1

If an adjoint identity reads

$$\boldsymbol{v} \cdot (\mathsf{C}\boldsymbol{x}) =(\mathsf{C}^+\boldsymbol{v})\cdot\boldsymbol{x},$$

where the adjoint vector $\boldsymbol{v}$ and the original vector $\boldsymbol{x}$ are column vectors in $\mathbb{C}^{n\times1}$, both can include complex numbers. And $\mathsf{C}$ is a unsymmetric complex-valued matrix, which can be non-square in general.

My question:

What should $\mathsf{C}^+$ be actually, i.e. transpose or conjugate transpose of $\mathsf{C}$? Thank you for any suggestion.

jsxs
  • 511
  • 2
    It should be conjugate transpose. You can check in the case $n=0$, in which case $v, C, x$ are simply complex numbers and the dot product of $u$ and $w$ is $u\cdot w = u \overline{w}$. Then $$v\cdot (Cx)= v \overline{Cx}= \overline{C} v\overline{x}= \overline{C} v\cdot x.$$ So conjugation is required. –  Sep 07 '18 at 10:12
  • Yes, I did mean $n=1$. I misread your $\mathbb{C}^{n\times 1}$ as $\mathbb{C}^{n+1}$. –  Sep 07 '18 at 12:47
  • That is the definition of the inner product for complex scalars. –  Sep 07 '18 at 16:08
  • Look at the answer to this question https://math.stackexchange.com/questions/2459814/what-is-the-dot-product-of-complex-vectors –  Sep 07 '18 at 16:10
  • @bangs do u mean complex vector? – jsxs Sep 07 '18 at 16:13
  • My comment applies to both (in the comment above with n=1, it is complex scalars). –  Sep 07 '18 at 16:14

0 Answers0