Knill-Laflamme condition derivation in Nielsen&Chuang: issue to understand a part of the proof

Question

I have some trouble to understand the proof in Nielsen&Chuang about Knill-Laflamme conditions.

The conditions:

Let $C$ be a quantum code and $P$ the projector onto $C$. Suppose $\mathcal{E}$ is a quantum operatiion with operation elements $\{E_i\}$. A necessary and sufficient condition for the existence of an error-correction operation $\mathcal{R}$ correcting $\mathcal{E}$ on $C$ is that:

$$P E_i^{\dagger} E_j P = \alpha_{ij} P$$

For some Hermitian matrix $\alpha$ of complex numbers.

To show the sufficient condition, he diagonalizes the matrix $\alpha$. Doing this, we end up with a simpler condition:

$$ P F_k^{\dagger} F_l P = d_{kl} P $$

Where $d_{kl}$ is the matrix element of a diagonal matrix.

The polar decomposition tells us that there exist a unitary $U_k$ such that:

$$F_k P = U_k \sqrt{P F_k^{\dagger} F_k P} = \sqrt{d_{kk}} U_k \sqrt{P} = \sqrt{d_{kk}} U_k P \tag{1}$$

My question

At this point, he says that the effect of $F_k$ is therefore to rotate the coding subspace into the subspace defined by the projector:

$$P_k \equiv U_k P U_k^{\dagger}$$

I am not sure to understand precisely this statement. I agree that $P_k F_k P = F_k P$ such that $P_k$ stabilizes the space $F_k C$. But how to be sure that $P_k$ is exactly a good projector. What I mean is that veryfing $P_k F_k P = F_k P$ is for sure not enough as $P_k=I$ would also do the job for instance.

Answer 1

I found the answer. It is just a basic linear algebra property to show but I take the same notations as in the question for clarity.

Let $P$ be a projector on $C$

As shown below in this post, a projector on the space $U_k C$ can always be found as $P_k = U_k P U_k^{\dagger}$.

From this property, we see that as $F_k P \propto U_k P$, a projector onto $Im(F_k P)$ is $P_k$ which is what is claimed by the book.

Proof

We just need to show that $Im(P_k)=U_k C$.

I first show $Im(P_k) \subset U_k C$:

Let $|y \rangle \in Im(P_k)$. We thus have: $|y\rangle=U_k P U_k^{\dagger} |x\rangle$ for some $|x \rangle$

And as $P U_k^{\dagger} |x\rangle \in C$ (because $P$ is a projector on $C$), we have $Im(P_k) \subset U_k C$

Now, I need to show $U_k C \subset Im(P_k)$:

Let $|y \rangle \in U_k C$:

For some $|x \rangle$, we have: $|y\rangle = U_k P |x \rangle$

Also, for some $|z \rangle$, we have $|x \rangle = U_k^{\dagger} |z \rangle$, thus in the end: $|y\rangle = U_k P U_k^{\dagger} |z \rangle \in Im(P_k)$

Conclusion: $Im(P_k)=U_k C$