Is it true you use adjoint for linear transforms while you use transpose for matrix representation of linear transforms? In one lectures on linear algebra, my professor wrote on the first day:
Let $L$ be in a linear map $L: X \to Y$, then $Im(L) \oplus Ker(L^*) = Y$
On the second day, he wrote: $L: X \to Y$, then $Im(L) \oplus Ker(L^T) = Y$
I am a little bit confused by the mixture of adjoint and transpose. I suspect the transpose is used by assuming that $L$ is the matrix representation of the transform, whereas adjoint is just a general symbol to denote the adjoint of a linear transform and does not induce a transpose.
Can someone clarify this concept for me? THanks!
 A: Suppose you have a linear transformation $L : X \to Y$ where $X$ and $Y$ are vector spaces endows with inner products $\langle \cdot,\cdot \rangle_X$ and $\langle \cdot,\cdot \rangle_Y$, respectively. Then the adjoint of $L$ is a linear map $L^* : Y \to X$ such that
$$
\langle x, L^*(y) \rangle_X = \langle L(x), y \rangle_Y
\quad \forall x \in X, \forall y \in Y.
$$
Now, if $X$ and $Y$ are real vector spaces and $L$ is represented by a matrix $A$ with respect to a choice of orthonormal bases, then $L^*$ is represented by the transpose $A^T$ with respect to the same bases.
On the other hand, this may not hold over other fields. For example if $X$ and $Y$ are complex vector spaces, then $L^*$ is represented by the Hermitian transpose $A^{\dagger} = \overline{A^T}$.
In conclusion, when we talk about the adjoint of a matrix $A$, which represents a linear map $L$ with respect to some choice of orthonormal bases, we mean the matrix representing the adjoint map $L^*$ with respect to the same bases, usually denoted by $A^*$.
