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Suppose we have an $n$-qubit system. Let $Y_i$ and $Z_j$ denote the Pauli-Y and Pauli-Z operators acting on the $i$th and $j$th qubits, respectively.
Suppose we have a finite set of tuples $E = \{(i,j)\}_{i\neq j}$. We can think of $E$ being an edge set of an undirected graph $G$ that has no loops and no double edges.
My question is, how do I formally argue that the following is true: $$\sum_{(i,j) \in E} Y_i Z_j \neq 0,$$ where $Y_iZ_j$ is the usual matrix product of $Y_i$ and $Z_j$.
The set of $n$-qubit Pauli matrices $\{P_i\}_{i=1,\dots,4^n}$ forms a (orthogonal) basis for the vector space of complex matrices.
In particular, the set of Pauli matrices is linearly independent. Hence, the equation $$ \sum_{i=1}^{4^n} x_i P_i = 0, $$ does only have a trivial solution $x_i=0$. This implies that the sum over any subset cannot be the zero matrix (this would correspond to a solution where at least one $x_i =1$).
Definition: A $n$-qubit Pauli matrix $P$ is any tensor product $$ P = \sigma_1\otimes\dots\otimes\sigma_n, $$ where $\sigma_j \in \{ \mathrm I, X, Y, Z\}$.
Orthogonality: Note that $X,Y,Z$ are traceless. Moreover, we have $\sigma^2 = I$ for any $\sigma\in \{ \mathrm I, X, Y, Z\}$ and the product $\sigma\sigma'$ is, up to a phase, either $X$, $Y$, or $Z$ if $\sigma,\sigma'\in \{ \mathrm I, X, Y, Z\}$ and $\sigma\neq\sigma'$. Hence $$ \mathrm{tr}( \sigma\sigma' ) = \begin{cases} 2 & \text{if } \sigma=\sigma' \\ 0 & \text{else} \end{cases}. $$ Hence, the 1-qubit Pauli matrices are orthogonal w.r.t. the Hilbert-Schmidt (trace) inner product. This readily generalizes to the $n$-qubit cases which shows that the $4^n$ Pauli matrices form an orthogonal basis for the $4^n$-dimensional vector space of complex matrices $\mathbb{C}^{2^n \times 2^n}$.
I realized that this can be checked rather easily. We compute $$\left( \sum_{(i,j) \in E} Y_i Z_j \right) |0\rangle^{\otimes n}$$ and see that it is the sum of a subset of standard basis vectors: $$\left( \sum_{(i,j) \in E} Y_i Z_j \right) |0\rangle^{\otimes n} = \sum_{(i,j) \in E} \mathbb{i} \bigotimes_{k=1}^n |\delta_{k,i}\rangle \neq 0,$$ where $\delta_{k,i} = 1$ for $k=i$ and $\delta_{k,i}=0$ for $k \neq i$. The expression $\bigotimes_{k=1}^n |\delta_{k,i}\rangle$ is a ket of a binary string whose entries are all zero except the $i$th entry.
