Why aren't transformations caused by measurements unitary? It is said, that when measured, a quantum system undergoes "wave function collapse", which is a non-unitary transformation.
Why? 
The wave function is
$\Psi = \alpha \left|0\right\rangle + \beta \left|1\right\rangle$
where
$\left|\alpha\right|^2 + \left|\beta\right|^2 = 1$
The probabilities sum after measurement is still 1, for example, if system collapsed to $\left|0\right\rangle$, then
$\left|1\right|^2 + \left|0\right|^2 = 1$ 
For example, if function was
$\Psi = \frac{1}{\sqrt{2}} \left|0\right\rangle + \frac{1}{\sqrt{2}} \left|1\right\rangle$
the transformation was
$ \left[
  \begin{array}{ c c }
     \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
     0 & 0
  \end{array} \right]
$
Isn't this transformation unitary?
 A: No.
As long as your state is $|\Psi \rangle = \alpha \left|0\right\rangle + \beta \left|1\right\rangle$, then as you said $\alpha$ and $\beta$ need to satisfy $\left|\alpha\right|^2 + \left|\beta\right|^2 = 1$, so say $\alpha = \beta = 1/\sqrt 2$.
If you perform a measurement and find that the system in the $\left|0\right\rangle$ state, then the new wavefunction will be $\Psi =\left|0\right\rangle$. You can write it as $\Psi = \alpha \left|0\right\rangle$ but because of normalisiation $|\alpha|^2$ needs to be 1, so $\alpha$ must be either 1 or a pure phase factor.
You had to change the normalisation by hand (changing $\alpha$ from $1/\sqrt 2$ to $1$). A unitary transformation on $|\Psi \rangle$ would affect only the kets and not the constants. The time evolution of any wavefunction is governed by the Schrodinger equation which, when solved, is effectively a unitary transformation -- unitary transformations leave the norm unchanged. The time evolution due to performing measurements, however, is something entirely different and does not follow the formalism of the Schrodinger equation.
A unitary transformation leaves the norm unchanged, since the norm of  $U|\Psi \rangle$ is $\langle \Psi |U^{\dagger}U|\Psi \rangle = \langle \Psi | \Psi \rangle$ if $U$ is unitary. In your specific case this is true, but only because you had to manually change the normalisation of the post-measurement wavefunction. If QM included measurement, then there should be a deterministic way of computing how $\alpha$ or $\beta$ would change. But it doesn't, so you need to re-normalise it.
A: Answering the question in the title: a measurement process is intrinsically non-unitary.
One way to see this is to realise that the unitarity of a process is equivalent to its being reversible.
A measurement process is intrinsically non-reversible, as some information gets lost. For example, measuring $(|0\rangle+|1\rangle)/\sqrt2$ in the computational basis, you can get either $|0\rangle$ or $|1\rangle$. The same outputs can be obtained measuring a different state, e.g. $(|0\rangle-|1\rangle)/\sqrt2$. This means that, given a measurement result (say $|0\rangle$), there is no way to know from what state it came from. Some information is lost in the process. It follows that the process cannot be described by a unitary matrix.
A: I assume that you are referring to the outcome of any observable $O$ acting on a state $\psi$, since the act of measuring something is interpreted as "averaging" such operators on some state. General observables are self-adjoint operators which need not be unitary. Perhaps the simplest example of an observable is a projection, i.e. an operator $P$ with the property that $P^*P = P$ (idempotent and self-adjoint). Suppose that, in your case, $P = |0\rangle\langle0|$. The outcome of a measurement of $P$ on your state $\Psi$, when repeated $N$ times, is $|\alpha^2|N$ times YES (and hence $(1-|\alpha|^2)N$ times no. Moreover the result of $P\Psi$ is $\alpha|0\rangle$, which isn't a normalised vector, simply because $P$ is not a unitary.
A: Measurement is not unitary, because it does not implement a unitary. It selects a unitary. 
We can certainly say that a preparation $|\psi_i\rangle$ and (renormalized) 
 measurement outcome $|\psi_f\rangle$ are related by a unitary transformation $$|\psi_f\rangle = U_{fi} |\psi_i\rangle,$$ where $U_{fi} \in U(2)$. However, we can only say this once a measurement has occurred. 
Before measurement, the possible outcomes are $|{+}\psi_f\rangle$ and $|{-}\psi_f\rangle$, so all we can say is that $|\psi_i\rangle$ and the not-yet-measured outcome $|\psi_f\rangle$ are related by either $U_{fi}^+$ or $U_{fi}^-$, where 
$$|{+}\psi_f\rangle = U_{fi}^+ |\psi_i\rangle \hspace{1em} \text{and} \hspace{1em} |{-}\psi_f\rangle = U_{fi}^- |\psi_i \rangle.$$
Formally, we can construct an equivalence class of unitaries $U_{fi}^+ \sim U_{fi}^-$ over the possible outcomes of the measurement process, so that $[U_{fi}^+] = [U_{fi}^-] \in U(2)/\mathbb Z_2$ represents the part of the relation between the initial and final state that is fixed by our experimental setup.
The point is that the measurement process implements a transition
$$
U(2)/\mathbb Z_2 \to U(2) \\
[U_{fi}^\pm] \mapsto U_{fi}^\pm
$$
that breaks the $\mathbb Z_2$ symmetry and selects a particular representative of $[U_{fi}^\pm]$. In other words, the unitary relation $U_{fi}^+$ or $U_{fi}^-$ between an initial state and measurement outcome is not a description of the measurement process. It is the result of the measurement process. 
