In linear algebra, why is the dual basis of a basis the *rows* of the matrix $A^{-1}$?

Question

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.

Answer 1

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.

Answer 2

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.