Decomposition of matrices with the Pauli matrices

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In this article, we introduce that any square matrix can be decomposed as the linear combination of the Pauli matrices.

Denote by $M_n(\mathbb{C})$ the set of all square matrices of order $n\in\mathbb{N}$ whose components value on $\mathbb{C}$ .
Furthermore, we write the 3 Pauli matrices as
\begin{align*}
X =
\begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix}, \quad
Y =
\begin{pmatrix}
0 & -i \\
i & 0
\end{pmatrix}, \quad
Z =
\begin{pmatrix}
1 & 0 \\
0 & -1
\end{pmatrix}.
\end{align*}
The following assertion holds obviously:
Proposition. $M_2(\mathbb{C})$ is a Hilbert space under the inner product; for $A,B\in M_2(\mathbb{C})$ ,
\begin{align*}
\langle A, B \rangle_{M_2(\mathbb{C})} = \frac{1}{2} \textrm{Tr}(B^*A) = \frac{1}{2}\sum_{j=1}^2 \langle Ae_j, Be_j\rangle_{\mathbb{C}^2},
\end{align*}
where $\{e_1,e_2\}$ is an orthonormal basis of $\mathbb{C}^2$ . Moreover, $\{I,X,Y,Z\}$ forms an orthonormal basis of $M_2(\mathbb{C})$ .

Consequently, we get the following:
Theorem. Let $n\in\mathbb{N}$ . Then, $\mathcal{B}=\{\bigotimes_{j=1}^n\sigma_j \mid \sigma_j=I,X,Y,Z,~j=1,2,\dots,n\}$ forms an orthonormal basis of the tensor product Hilbert space $\bigotimes^n M_2(\mathbb{C}) \simeq M_{2^n}(\mathbb{C})$ . The members $\bigotimes_{j=1}^n\sigma_j \in \mathcal{B}$ are called $n$ -qubit Pauli matrices.

Thus, for any $A\in\bigotimes^n M_2(\mathbb{C})$ , we get the decomposition $A=\sum_{\sigma\in\mathcal{B}}\langle A, \sigma\rangle\sigma$ , where the inner product represents the one of $\bigotimes^n M_2(\mathbb{C})$ .

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Decomposition of matrices with the Pauli matrices