Finding a basis for the subspace of P_2

Apr 2009
48
3
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.
 

dwsmith

MHF Hall of Honor
Mar 2010
3,093
582
Florida
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
Jun 2009
1,266
498
Canada
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.
 
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Apr 2009
48
3
\(\displaystyle a+b+ax+bx^2=a(1+x)+b(1+x^2)\) 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 2010
43
1
If the space is P2, why would you expect to have 3 basis functions?
 
Oct 2009
4,261
1,836
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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