Inner products and basis of orthogonal complement

Mar 2010
15
0
Hey guys, I've been thinking about this problem for some time now, but I'm not really sure how to proceed to do it. I think the inner product stuff is throwing me off.

Let V be the vector space of all polynomials of degree 2 or less on the unit interval and define

\(\displaystyle \langle f,g \rangle = \int^1_0 fg\,dt \)

Find a basis for the orthogonal complement of {t-1,t^2}
 
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Bruno J.

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Jun 2009
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I don't see what \(\displaystyle 2\times 2\) matrices have to do with this problem!
 

HallsofIvy

MHF Helper
Apr 2005
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Hey guys, I've been thinking about this problem for some time now, but I'm not really sure how to proceed to do it. I think the inner product stuff is throwing me off.

Let V be the vector space of all 2x2 matrices of all polynomials of degree 2 or less on the unit interval and define

\(\displaystyle \langle f,g \rangle = \int^1_0 fg\,dt \)

Find a basis for the orthogonal complement of {t-1,t^2}
If you are really talking about 2 by 2 matrices that have polynomials of degree 2 or less as entries, then "\(\displaystyle \int_0^1 fg dt\)" makes no sense. And \(\displaystyle \{t- 1, t^2\}\) is not a basis for any subset of that.

I suspect you mean simply "the vector space of all polynomials of degree 2 or less".

In that case, any such "vector" can be written as \(\displaystyle at^2+ bt+ c\) and any vector in the "orthogonal complement of \(\displaystyle \{t-1,t^2\}\)" must satisfy \(\displaystyle \int_0^1 (at^2+ bt+ c)(t- 1)dt= 0\) and \(\displaystyle (at^2+ bt+ c)(t^2) dt= 0\). Do those integrals to get equations for a, b, and c. Since those are two equations, you will probably be able to solve for two of them in terms of the third. Write replace those two by their expression in terms of the third to get \(\displaystyle ax^2+ bx+ c\) in terms of just one of a, b, or c. Finally, factor that a, b, or c out leaving just a number times a polynomial. That polynomial will be the single vector in the basis.
 
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dwsmith

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Mar 2010
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If you are really talking about 2 by 2 matrices that have polynomials of degree 2 or less as entries, then "\(\displaystyle \int_0^1 fg dt\)" makes no sense. And \(\displaystyle \{t- 1, t^2\}\) is not a basis for any subset of that.

I suspect you mean simply "the vector space of all polynomials of degree 2 or less".

In that case, any such "vector" can be written as \(\displaystyle at^2+ bt+ c\) and any vector in the "orthogonal complement of \(\displaystyle \{t-1,t^2\}\)" must satisfy \(\displaystyle \int_0^1 (at^2+ bt+ c)(t- 1)dt= 0\) and \(\displaystyle (at^2+ bt+ c)(t^2) dt= 0\). Do those integrals to get equations for a, b, and c. Since those are two equations, you will probably be able to solve for two of them in terms of the third. Write replace those two by their expression in terms of the third to get \(\displaystyle ax^2+ bx+ c\) in terms of just one of a, b, or c. Finally, factor that a, b, or c out leaving just a number times a polynomial. That polynomial will be the single vector in the basis.

I wanted to do this problem. Here is what I obtained.

\(\displaystyle c\begin{bmatrix}
50\\
-52\\
9
\end{bmatrix}\)
 
Mar 2010
15
0
sorry about that, yeah i think I was helping a friend out with another problem concerning 2x2 matrices and i subconsciously typed that in, OP has been fixed.
 
Mar 2010
15
0
Also, I was wondering if I would get the same answer if I did the gram-schmidt process on the set? Or would I not because that would simply be the orthogonal basis, not the basis for the orthogonal complement. If I did want to use gram-schmidt, I'd have to find the orthogonal complement first right?
 

Bruno J.

MHF Hall of Honor
Jun 2009
1,266
498
Canada
Also, I was wondering if I would get the same answer if I did the gram-schmidt process on the set? Or would I not because that would simply be the orthogonal basis, not the basis for the orthogonal complement. If I did want to use gram-schmidt, I'd have to find the orthogonal complement first right?
You could use G-S to find an orthogonal basis for the subspace generated by \(\displaystyle \{t-1, t^2\}\), and then extend this basis to an orthogonal basis for the whole space; the extra vector(s) in the basis would generate the orthogonal complement. (Prove it!)
 
Mar 2010
15
0
Alright..I'll have to do that proof another day :)

Anyways, so I would have a single vector in my basis for the orthogonal complement? I got {50t^2 -52t +9} to be the basis for my orthogonal complement. For some reason I feel wierd having just one vector be a spanning and linearly independent set for the set I started with.

Thanks though guys.
 

HallsofIvy

MHF Helper
Apr 2005
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7,909
You need to start thinking more in terms of "dimension". The set of all polynomials of degree 2 or less, \(\displaystyle at^2+ bt+ c\), has dimension 3 (\(\displaystyle \{t^2, t, 1\}\) is a basis). It should be easy to see that \(\displaystyle t- 1\) and \(\displaystyle t^2\) are independent so they span a 2 dimensional subspace. The "orthogonal complement" of that subspace must have dimension 3- 2= 1 so a basis must have only one vector.