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!

  • $\begingroup$ That's typically how I interpret it, but I've seen others use the words "transpose of a linear transformation" and "adjoint of a matrix" (meaning transpose, not adjugate) before. It seems the words are somewhat interchangeable. $\endgroup$ – user137731 Oct 30 '15 at 15:16

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 $A^*$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.