Finding the basis for the annihilator

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.

• 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.
– amd
Apr 18 '19 at 17:28

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*}