Find the adjoint of the transformation given the expression of the transformation

Question

It's a very simple question but still rounds in my head, say we are given a linear transformation that is described as so: $$f(\mathbf{x})=\mathbf{x}+c \mathbf{u}(\mathbf{v},\mathbf{x})$$ being the notation $(,)$ the scalar product between two vectors, $c\in \mathbb{C}$ a constant, and $(\mathbf{u},\mathbf{v})\neq 0$. How do I find the adjoint transformation $f^* (\mathbf{x})$? The solution of the book is $f^* (\mathbf{x})=\mathbf{x}+\overline{c}\mathbf{v}(\mathbf{u},\mathbf{x})$ but how do I get there?

Answer 1

$ \newcommand\v\mathbf $I assume conjugate linearity in the first argument. By definition $f^*$ is the unique linear transformation such that $$ (f^*(\v x), \v y) = (\v x, f(\v y)) $$ for all $\v x, \v y$. So let us compute the RHS: $$\begin{aligned} (\v x, f(\v y)) &= (\v x,\,\v y + c\v u(\v v,\v y)) \\ &= (\v x, \v y) + c(\v v, \v y)(\v x, \v u) \\ &= (\v x, \v y) + (\bar c\v v(\v u, \v x),\, \v y) \\ &= (\v x + \bar c\v v(\v u, \v x),\, \v y). \end{aligned}$$ In the third line we've use conjugate linearity and the fact that $\overline{(\v x, \v u)} = (\v u, \v x)$. Now we see that we must have $$ (f^*(\v x), \v y) = (\v x + \bar c\v v(\v u, \v x),\, \v y) $$ for all $\v y$, and so it follows that $$ f^*(\v x) = \v x + \bar c\v v(\v u, \v x). $$

Answer 2

I assume that $(\cdot,\cdot)$ is a hermitian inner product. Then $(\mathbf v,\mathbf x)=\mathbf v^*\mathbf x$. In matrix form, we let $I$ denote the identity matrix. Then $f$ is represented by the matrix $$A=I+c\mathbf u\mathbf v^*$$ (check by multiplying by an arbitrary $\mathbf x$). Then we have $$A^* = I +\bar c (\mathbf u\mathbf v^*)^* = I+\bar c(\mathbf v\mathbf u^*).$$