Help on the notation of the 2-bit operation: $\frac 12{(I+Z_1Z_0)}$ This was adapted from Quantum Mechanichs for Computer Scientists.
On this equation, we have $I$ representing the identity operator, with $Z$ being defined for 1 bit as:

$$ Z|0⟩ = |0⟩, \quad Z|1⟩ = -|1⟩\,.$$

The operation $\frac 12{(I+Z_1Z_0)}$ acts as the identity for the two-bit states $|00⟩$ and $|11⟩$  while returning $0$ for the states $|01\rangle$ or $|10\rangle$.

I'm assuming the operator addition is the linear algebra definition, but I can't understand the $\frac 12$ fraction and its effect on the operation.

Could you point me to what concept I am missing, so I can derive the resulting states of this operation?
The operation $I$ on two bits can be written as the identity matrix,
$$
I = 
\begin{pmatrix} 1 &amp; 0 &amp; 0 &amp;0 \\ 0 &amp; 1 &amp; 0 &amp;0 \\ 0 &amp; 0 &amp; 1 &amp;0 \\ 0 &amp; 0 &amp; 0 &amp;1
\end{pmatrix}\,.
$$

The operation $Z_0Z_1$ that performs $Z$ on both qubits can be written as 
$$
Z_0Z_1 = 
\begin{pmatrix} 1 &amp; 0 &amp; 0 &amp;0 \\ 0 &amp; -1 &amp; 0 &amp;0 \\ 0 &amp; 0 &amp; -1 &amp;0 \\ 0 &amp; 0 &amp; 0 &amp; 1
\end{pmatrix}\,.
$$

Now, the operation that you ask about, $\frac12(I+Z_0Z_1)$ is simply the operation described by 

$$
\frac12(I+Z_0Z_1) = 
\frac12 \begin{pmatrix} 1 &amp; 0 &amp; 0 &amp;0 \\ 0 &amp; 1 &amp; 0 &amp;0 \\ 0 &amp; 0 &amp; 1 &amp;0 \\ 0 &amp; 0 &amp; 0 &amp;1
\end{pmatrix} +
\frac12 \begin{pmatrix} 1 &amp; 0 &amp; 0 &amp;0 \\ 0 &amp; -1 &amp; 0 &amp;0 \\ 0 &amp; 0 &amp; -1 &amp;0 \\ 0 &amp; 0 &amp; 0 &amp; 1
\end{pmatrix}
= 
\begin{pmatrix} 1 &amp; 0 &amp; 0 &amp;0 \\ 0 &amp; 0 &amp; 0 &amp;0 \\ 0 &amp; 0 &amp; 0 &amp;0 \\ 0 &amp; 0 &amp; 0 &amp;1
\end{pmatrix}\,.
$$
This indeed performs as you describe. If the $1/2$ wasn't there, we would get that $|0\rangle$ becomes $2|0\rangle$. Note that his operator is not unitary (it returns "$0$" for $|01\rangle$ and $|10\rangle$, which means the qubit "disappears", this shouldn't be allowed by quantum mechanics.)