What is Transpose map?

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.

Example 2.11 (Rank read on both sides)

Let

Column rank: the third row is the sum of the first two, so $\operatorname{rk} A \leq 2$; columns $1$ and $2$ are free: $\operatorname{rk} A = 2$. The transpose’s kernel: solving $A^{\mathsf T}y = 0$ gives $y \in \R\,(1, 1, -1)$, so $\ker A^{\mathsf T}$ has dimension $1 = 3 - 2$: exactly $(\operatorname{im} A)^\circ$ under the identification of $(\R^3)^*$ with row vectors, as Proposition 2.10 asserts — the single relation “row$_3$ = row$_1$ + row$_2$” is the annihilator of the column space. Row rank ($2$ free rows) and column rank agree not by accident but because both equal $\operatorname{rk} A = \operatorname{rk} A^{\mathsf T}$.