# Can a quantum error-correcting code really correct any linear combination of correctable errors?

It appears to me that in the survey by Gottesman (around Thm 2) as well as the book by Nielsen and Chuang (Thm 10.2) it is suggested that if a QEC code corrects errors $$A$$ and $$B$$ then it also corrects any linear combination of errors (in particular by Gottesman); the sources can be found here:

Gottesman: https://arxiv.org/abs/0904.2557 Nielsen, Chuang: http://mmrc.amss.cas.cn/tlb/201702/W020170224608149940643.pdf

A simple QEC code like Shor's 9-qubit code can correct arbitrary single-qubit errors bc it can correct the Pauli errors if they occur on the same qubit, but clearly it cannot correct more than one error if they occur in the wrong places (e.g. two bitflip errors in the same block). But such an error would be a linear combination of a bitflip error X_1 hitting the first and a bitflip error X_2 hitting the second qubit in the code. What am I missing here?

Let's make it even simpler by using the $$3$$-qubit bit-flip code. That code corrects the errors $$E_{1} = XII$$,$$E_{2} = IXI$$ and $$E_{3} = IIX$$.
The 'theorem' states that this code can then also correct any error which is a linear combination $$E_{l}$$ of these errors: $$\begin{equation} E_{l} = \alpha I + \beta E_{1} + \gamma E_{2} + \delta E_{3} \end{equation}$$
Note, however, that the error you describe (a bit flip on the first and the second qubit, let's call it $$K$$) is described by the operator $$K = XXI$$, which is not a linear combination of $$E_{1}, E_{2} \& E_{3}$$. In other words: $$\begin{equation} XXI \not = XII + IXI. \end{equation}$$
If you view the collection of all operators/errors as a space, then $$E_{1} = XII$$,$$E_{2} = IXI$$ and $$E_{3} = IIX$$ form a basis for a subspace of that entire space; the theorem is that every element of that subspace is then also correctable (i.e. you only need to make sure that you can correct the elements from a basis, and the rest of the space comes for free. Any operator outside that space will be non-correctable.
$$K$$ is what we call a correlated error: the flips on the first and second qubits are correlated. These errors (also called higher-weight errors) are normally non-correctable by QECC's, and therefore they need to be circumvented at all cost (through fault-tolerance and the likes).