Is a linear operator applied to itself still linear? I'm not sure if this is a really obvious result, but I'd like to make sure I've not got it wrong. 
Here is my attempt:
Take the vector space V to be a (complex) Hilbert Space. For $x,y$ in $V$ and $a,b, $in the complex numbers, if we take $A$ to be a linear operator then $A(ax+by) = aA(x)+bA(y)$.
Consider $A^{2}(ax+by)
= A(A(ax+by))
=A(aA(x)+bA(y))
aA^{2}(x)+bA^{2}(y)$
Assuming I am allowed to do this, this leads me to think that a linear operator applied to itself IS linear, and that you could do it several times and it would still be linear?
 A: The linear operator is not applied to itself but rather composed with itself. And then yes, it is true and obvious result with the proof you give. Actually, even any composition of linear maps is still linear. This is really a typical situation as morphisms of any category are closed under composition – composition of continuous maps is continuous, composition of Lipschitz maps is Lipschitz, composition of group homomorphisms is a group homomorphism, etc.
A: Sure, and to see why just rewrite that line in your third paragraph as
\begin{align}
A^2(ax+by)&=A(A(ax+by))\\
&=A(a(Ax)+b(Ay))\\
&=A(av+bw) \qquad \text{with }v:=Ax,\ w:=Ay\\
&=a(Av)+b(Aw)\\
&=ap+bq \qquad \text{with }p:=Av,\ q:=Aw
\end{align}
and compare that to
\begin{align}
aA^2x+bA^2y&=a(A(Ax))+b(A(Ay))\\
&=a(Av)+b(Aw)  \qquad \text{with }v:=Ax,\ w:=Ay\\
&=ap+bq  \qquad \text{with }p:=Av,\ q:=Aw
\end{align}
Thus, $A^2=AA$ is linear when $A$ is linear.
Extending this argument you can see that if $A$ is linear then $p(A)$ is linear where $p$ is any polynomial. Here we just happen to be investigating $p(A)=A^2$.
