# Inner products and basis of orthogonal complement

• May 15th 2010, 01:34 PM
Dave2718
Inner products and basis of orthogonal complement
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

$\langle f,g \rangle = \int^1_0 fg\,dt$

Find a basis for the orthogonal complement of {t-1,t^2}
• May 15th 2010, 02:03 PM
Bruno J.
I don't see what $2\times 2$ matrices have to do with this problem!
• May 15th 2010, 03:48 PM
HallsofIvy
Quote:

Originally Posted by Dave2718
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

$\langle f,g \rangle = \int^1_0 fg\,dt$

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 " $\int_0^1 fg dt$" makes no sense. And $\{t- 1, t^2\}$ 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 $at^2+ bt+ c$ and any vector in the "orthogonal complement of $\{t-1,t^2\}$" must satisfy $\int_0^1 (at^2+ bt+ c)(t- 1)dt= 0$ and $(at^2+ bt+ c)(t^2) dt= 0$. 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 $ax^2+ bx+ c$ 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.
• May 15th 2010, 04:18 PM
dwsmith
Quote:

Originally Posted by HallsofIvy
If you are really talking about 2 by 2 matrices that have polynomials of degree 2 or less as entries, then " $\int_0^1 fg dt$" makes no sense. And $\{t- 1, t^2\}$ 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 $at^2+ bt+ c$ and any vector in the "orthogonal complement of $\{t-1,t^2\}$" must satisfy $\int_0^1 (at^2+ bt+ c)(t- 1)dt= 0$ and $(at^2+ bt+ c)(t^2) dt= 0$. 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 $ax^2+ bx+ c$ 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.

Spoiler:

$c\begin{bmatrix}
50\\
-52\\
9
\end{bmatrix}$

• May 15th 2010, 06:24 PM
Dave2718
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.
• May 15th 2010, 06:29 PM
Dave2718
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?
• May 15th 2010, 10:18 PM
Bruno J.
Quote:

Originally Posted by Dave2718
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 $\{t-1, t^2\}$, 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!)
• May 16th 2010, 03:18 AM
Dave2718
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.
• May 16th 2010, 04:17 AM
HallsofIvy
You need to start thinking more in terms of "dimension". The set of all polynomials of degree 2 or less, $at^2+ bt+ c$, has dimension 3 ( $\{t^2, t, 1\}$ is a basis). It should be easy to see that $t- 1$ and $t^2$ 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.