 \\ a_{31} &amp; a_{32} &amp; 1\end{array} \right] [ \begin{array}{ccc} a_{0} \\ a_{1} \\ a_{2}\end{array} ] = 0<br /> <br />

 

[tiny-tim]

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.

 

[ulysses1729]

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.

 

[HallsofIvy]

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.