# Finding a basis for the subspace of P_2

#### Ares_D1

Problem Statement:
Let $$\displaystyle p(x), q(x) \in P_2$$. You may assume that

<$$\displaystyle p(x), q(x)$$> $$\displaystyle = \int_{-1}^1 p(x)q(x)dx$$

defines an inner product on $$\displaystyle P_2$$

Find a basis for the subspace of $$\displaystyle P_2$$:

$$\displaystyle V = \{ a + b + ax + bx^2 | a,b \in R\}$$

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I'm sure this is a very easy question, just not sure how to go about it. Would we not just use as the basis the standard basis of $$\displaystyle P_2$$, i.e.,
$$\displaystyle \{1, x, x^2\}$$?

Note: I'm not sure if the inner product is important for answering this specific question, as there is another question that follows to which it may apply instead.

#### dwsmith

MHF Hall of Honor
Problem Statement:
Let $$\displaystyle p(x), q(x) \in P_2$$. You may assume that

<$$\displaystyle p(x), q(x)$$> $$\displaystyle = \int_{-1}^1 p(x)q(x)dx$$

defines an inner product on $$\displaystyle P_2$$

Find a basis for the subspace of $$\displaystyle P_2$$:

$$\displaystyle V = \{ a + b + ax + bx^2 | a,b \in R\}$$

-----------------------

I'm sure this is a very easy question, just not sure how to go about it. Would we not just use as the basis the standard basis of $$\displaystyle P_2$$, i.e.,
$$\displaystyle \{1, x, x^2\}$$?

Note: I'm not sure if the inner product is important for answering this specific question, as there is another question that follows to which it may apply instead.
For some reason, I am having a mental block but I believe all you need to do is solve $$\displaystyle A\mathbf{x}=\mathbf{0}$$.

$$\displaystyle \begin{bmatrix} 0 & 1 & 0\\ 1 & 0 & 0\\ 1 & 1 & 0 \end{bmatrix}$$ so if I am correct we have an overdetermined system.

$$\displaystyle rref=\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 0 \end{bmatrix}\to a=0, \ b=0$$

Basis:
{$$\displaystyle 0$$} which is also equal to the null space and ker(V); therefore, the range of V is {$$\displaystyle 1,x,x^2$$}

Exercise extreme caution with what I have since I can't think right now.

#### Bruno J.

MHF Hall of Honor
Problem Statement:
Let $$\displaystyle p(x), q(x) \in P_2$$. You may assume that

<$$\displaystyle p(x), q(x)$$> $$\displaystyle = \int_{-1}^1 p(x)q(x)dx$$

defines an inner product on $$\displaystyle P_2$$

Find a basis for the subspace of $$\displaystyle P_2$$:

$$\displaystyle V = \{ a + b + ax + bx^2 | a,b \in R\}$$

-----------------------

I'm sure this is a very easy question, just not sure how to go about it. Would we not just use as the basis the standard basis of $$\displaystyle P_2$$, i.e.,
$$\displaystyle \{1, x, x^2\}$$?

Note: I'm not sure if the inner product is important for answering this specific question, as there is another question that follows to which it may apply instead.
$$\displaystyle a+b+ax+bx^2=a(1+x)+b(1+x^2)$$ and therefore $$\displaystyle V = \mbox{span }(1+x,1+x^2)$$. I'll let you show that $$\displaystyle 1+x, 1+x^2$$ are in fact linearly independent.

The inner product part is irrelevant to this problem.

Ares_D1

#### Ares_D1

$$\displaystyle a+b+ax+bx^2=a(1+x)+b(1+x^2)$$ and therefore $$\displaystyle V = \mbox{span }(1+x,1+x^2)$$. I'll let you show that $$\displaystyle 1+x, 1+x^2$$ are in fact linearly independent.

The inner product part is irrelevant to this problem.
Of course. Thanks for clearing that up.

Out of curiosity though; obviously $$\displaystyle \{1, x, x^2\}$$ is still a basis for this subspace, no? So then, in finding an orthonormal basis for V (via Gram-Schmidt procedure), would there not be multiple, equally valid solutions which arise from starting with either basis?

#### GeoC

If the space is P2, why would you expect to have 3 basis functions?

#### tonio

Of course. Thanks for clearing that up.

Out of curiosity though; obviously $$\displaystyle \{1, x, x^2\}$$ is still a basis for this subspace, no?

If you write "obviously" then why the question "no?" But the answer is no: that is not a basis for that subspace, whose elements are characterized for having

their free coefficient equal to the sum of the linear and the quadratic coefficients. The element $$\displaystyle x$$ does not share this characteristic.

Tonio

So then, in finding an orthonormal basis for V (via Gram-Schmidt procedure), would there not be multiple, equally valid solutions which arise from starting with either basis?
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