Pulling measurements backward
Take a linear transformation T: V -> W. Now grab a measurement g on the target, g in W*. You can build a measurement on the source by first applying T, then g: the composite g∘T is a functional on V. The rule g -> g∘T is the transpose of the linear map T, written T*: W* -> V*. Crucially it points the opposite way from T.
Why this is THE transpose
Choose bases for V and W, and dual bases for V* and W*. Write T as a matrix A. Then the matrix of T* in the dual bases is exactly A^T — the ordinary matrix transpose you learned in Vol I. So the mysterious 'flip the matrix over its diagonal' operation is really 'the same map, acting on measurements instead of vectors'. The coordinate-free definition explains what the bookkeeping always meant.
Check (T*g)(v) = g(Tv) in coordinates.
T: R^2 -> R^2, A = [ 2 1 ; 0 3 ].
Take g = row [1 4] in W*.
Left: T*g = g composed with T = (row g)*(matrix A)
= [1 4] * [2 1; 0 3] = [2 13] (a row in V*)
Then (T*g)(v) for v=[5;1]: [2 13]*[5;1] = 10+13 = 23.
Right: Tv = A*[5;1] = [11; 3]. g(Tv) = [1 4]*[11;3] = 11+12 = 23.
Equal ✓. And note: T*g = [1 4]*A, i.e. the matrix of T* acting
on rows is A^T acting on columns.The payoff: row rank = column rank
The transpose links kernels and annihilators by the kernel-image duality: ker(T*) = (im T)^0 and im(T*) = (ker T)^0. Counting dimensions with the annihilator law gives rank of the transpose = rank(T*) = rank(T). Since rank(T) is column rank and rank(T*) corresponds to row rank, this proves the Vol I fact you took on faith: a matrix's row rank equals its column rank.