Showing that a set of errors is correctable (Knill-Laflamme conditions)? I am confused about how to apply the Knill-Laflamme Quantum Error-correction conditions, which are the following: 

A code $C \leq H$ is correctable for $\mathcal{E} = \sum_{i=1}^{n}E_i \rho E_i^*$ if and only if $P_cE_i^*E_jP_c = \lambda_{ij}P_c$ for some scalar matrix $[\lambda_{ij}] \in M_n(\mathbb{C})$.

How does one apply this to an actual example, like the three qubit bit flip channel? 
I would like to show that $C = \text{span}\{|000>, |111>\}$ is correctable for $\mathcal{E}(\rho) = \frac{1}{4}(\rho + X_1\rho X_1^* + X_2 \rho X_2^* + X_3 \rho X_3^*)$. However, I cannot figure out how to put the theory into practice. How do I actually do this computationally?
 A: Assuming you're using notation in accordance with what's used in chapter 10 of Nielsen and Chuang, then you can see immediately from the form of your noise $\mathcal{E}(\rho)$ that your corresponding errors that constitute $\mathcal{E}$, being the set $\{E_i\}$, are in fact the bit-flip operators $\{X_i\}$ for $i \in \{1,2,3\}$
With that, you need only verify that $P_cE^*_iE_jP_c = \lambda_{ij}P_c$ for your projector. 
A: To get a better understanding of how the theorem applies, it is useful to look at the general setting that is used in error correction theory. First of all, one has a Hilbert space $H$ of which the code $C$ is a subspace. Within the space of linear operators from $H \rightarrow H$ with the inner product $(A,B)=Tr(A^*B)$ one has the group called the error group 
$\varepsilon =\left \{ U_{a}V_{b}: a,b \in S \right \}$,
where $S$ are the binary string of length n and the operators $U_{a}$ and $V_{b}$ correspond (loosely speaking) to the operators that flip bits ($U_{a}$) and shift phase ($V_{b}$). Any group of operators generating errors is a subgroup A of $\varepsilon$. In the case you describe the setting is:
$n=3$
$A = \left \{ U_{a}:  w\left ( a \right ) = 1 \right \}$,
where $w\left ( a \right )$ represents the number of positions at which the bit string $\neq 0$. So since we have $n = 3$ our subgroup becomes
$\left \{ X_{1}, X_{2},X_{3} \right \}$ and our code is
$C=\left \{  \left | 000 \right \rangle, \left | 111 \right \rangle \right \}$. The Knill-Laflamme Error-correction conditions come in 2 fomrulations, one that features the basis of the code $C$ and the other uses the projection operator onto $C$ which I will call $P$. In this case both can be used to check that $C$ is indeed a $A$-error correcting code. Let's see how the projection variant would be verified. As linear operator is defined by it's action on the basis, we only need to check it's action on all the basis vectors of $H$. 
For $ \left | \phi  \right \rangle \in C^{\perp }$ the defintion of the projectors gives $0=0$. Thus the condition is satisfied.
Now assume that $ \left | \phi  \right \rangle \in C$ we have the calculations:
$ PX_{i}X_{j} \left | 000  \right \rangle = \delta_{i,j}\left | 000  \right \rangle$,
$ PX_{i}X_{j} \left | 111  \right \rangle = \delta_{i,j}\left | 111  \right \rangle$.
This calculation shows that $\lambda _{i,j} = \delta_{i,j}$. The key thing is that this is independent of the basis vectors of $C$. The condition would fail if for instance would find out that your function $\lambda _{i,j}$ is dependent of the  basis element that you put into your calculation.
