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!
 A: 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 $\langle T^*x,y\rangle= \overline{\langle y , T^*x \rangle} = \overline{\langle Ty, x \rangle} = \langle x , Ty \rangle$.
A: I am assuming that you are working in some Hilbert space $\mathbb{H}$. The key fact is that any continuous linear functional $f:\mathbb{H} \to \mathbb{C}$ can be represented by a unique element $\phi \in \mathbb{H}$ in the sense that $f(x) = \langle \phi, x \rangle$ for all $x \in \mathbb{H}$ (and conversely any element $\phi \in \mathbb{H}$ determines a unique continuous linear functional on $\mathbb{H}$). A little work shows that $\|f\| = \|\phi\|$. The big deal above is the uniqueness.
Now suppose $T: \mathbb{H} \to \mathbb{H}$ is a continuous linear operator and $y \in \mathbb{H}$. Then $f_y(x) = \langle y, Tx \rangle$ is a continuous linear functional, and can be represented by some $\phi_{y} \in \mathbb{H}$, ie, $f_y(x) = \langle \phi_{y}, x \rangle$. It is easy to show using uniqueness that $\phi_{\lambda y} = \lambda \phi_y$ and $\phi_{y_1+y_2} = \phi_{y_1}+\phi_{y_2}$, hence the mapping $y \mapsto \phi_y$ is linear. Furthermore, since $|f_y(x)| = |\langle \phi_{y}, x \rangle|  \leq \|x\| \|y\| \|T\|$, choosing $x=\frac{\phi_y}{\|\phi_y\|}$ (or zero, if $\phi_y=0$) shows that $\| \phi_y \| \leq \|T \| \|y\|$, hence the mapping $y \mapsto \phi_y$ is bounded, and hence continuous. Instead of writing $y \mapsto \phi_y$, we now use the more usual notation $T^* y = \phi_y$. We have shown that $T^*$ is linear, continuous, and completely defined by the requirement that $\langle T^*y, x \rangle = \langle y, Tx \rangle$ for all $x,y \in \mathbb{H}$.
All other properties follow from this requirement and properties of the inner product.
For example, to show additivity: $\langle (S+T)^*y, x \rangle = \langle y, (S+T)x \rangle = \langle y, Sx \rangle + \langle y, Tx \rangle = \langle S^*y, x \rangle + \langle T^*y, x \rangle = \langle (S^*+T^*)y, x \rangle$, hence $(S+T)^* = S^*+T^*$.
For the adjoint of the adjoint:
$\langle (T^*)^*y, x \rangle = \langle y, T^*x \rangle = \overline{\langle T^*x,y \rangle} = \overline{\langle x,Ty \rangle} = \langle Ty,x \rangle$, hence $(T^*)^* = T$.
Now try the rest.
A: For the $(T^*)^*=T$ problem:
$$\langle (T^*)^*v, w \rangle=\langle v, T^*w \rangle=\langle Tv, w \rangle$$
by definition of the adjoint.
A: 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 $T^{\ast\ast} = T$.
