0

So I am trying to get my head around the concept of a dual basis. I think I have an okay idea of it, but my problem lays at finding a dual basis given a basis $B$. So I have given that $B = ((2,1),(3,1))$, which is a basis for $V=\mathbb{R}^2$. Then, apparently, $B^*=(-1x_1+3x_2,x_1-2x_2)$ is an ordered dual basis for $B$.

I know that given a vector $\textbf{x} \in V$ we can get the function $\phi_k(\textbf{x})=([\textbf{x}]_B)_k$ (evaluating it's coordinates in the basis B and taking the coefficient that corresponds to the k-th position). If we then take $B^*=(\phi_1,\phi_2)$, we have an ordered dual basis to $B$.

However, I don't understand how we can compute $\phi_1$ in such an instance, like is done in the example I've given. Could anybody help me out?

Thanks,

K. Kamal

ArianJ
  • 670
  • 5
  • 21

1 Answers1

2

The condition $\phi_i(\mathbf v_j)=\delta_{ij}$, where $\mathbf v_i\in B$, gives you a system of linear equations in the coefficients of the $\phi_i$. If we let $\beta_i$ be the row vector consisting of the coefficients of $\phi_i$, so that $\phi_i(\mathbf x) = \beta_i[\mathbf x]_B$, this system can be written as the matrix equation $$\begin{bmatrix}\beta_1 \\ \beta_2 \\ \vdots \\ \beta_n\end{bmatrix}\begin{bmatrix}\mathbf v_1 & \mathbf v_2 & \cdots & \mathbf v_n\end{bmatrix}=I_n.$$ From this it should be obvious that the $\beta_i$ are the rows of $\begin{bmatrix}\mathbf v_1 & \mathbf v_2 & \cdots & \mathbf v_n\end{bmatrix}^{-1}$. This matrix, in turn, is the change-of-basis matrix from the standard basis to $B$.

amd
  • 53,693