I don't see what matrices have to do with this problem!
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
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 " " makes no sense. And 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 and any vector in the "orthogonal complement of " must satisfy and . 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 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.
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?
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.
You need to start thinking more in terms of "dimension". The set of all polynomials of degree 2 or less, , has dimension 3 ( is a basis). It should be easy to see that and 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.