# Expansion of multi-qubit density matrix in the Pauli matrix basis

The single qubit density matrix can be expanded as $$\rho=\frac{tr(\rho)I+tr(X\rho)X+tr(Y\rho)Y+tr(Z\rho)Z}{2}$$ which can be shown as,

$$\rho$$ is a positive operator with $$tr(\rho)=1$$, ie., $$\rho=\begin{bmatrix}u&v+iw\\v-iw&1-u\end{bmatrix}$$ and $$X=\begin{bmatrix}1&0\\0&1\end{bmatrix},Y=\begin{bmatrix}0&-i\\i&0\end{bmatrix},Z=\begin{bmatrix}1&0\\0&-1\end{bmatrix}$$ are the Pauli matrices. $$X\rho=\begin{bmatrix}v-iw&1-u\\u&v+iw\end{bmatrix}\implies tr(X\rho)=2v\\ Y\rho=\begin{bmatrix}-iv-w&-i-iu\\iu&iv-w\end{bmatrix}\implies tr(Y\rho)=-2w\\ Z\rho=\begin{bmatrix}u&v+iw\\-v+iw&-1+u\end{bmatrix}\implies tr(Z\rho)=2u-1\\ 2\rho=\begin{bmatrix}2u&2v+i2w\\2v-i2w&2-2u\end{bmatrix}=\begin{bmatrix}1+(2u-1)&2v+i2w\\2v-i2w&1-(2u-1)\end{bmatrix}\\ =I+2vX+(-2w)Y+(2u-1)Z\\ =tr(\rho)I+tr(X\rho)X+tr(Y\rho)Y+tr(Z\rho)Z\\ \implies \boxed{\rho=\frac{tr(\rho)I+tr(X\rho)X+tr(Y\rho)Y+tr(Z\rho)Z}{2}}$$

How do I prove that the density matrix for $$n$$ qubits can be written as, $$\rho=\sum_{\vec{v}}\frac{tr(\sigma_{v_1}\otimes\sigma_{v_2}\otimes\cdots\otimes\sigma_{v_n}\rho)\sigma_{v_1}\otimes\sigma_{v_2}\otimes\cdots\otimes\sigma_{v_n}}{2^n}$$

My Attempt

Lets consider a $$2$$ qubit case,

We need to prove $$\rho=\sum_{i,j} \frac{tr(\sigma_i\otimes\sigma_j\rho)\sigma_i\otimes\sigma_j}{4}$$

Lemma: Let $$V$$ and $$W$$ be normed spaces. If $$\{v_i\}\in V$$ and $$\{w_j\}\in W$$ are linearly independent in $$V$$ and $$W$$ respectively, then $$\{v_i\otimes w_j\}$$ is linearly independent in the algebraic tensor product $$V\otimes W.$$

Since we can define $$\mathbb{C}^{16}$$ an isometric to the matrix space $$M_{4\times 4}(\mathbb{C})$$, lemma can be used to prove $$\sigma_i\otimes\sigma_j$$ form an orthogonal basis for the $$2$$ qubit density matrix, i.e., $$\rho=\sum_{i,j} c_{i,j}\sigma_i\otimes\sigma_j$$

So, how do I prove that the coefficients $$c_{i,j}=tr(\sigma_i\otimes\sigma_j\rho)$$ ?

"Trace out" $$\sigma_i \otimes \sigma_j$$
$$\rho = \sum c_{ij}\sigma_i \otimes \sigma_j$$
$$\rho \ \sigma_m \otimes \sigma_n = \sum c_{ij} \ \sigma_i \sigma_m \otimes \sigma_j \sigma_n$$
\begin{align} \mathrm{Tr \{\rho \ \sigma_m \otimes \sigma_n\}} &=\mathrm{Tr}(\sum c_{ij} \ \sigma_i \sigma_m \otimes \sigma_j \sigma_n)\\ &=\sum \mathrm{Tr}(c_{ij} \ \sigma_i \sigma_m \otimes \sigma_j \sigma_n)\\ &=\sum c_{ij} \ \mathrm{Tr}(\sigma_i \sigma_m \otimes \sigma_j \sigma_n)\\ &=\sum c_{ij} \ \mathrm{Tr}(\sigma_i \sigma_m)\mathrm{Tr}(\sigma_j \sigma_n)\\ &= 4\sum c_{ij} \delta_{im} \delta_{jn}\\ &=4c_{mn} \end{align}
$$c_{mn} = \frac{\mathrm{Tr \{\rho \ \sigma_m \otimes \sigma_n\}}}{4}$$
Where I have used the property $$\mathrm{Tr}(\sigma_a \sigma_b) = 2\delta_{ab}$$, where $$\sigma_a$$ belongs to the set $$\{1_2, \sigma_1, \sigma_2, \sigma_3 \}$$, i.e. identity followed by the Pauli matrices.