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I am revising Linear Algebra, specifically dual bases, and am stuck on this part of the question.

You are given that V is a real vector space, and U is a subspace of V. You are then told that {$x_1$,$x_2$, $x_3$,$x_4$} is a basis for V. You are then told that U is the subspace spanned by: $$x_1+2x_2+3x_3+4x_4$$ and $$5x_1+6x_2+7x_3+8x_4$$You are then asked to find a basis for $U^0$ (the annihilator of U) in terms of the dual basis, {$x_1^*$,$x_2^*$,$x_3^*$,$x_4^*$}.

I thought that we should be looking for a functional f such that $$f(x_1+2x_2+3x_3+4x_4)$$ and $$f(5x_1+6x_2+7x_3+8x_4)$$ are 0. But then, I struggle to see how I should go about finding this function, or the group of functions that satisfy that, and then from there how I express these using the dual basis.

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  • $\begingroup$ A hint that’s the key idea in Nick’s answer, below: Relative to dual bases, evaluating $f(x)$ amounts to computing a dot product. $\endgroup$
    – amd
    Apr 18 '19 at 17:28
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I'm going to write $u_1$ and $u_2$ for the basis vectors of $U$ you gave above. So

$$ \begin {align*} u_1 &= x_1 + 2x_2 + 3x_3 + 4x_4 \\ u_2 &= 5x_1 + 6x_2 + 7x_3 + 8x_4 \end{align*} $$

Let $f \in V^*$ be the functional(s) you are looking for. Write $f$ in the dual basis as

$$ f = \sum_i c_i x_i^* $$

By definition, $x_i^*(x_i) = 1$ and $x_i^*(x_j) = 0$ when $i \neq j$. So plug in $u_1$ and $u_2$ into $f$ to get some equations for the coefficients $c_i$:

$$ \begin{align*} f(u_1) &= c_1 + 2c_2 + 3c_3 + 4c_4 \\ f(u_2) &= 5c_1 + 6c_2 + 7c_3 + 8c_4 \end{align*} $$

You want these to both be zero. So solve the linear system to get the relations:

$$ \begin{align*} c_1 &= c_3 + 2c_4 \\ c_2 &= -2c_3 - 3c_4 \end{align*} $$

This means the generic $f$ which solves the equations depends on 2 parameters ($c_3$ and $c_4$), and looks like:

$$ f = (c_3+2c_4)x_1^* - (2c_3+3c_4)x_2^* + c_3 x_3^* + c_4 x_4^* $$

Factoring out the $c_3$ and $c_4$ you see that a basis for $U^0$ consists of the two functionals:

$$ \begin{align*} f_1 &= x_1^* - 2x_2^* + x_3^* \\ f_2 &= 2x_1^* - 3x_2^* + x_4^* \end{align*} $$

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