# Dual basis of a two dimensional vector space

I am posed with the following question.

Let $\left(V:F\right)$ be a two dimensional vector space. $\beta = \{x_1, x_2\}$ is a basis of $V$ and $\beta^* = \{\phi_1, \phi_2\}$ is the dual basis of $V$. If $\beta^{'} = \{x_1+2x_2\, 3x_1+4x_2\}$ is also a basis of $V$, find the dual basis of this in terms of $\phi_1$ and $\phi_2$.

What would you suggest for this question?

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Are you sure you didn't mean "the dual basis for $V^*$"? The dual basis lives in the dual space, unless you have an inner product you're not telling us about. – Justin Campbell Mar 28 '11 at 19:29
I'm having a hard time following you, I'm just trying to learn this subject and I got this question from a friend. The answer's also written there, ${\beta'}^{\ast} = \{-2\phi_1+\phi_2,\frac{-3}{2}\phi_1+\frac{1}{2}\phi_2\}$ – hattenn Mar 28 '11 at 20:19
Which part gives you trouble? I'd be glad to elaborate. Also, your friend's answer is incorrect. – Justin Campbell Mar 28 '11 at 20:58
– yoyo Mar 28 '11 at 21:06

See my comment above: $\beta^*$ is a basis for the dual space $V^*$, not $V$. With that in mind: write $\gamma = \{ x_1 + 2x_2,3x_1 + 4x_2 \}$ for the second basis (of $V$), since you already used $\beta$. Write $\gamma^* = \{ \psi_1,\psi_2 \}$ in terms of $\beta^*$, as in $\psi_1 = a_1 \phi_1 + a_2 \phi_2, \ \psi_2 = b_1 \phi_1 + b_2 \phi_2$. Then use the definition of "dual basis" to see that you really need to solve the system of linear equations \begin{align*} a_1 + 2a_2 &= 1, \\ 3a_1 + 4a_2 &= 0, \\ b_1 + 2b_2 &= 0, \\ 3b_1 + 4b_2 &= 1. \end{align*}
The dual basis to $\beta'$ consists of functionals $\mathbf{f}_1$ and $\mathbf{f}_2$ such that $\mathbf{f}_1(x_1+2x_2) = 1$ and $\mathbf{f}_1(3x_1+4x_2) = 0$; and $\mathbf{f}_2(x_1+2x_2) = 0$ and $\mathbf{f}_2(3x_1+4x_2) = 1$.
You already have functionals $\phi_1$ and $\phi_2$ satisfying $\phi_i(x_j) = \delta_{ij}$ (Kronecker's delta). And you know that you can write $\mathbf{f}_1$ and $\mathbf{f}_2$ in terms of $\phi_1$ and $\phi_2$ (because the latter form a basis). So you know you can find scalars $\alpha_1,\beta_1$, $\alpha_2,\beta_2$ such that \begin{align*} \mathbf{f}_1 &= \alpha_1\phi_1 + \beta_1\phi_2\\ \mathbf{f}_2 &= \alpha_2\phi_1 + \beta_2\phi_2. \end{align*} What is the value of $\alpha_1\phi_1+\beta_1\phi_2$ on $3x_1+4x_2$? Well, \begin{align*} (\alpha_1\phi_1+\beta_1\phi_2)(3x_1+4x_2) &= \alpha_1\phi_1(3x_1+4x_2) + \beta_1\phi_2(3x_1+4x_2)\\ &= \alpha_1\Bigl( 3\phi_1(x_1) + 4\phi_1(x_2)\Bigr) + \beta_1\Bigl(3\phi_2(x_1) + 4\phi_2(x_2)\Bigr)\\ &= \alpha_1\Bigl(3(1) + 4(0)\Bigr) + \beta_1\Bigl( 3(0) + 4(1)\Bigr)\\ &= 3\alpha_1 + 4\beta_1. \end{align*} Similarly, \begin{align*} (\alpha_2\phi_1 + \beta_2\phi_2)(3x_1+4x_2) &=3\alpha_2 + 4\beta_2\\ (\alpha_1\phi_1+\beta_1\phi_2)(x_1+2x_2) &= \alpha_1+2\beta_1\\ (\alpha_2\phi_1+\beta_2\phi_2)(x_1+2x_2) &= \alpha_2+2\beta_2. \end{align*} But you know what values you want; you want: \begin{align*} 1 = \mathbf{f}_1(x_1+2x_2) &= (\alpha_1\phi_1 + \beta_1\phi_2)(x_1+2x_2)\\ &= \alpha_1 + 2\beta_1;\\ 0 = \mathbf{f}_1(3x_1+4x_2) &= (\alpha_1\phi_1 + \beta_1\phi_2)(3x_1+4x_2)\\ &= 3\alpha_1 + 4\beta_1;\\ 0 = \mathbf{f}_2(x_1+2x_2) &= (\alpha_2\phi_1 + \beta_2\phi_2)(x_1+2x_2)\\ &= \alpha_2 + 2\beta_2;\\ 1 = \mathbf{f}_2(3x_1+4x_2) &= (\alpha_2\phi_1+\beta_2\phi_2)(3x_1+4x_2)\\ &= 3\alpha_2 + 4\beta_2. \end{align*} So now you have a system of four equations in four unknowns: $$\begin{array}{rcccccccl} \alpha_1 & + & 2\beta_1 & & & & &=& 1\\ 3\alpha_1 & + & 4\beta_1 && & & & = &0\\ & & & & \alpha_2 & + & 2\beta_2 &=& 0\\ & & & & 3\alpha_2 &+& 4\beta_2 &=& 1. \end{array}$$ Solving it will give you the coefficients that express the dual basis of $\beta'$ in terms of $\phi_1$ and $\phi_2$.