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My text as the following problem:

Let $V$ be the Euclidean space $P_3$ with the inner product defined in Example 2. Let $W$ be the subspace of $P_3$ spanned by $[t-1,t^2]$ Find a basis for $W^{\perp}$.

Inner product defined in Example 2: Let $V = P$; if $p(t)$ and $q(t)$ are polynomials in $P$, we define $\int_0^1 p(t)q(t)\,\mathrm{d}t$ as the inner product.

I am having trouble trying to figure out what I am supposed to do with this information. I assume $p(t) = t-1$ and $q(t) = t^2$ ? Then I can find the inner product by just evaluating it.

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You have an inner product space. $W^{\perp}$ is the collection of all polynomial $r(t)$ in $P_3$ such that $\langle w(t),r(t)\rangle=0$ for all $w(t)\in W$. Verify that this holds if and only if $\langle t-1,r(t)\rangle=\langle t^2,r(t)\rangle=0$, and use that to get information about what such an $r(t)$ will look like; this will tell you what $W^{\perp}$ is, and then you can use that to find a basis. –  Arturo Magidin Nov 29 '11 at 22:50
    
You need to find a basis for all polynomials $p(t)\in P_3$ such that $$ \int_0^1p(t)(t-1)\,dt=\int_0^1p(t)t^2\,dt=0. $$ –  Joe Johnson 126 Nov 29 '11 at 22:51
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So if I understand it correctly I start with $p(t) = at^3 + bt^2 + ct + d$ then I create two equations like what Johnson said above with $p(t) = t - 1$ and $p(t) = t^2$ in the other. Which will give a set of linear equations should be four I think. If that's right let me see what I come with I guess. –  eWizardII Nov 29 '11 at 23:51
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1 Answer 1

You look for a third polynomial which is orthogonal to both $\{(t-1),t^2\}$, within the space of polynomials of degree 3 - which has dimension 4. A standard base for $P_3$ would be $\{1,t,t^2,t^3\}$. Now the task to be done is:

that should solve the problem (hopefully)

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The exercise doesn't seem to require to find an orthonormal basis of $W^\perp$ so the second step is unnecessary –  Ulysse Mizrahi May 17 '13 at 10:22
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