Interpreting dual basis vectors given basis I was searching for how to find the elements of the dual basis for some vector space $V$ given the basis for $V$ and I found this (Finding dual basis to some given basis). I'm now trying to compute some simple examples and I have difficulties in interpreting the results.
Say I have the plane $\mathbb{R}^2$ and vectors $(1,0)^T$ and $(1,1)^T$, then according to the links answer the dual basis should be the columns of $$\left(\begin{pmatrix}1&1\\ \:0&1\end{pmatrix}^{-1}\right)^T = \begin{pmatrix}1&0\\ -1&1\end{pmatrix}$$
but these are vectors and not functions from $\mathbb{R}^2 \to \mathbb{R}$. How should this be interpreted correctly?
 A: If you represent elements of $V$ by column vectors, then you should represent elements of $V^*$ by row vectors (and the pairing of an element of $V^*$ with an element of $V$ is given by multiplication of the row by the column to get a scalar).
We start with a basis $\{v_1,\dots,v_n\}$ and we want the dual basis $\{\phi_1,\dots,\phi_n\}$. This is determined by the conditions $\phi_i(v_j) = \delta_{ij}$ ($1$ if $i=j$, $0$ otherwise).
If your original vectors $v_1,\dots,v_n$ are inserted as the columns of $A$, then we seek a matrix $B$ so that $BA = I$. The rows of $B$ will then be the dual basis vectors. You recognize $B$ as the inverse matrix $A^{-1}$.  The link you found takes the transpose in order to write the dual vectors again as column vectors.
In the case of the example, we have $v_1=\begin{bmatrix} 1\\0\end{bmatrix}$ and $v_2 =\begin{bmatrix} 1\\1\end{bmatrix}$, which gives us
$$A=\begin{bmatrix} 1&1\\0&1\end{bmatrix}.$$
Then
$$B=A^{-1}=\begin{bmatrix} 1&-1\\0&1\end{bmatrix},$$
so the dual basis vectors will be $\phi_1 = \begin{bmatrix} 1&-1\end{bmatrix}$ and $\phi_2 = \begin{bmatrix}0&1\end{bmatrix}$. Check that $\phi_1(v_1)=\phi_2(v_2)=1$ and $\phi_1(v_2)=\phi_2(v_1)=0$.
A: The vectors in a (finite dimensional) real vector space are column vectors. The vectors in the dual space are linear functions from $V$ to the real numbers. But, you can also represent them as row vectors. When you do this, the function represented by $\varphi = (a_1 \cdots a_n) \in V^*$ is:
$$f_{\varphi}(v) = \varphi v$$
This uses matrix multiplication. So if $v = (b_1 \cdots b_n)^T$ then the output will be:
$$a_1 b_1 + \cdots + a_n b_n$$
