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This is a homework problem from my linear algebra class.

Let $g(t) = 1$. Using the inner product [defined in the textbook], find the orthogonal complement of Span(g) in $$ \mathcal{P}_1 \subset \mathcal{C}^0([0,1]).$$

In the book, the inner product is defined as $$\langle f,g \rangle = \int_a^bf(t)g(t)dt,$$ so, by my logic, this problem should be solvable by integrating $\int_0^1f(t)g(t)dt$, which equals $\int_0^1f(t)dt$ because $g(t) = 1$, which equals $\tfrac{1}{2}f^2(t) \Big|_0^1$. At this point, I'm at a loss at what to do, because it doesn't seem like that manipulation was helpful at all. There are no examples and the book and we didn't do any in class. Is integration the wrong thing to do here? How should I be thinking about this instead?

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    $\begingroup$ $f\in span(g)^\perp\iff <f,g>=0\iff \int_0^1 f(x)dx=0.$ And that's all ! $\endgroup$
    – Surb
    Oct 31, 2016 at 7:01

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You wrote $\int_0^1f(t)dt=f^2(t) \Big|_0^1$. This is wrong ! For example take $f(t)=t+65$

If $\mathcal{P}_1$ denotes the functions of the form $f(t)=at+b$, then the orthogonal complement you are looking for consist of all such functions with $\int_0^1(at+b)dt=0$

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  • $\begingroup$ How do I know they have the form at+b? $\endgroup$
    – coffeeprincess
    Oct 31, 2016 at 7:05
  • $\begingroup$ Above you have not telled us what is the meaning of $\mathcal{P}_1$. Is $\mathcal{P}_1$ the set of all polynomial of degree $\le 1$ ? $\endgroup$
    – Fred
    Oct 31, 2016 at 7:22

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