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I have problems finding a dual basis. (exercise b)

$B:=(\sin(x),\cos(x)) \in \mathbb{R}^\mathbb{R}$ is a basis of a subspace spanned by sine and cosine. $V:=\langle\sin(x),\cos(x)\rangle \subseteq \mathbb{R}^\mathbb{R}$.

Let $(\alpha, \beta) \in (V^*)^2$ with

$\alpha: V\rightarrow \mathbb{R}: p \mapsto p(\pi/2)$ and
$\beta: V\rightarrow \mathbb{R}: p \mapsto p(0)$

Show that:
a) $\alpha$ and $\beta$ are linear functionals
b) $(\alpha, \beta)$ is dual basis of $B$.

My approach:
Since $p \in V$, p must have the form $ p = a\cdot\sin(x)+b\cdot\cos(x)$.
So for $\alpha$, p maps to $a\cdot\sin(\pi/2)+b\cdot\cos(\pi/2) = a$ and for $\beta$, p maps to $a\cdot\sin(0)+b\cdot\cos(0) = b$.

a)
$\alpha$ is a linear functional since $p(\pi/2)(k\cdot(x+y)) = a\cdot(k\cdot(x+y)) =ka(x)+ka(y)= ka\cdot p(\pi/2)(x)+ka\cdot p(\pi/2)(y)$
$\beta$ is a linear functional since $p(0)(k\cdot(x+y)) = b\cdot(k\cdot(x+y)) =kb(x)+kb(y)= kb\cdot p(0)(x)+kb\cdot p(0)(y)$

b)
For a dual basis I need that $\alpha(\sin(x)) = 1, \alpha(\cos(x)) = 0, \beta(\sin(x)) = 0, \beta(\cos(x)) = 1$.
But with my $\alpha$ and $\beta$ it would not work. So where is my mistake?

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    Which of the equations you list in b) is not true?2012-10-23
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    I do not see which equations are not true. With $e^j(e_i), e^j \in (V^*), e_i \in B$ these equations should follow.2012-10-23
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    You list four equations, all of which have to be true for $(\alpha, \beta)$ to be a dual basis. You then say "But with my $\alpha$ and $\beta$ it would not work." So you must think one of these equations is false. Which?2012-10-23
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    I think that $\alpha(\sin(x)) = a \cdot(\sin(x)) = 1$ and $\alpha(\cos(x)) = a \cdot(\cos(x)) = 0$ cannot be true the same time because for $a = csc(x)$ the first equation would be true but the second not.2012-10-23
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    What is this "$a$" that $\alpha (\sin(x))$ depends on? There's no $a$ in the definition of $\alpha$, or of $\sin(x)$.2012-10-23
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    But p is a linearcombination of the span $\langle \sin(x), \cos(x) \rangle$ so I used $a, b$ for coefficients2012-10-23
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    @monoid: $p$ is equal to $\sin$. There is no "a".2012-10-23
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    So I have to solve $\sin(x) = 1, \cos(x) = 0$ and $\sin(x) = 0, \cos(x) = 1$ for $x$?2012-10-23
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    @monoid: There is no "x" either. I think you're getting lost in the symbols. Try to step back and look at the bigger picture. You should also clean up your notation - writing "$p = a\sin x+b\cos x$" is incorrect!2012-10-23

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Your mistake is that you confuse what is a function and what is an element:

$\alpha(\sin(x))$ is not meant as take $x\in \mathbb{R}$, then map it to $\sin(x)$ and then apply $\alpha$ to it.

If you look at the definition of $\alpha$ this is a (linear) map from $V$ to $\mathbb{R}$, so it takes the element of $V$, $f(x):=\sin(x)$ to $\alpha(f)$. And if you look in the definition of $\alpha$ it takes $f$ to $\alpha(f)=f(\pi/2)=\sin(\pi/2)=1$.

I guess you can figure the other equalities out yourself.

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    Okidoki, after getting out of bed to solve this, I got this: a) $\alpha$ is a linear functional since $\alpha(f+g)=(f+g)(\pi/2)=f(\pi/2)+g(\pi/2)=\alpha(f)+\alpha(g)$ $\alpha(kf)=(kf)(\pi/2)=k \cdot f(\pi/2)=k \cdot \alpha(f)$ arbitrary $f, g \in V$, analogously for $\beta$ b) $(\alpha, \beta)$ is dual basis since: Let $f(x):=\sin(x), g(x):=\cos(x)$, $\alpha(f) = f(\pi /2) = \sin(\pi/2) = 1$, $\alpha(g) = g(\pi /2) = \cos(\pi/2) = 0$, $\beta(f) = f(0) = \sin(0) = 0$, $\beta(g) = g(0) = \cos(0) = 1$2012-10-23
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    @monoid : correct.2012-10-24