= Stabilizer subspace
The <stabilizer subspace> associated with a <stabilizer group> $\mathcal S$ on $n$ <qubits> is its simultaneous $+1$ <eigenspace>:
$$
V_{\mathcal S}=\{|\psi\rangle:g|\psi\rangle=|\psi\rangle\text{ for every }g\in\mathcal S\}.
$$
Assume that $\mathcal S$ is a commuting subgroup of the <Pauli group> and excludes $-I$. Its elements are then <Hermitian operators> with <eigenvalues> $\pm1$. If $r$ generators $g_1,\ldots,g_r$ are independent modulo scalar phases, then
$$
\dim V_{\mathcal S}=2^{n-r}.
$$
The exclusion of $-I$ is essential: the equation $-I|\psi\rangle=|\psi\rangle$ has only the zero solution.
A direct proof uses the <orthogonal projector>
$$
P_{\mathcal S}=\prod_{j=1}^r\frac{I+g_j}{2}=\frac1{2^r}\sum_{g\in\mathcal S}g.
$$
All nonidentity <Pauli operators> have zero <trace>, so $\operatorname{Tr}P_{\mathcal S}=2^{n-r}$. The <trace> of an <orthogonal projector> equals the dimension of its image. <Quantum Error Correction by John Preskill>[Section 7.9.1, equations 7.117–7.120, printed pp. 35–36; projector factorization also equation 7.160]{ref}
For example, on two <qubits>, the single generator $Z\otimes Z$ fixes the two-dimensional space spanned by $|00\rangle$ and $|11\rangle$. Adding $X\otimes X$ leaves the one-dimensional space spanned by the <Bell state> $(|00\rangle+|11\rangle)/\sqrt2$. This dimension count determines how many logical <qubits> a <stabilizer code> can encode.
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