Adjoint Operators and Inner Product Spaces My linear algebra textbook gives the definition of the Adjoint Operator and then says, 


  You should verify the following properties:
  
  
  Additivity: $(S + T)^* = S^* + T^*$
  Conjugate homogeneity: $(aT)^* = \overline{a}\,T^*$
  Adjoint of adjoint: $(T^*)^* = T$
  Identity: $I^* = I$, where $I$ is the identity operator on $V$.
  


I've stared at the pages for a couple hours now. How do you verify this? 

Here's my attempt at a proof for Adjoint of Adjoint: (T*)* = (T*v, w)* = (v, Tw)* = (Tv, w) = T

Is that correct reasoning?

BTW, this is NOT homework. Just reading for pleasure.

Thanks!
The adjoint of a transformation is defined as the unique transformation $T^{*}$ so that $$ \langle Tx, y \rangle = \langle x , T^{*}y \rangle $$ for every $x$ and $y$. 

So to prove any of your equalities above you simply need to show that the transformation you want to be the adjoint satisfies this property. Generally you can do this using the properties of an inner product. 

For example, to prove $(T^{*})^{*} = T$ you need to show that for any $x,y$ $$ \langle T^{*}x , y \rangle = \langle x , Ty \rangle.$$ 
This follows because $\langle T^*x,y\rangle= \overline{\langle y , T^*x \rangle} = \overline{\langle Ty, x \rangle} = \langle x , Ty \rangle$.
The adjoint of a transformation is defined as the unique transformation $T^\ast$ so that
$$\langle Tx,y\rangle = \langle x,T^\ast y\rangle$$
for every $x$ and $y$. So

$$\langle T^{\ast\ast} x, y\rangle = \langle x, T^\ast y\rangle = \overline{\langle T^\ast y, x\rangle} = \overline{\langle y, Tx\rangle} = \langle Tx,y\rangle.$$

And we'll get that $\langle T^{\ast\ast}x, y\rangle = \langle Tx, y\rangle$ implies that $T^{\ast\ast} = T$.