# 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?

• 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. Mar 28, 2011 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, ${\beta'}^{\ast} = \{-2\phi_1+\phi_2,\frac{-3}{2}\phi_1+\frac{1}{2}\phi_2\}$ Mar 28, 2011 at 20:19
• Which part gives you trouble? I'd be glad to elaborate. Also, your friend's answer is incorrect. Mar 28, 2011 at 20:58
• – yoyo
Mar 28, 2011 at 21:06

Note: This is essentially the same as Justin's answer, only expanded and explained with more details. You'll recognize my final equations as the same as Justin's final equations, after suitable renaming of the variables.

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 $\phi_2$.

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*}