1. Jul 15, 2009

### mnb96

Hi,
in the text I am reading I found the following implicit definition of an adjoint transformation:

$$\overline{f}( \textbf{a}) \ast \textbf{b} = \textbf{a} \ast f(\textbf{b})$$

then it is said that $$\overline{f^{-1}} = (\overline{f})^{-1}$$. Basically the inverse and ajoint are interchangeable, and this property is (supposed to be) easily shown from the definition above.
Unfortunately I have problems figuring out how to prove it. Any ideas?

2. Jul 15, 2009

### jimmypoopins

it's been a really long time since i did anything math related, but i think if you * the inverse of everything to both sides (i.e. step 1, * (f^bar(a))^-1 to both sides, step 2 * b^1 to both sides, step 3 * a^1 to both sides, step 4 * (f(b))^-1 to both sides) you should end up with

$$\textbf{a}^{-1} \ast f(\textbf{b})^{-1} = \textbf{b}^{-1} \ast \overline{f}( \textbf{a})^{-1}$$

does that give you what you want? sorry, it's been a long time since i've done this type of stuff. i think this is probably wrong, but it may give you a step in the right direction.

3. Jul 15, 2009

### mnb96

When you say, step1 * f^bar(a))^-1 to both sides do you mean the following?

$$\overline{f}\\^{-1}(\textbf{a}) \ast \overline{f}( \textbf{a}) \ast \textbf{b} = \overline{f}\\^{-1}(\textbf{a}) \ast \textbf{a} \ast f(\textbf{b})$$

4. Jul 15, 2009

### jimmypoopins

actually i meant this:

$$(\overline{f}\\(\textbf{a}))^{-1} \ast \overline{f}( \textbf{a}) \ast \textbf{b} = (\overline{f}\\(\textbf{a}))^{-1} \ast \textbf{a} \ast f(\textbf{b})$$

i'm not that great with latex, or i would have tried to explain it better; my apologies. i think the equation in my post is actually incorrect now that i look at it, but if you follow the method i described i think it gets you somewhere.

after you simplify you get

$$\textbf{b} = (\overline{f}\\(\textbf{a}))^{-1} \ast \textbf{a} \ast f(\textbf{b})$$

and then you continue to * inverses on both sides (i don't want to say multiply because * is an operator not always multiplication... is there a better word for that?)

5. Jul 16, 2009

### mnb96

I found a solution when the operation * is cancellative (which I can assume to be my case):

$$\overline{f^{-1}}(a) * f(b) = a * f^{-1}(f(b)) = a * b = \overline{f}(\overline{f}^{-1}(a)) * b = \overline{f}^{-1}(a) * f(b)$$

However if you check from wikipedia http://en.wikipedia.org/wiki/Hermitian_adjoint#Properties this seems to be a property of linear operators, but I wonder how you can prove it without assuming cancellativity ($$y*x = z*x \\ \Rightarrow \\ y=z$$)