There seems to be two kinds of transposes of a linear mapping:
If $f: V→W$ is a linear map between vector spaces $V$ and $W$ with nondegenerate bilinear forms, we define the transpose of $f$ to be the linear map $^tf : W→V$, determined by $B_V(v,{}^tf(w))=B_W(f(v),w) \quad \forall\ v \in V, w \in W.$ Here, $B_V$ and $B_W$ are the bilinear forms on $V$ and $W$ respectively. The matrix of the transpose of a map is the transposed matrix only if the bases are orthonormal with respect to their bilinear forms.
I was wondering what "the bases are orthonormal with respect to their bilinear forms" means specifically?
Does the transpose depends on the choice of the nondegenerate bilinear forms $B_W$ and $B_V$, given that there can be many choices?
If $f : V → W$ is a linear map, then the transpose (or dual) $f^* : W^* → V^*$ is defined by $f^*(\varphi) = \varphi \circ f \, $ for every $φ ∈ W^*$. If the linear map $f$ is represented by the matrix $A$ with respect to two bases of $V$ and $W$, then $f^*$ is represented by the transpose matrix $A^T$ with respect to the dual bases of $W^*$ and $V^*$.
Is the this a different definition from the previous one, or are they essentially the same? Can they be directly related in some way?
In Fundamental theorem of linear algebra, for each matrix $A \in \mathbf{R}^{m \times n}$,
$\mathrm{ker}(A) = (\mathrm{im}(A^T))^\perp$, that is, the nullspace is the orthogonal complement of the row space
$\mathrm{ker}(A^T) = (\mathrm{im}(A))^\perp$, that is, the left nullspace is the orthogonal complement of the column space.
Can the theorem be rephrased in terms of a linear mapping instead of a matrix? Which of the above two definitions of transpose of a linear mapping is used in the rephrase to replace $A^T$?
Thanks and regards!