# Finding a basis for the subspace of P_2

• May 23rd 2010, 05:58 AM
Ares_D1
Finding a basis for the subspace of P_2
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.
• May 23rd 2010, 11:26 AM
dwsmith
Quote:

Originally Posted by 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 $\displaystyle P_2$:

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

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

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.
• May 23rd 2010, 12:34 PM
Bruno J.
Quote:

Originally Posted by 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 $\displaystyle P_2$:

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

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

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.
• May 23rd 2010, 04:18 PM
Ares_D1
Quote:

Originally Posted by Bruno J.
$\displaystyle a+b+ax+bx^2=a(1+x)+b(1+x^2)$ 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?
• May 23rd 2010, 04:47 PM
GeoC
If the space is P2, why would you expect to have 3 basis functions?
• May 23rd 2010, 06:01 PM
tonio
Quote:

Originally Posted by Ares_D1
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?

.