Dual space, dual basis কী?

Example 2.2

On $K_n[X]$ with basis $(1, X, \dots, X^n)$: the dual basis is $P \mapsto \frac{P^{(k)}(0)}{k!}$ (Taylor coefficients). Another basis of the dual: the evaluations $P \mapsto P(x_i)$ at $n + 1$ distinct points — its “pre-dual” basis in $K_n[X]$ is exactly the family of Lagrange polynomials $L_i$ (Year 1 volume), since $L_i(x_j) = \delta_{ij}$. Interpolation is duality.

Example 2.4 (A dual basis of $\R^2$, fully computed)

For the basis $b_1 = (1, 1)$, $b_2 = (1, -1)$ of $\R^2$: the dual basis $(b_1^*, b_2^*)$ must satisfy $b_i^*(b_j) = \delta_{ij}$. Writing $b_1^*(x, y) = \alpha x + \beta y$, the conditions $\alpha + \beta = 1$ and $\alpha - \beta = 0$ give

Sanity checks: $b_1^*$ is not $e_1^* + e_2^*$ evaluated naively — the dual basis depends on the whole basis, not on each vector separately (replacing $b_2$ by $(0, 1)$ changes $b_1^*$ into $x \mapsto x$). And the expansion formula works: $(x, y) = \frac{x+y}2\,b_1 + \frac{x-y}2\,b_2$, the even/odd decomposition of a pair — dual bases are coordinate extractors, and this one extracts symmetric and antisymmetric parts.

Example 2.9 (The transpose, entry by entry)

Let $u \colon \R^2 \to \R^3$ have matrix $A = \left(\begin{smallmatrix} 1 & 2\\ 0 & 1\\ 3 & 0\end{smallmatrix}\right)$ in the canonical bases. For $\psi = b_1f_1^* + b_2f_2^* + b_3f_3^* \in (\R^3)^*$, compute $u^{\mathsf T}(\psi) = \psi \circ u$ on the basis of $\R^2$:

So $u^{\mathsf T}(\psi) = (b_1 + 3b_3)\,e_1^* + (2b_1 + b_2)\,e_2^*$, and in the dual bases the matrix of $u^{\mathsf T}$ is

the abstract transpose is the flipped matrix, with no computation left to believe on faith. Note the mechanism: the $j$-th column of $A$ became the $j$-th row of the new matrix because $\psi \circ u$ reads $u$’s outputs through $\psi$’s coefficients.