In linear algebra, why is the dual basis of a basis the *rows* of the matrix $A^{-1}$? When finding the dual basis, we first find the transition matrix from the basis to the other (dual) basis. So if I have some basis in $\mathbf{R^2}$ $\{(a, b), (c, d)\}$, and I want to find the dual basis in terms of the standard basis, the transition matrix would be 
$A =\begin{bmatrix}
    a & c \\
    b & d
\end{bmatrix}$
Lastly, we find $A^{-1}$. This gives us the dual basis vectors as the rows.
$1.$ Is my understanding up to this point correct? 
$2.$ In linear algebra, why is the dual basis of a basis the rows of the matrix $A^{-1}$? After all, it's the columns that are the vectors.
I would greatly appreciate it if people could please take the time to clarify this.
 A: I try to share my ideas though I'm not sure if I had correctly understand your question XD.
If I write your $A$ as 
$A =\begin{bmatrix}
v_1& v_2
\end{bmatrix},$
where $v_1,v_2$ are vectors, then their dual basis $w_1,w_2$ are the functionals such that $w_i(v_j)=\delta_{ij}.$ Then in the matrix type, it would become
$$\begin{bmatrix}
    w_1\\
    w_2
\end{bmatrix}\begin{bmatrix}
v_1& v_2
\end{bmatrix}
=\begin{bmatrix}
    w_1\cdot v_1&w_1\cdot v_2\\
    w_2\cdot v_1&w_2\cdot v_2
\end{bmatrix}=I$$
so you need to find the inverse of $A$ and take its row vectors.
A: This works for every ${\Bbb R}^n$ and in fact for every finite dimensional vector space $V$ over any field $K$.
To fix ideas, consider ${\Bbb R}^n$ with its standard basis $\{e_i\}$, and the dual $({\Bbb R}^n)^\ast$ with the dual basis $\{e_i^\ast\}$.
Given a basis $\{v_i\}$, let $A=(a_{ij})$ be the matrix whose $j$-th column is the components of $v_j$ in terms of the standard basis.
Expressing the dual basis $\{v_i^\ast\}$ in terms of the standard dual basis amounts to finding expressions
$$
v_i^\ast=\sum_{j=0}^nb_{ij}e_j^\ast.
$$
But just because $v_i^\ast(v_j)=\delta_i^j$these coefficients $b_{ij}$ must satisfy the relations
$$
\sum_{\ell=1}^nb_{i\ell}a_{\ell j}=\delta_i^j,
$$
which makes clear that they are indeed the rows of the inverse matrix $A^{-1}$ because the rows of that matrix satisfy that property.
