Pauli decomposition of two-by-two matrices

The expression exists and is unique because $\{\mathbf{1},\sigma_x,\sigma_y,\sigma_z\}$ is a basis. To obtain the coefficients, take the trace of $A$ multiplied by each basis element: $\tr A=2a_0$ because the $\sigma_j$ are traceless, and $\tr(\sigma_jA)=a_0\tr\sigma_j+\sum_ka_k\tr(\sigma_j\sigma_k)=2a_j$. Since the matrices $\mathbf{1}$ and $\sigma_j$ are Hermitian, $A^\dagger=a_0^*\mathbf{1}+\sum_ja_j^*\sigma_j$. By uniqueness of the decomposition, $A^\dagger=A$ is equivalent to $a_0^*=a_0$ and $a_j^*=a_j$.

We have $a_0=\frac12(3+1)=2$. For the other coefficients, it is enough to calculate the diagonal entries of the products. For $\sigma_z$, $\tr(\sigma_zA)=3-1=2$, so $a_z=1$. For $\sigma_x$, $\tr(\sigma_xA)=A_{21}+A_{12}=(1+i)+(1-i)=2$, so $a_x=1$. For $\sigma_y$, the diagonal entries of $\sigma_yA$ are $-iA_{21}=-i(1+i)=1-i$ and $iA_{12}=i(1-i)=1+i$, hence $\tr(\sigma_yA)=2$ and $a_y=1$. Thus

$$A=2\,\mathbf{1}+\sigma_x+\sigma_y+\sigma_z,$$

as may be checked by adding the matrices: the off-diagonal entries are $1-i$ and $1+i$, and the diagonal entries are $2\pm1$.

Set $K=\mathbf n\cdot\boldsymbol\sigma$, which satisfies $K^2=|\mathbf n|^2\mathbf{1}=\mathbf{1}$ and $\tr K=0$. Its eigenvalues satisfy $\lambda^2=1$, and their sum is zero: they are $+1$ and $-1$. The operators $P_\pm=\frac12(\mathbf{1}\pm K)$ satisfy $P_\pm^2=\frac14(\mathbf{1}\pm2K+K^2)=P_\pm$, $P_+P_-=\frac14(\mathbf{1}-K^2)=0$, $P_++P_-=\mathbf{1}$, and they are Hermitian: they are the orthogonal projectors onto the eigenspaces of $K$, because $KP_\pm=\frac12(K\pm\mathbf{1})=\pm P_\pm$. Since $A=a_0\mathbf{1}+|\mathbf a|K$, we obtain $A=(a_0+|\mathbf a|)P_++(a_0-|\mathbf a|)P_-$. For the previous matrix, $\mathbf a=(1,1,1)$, $|\mathbf a|=\sqrt3$, and the eigenvalues are $2\pm\sqrt3$. Their sum is $4=\tr A$ and their product is $4-3=1=\det A=3-|1-i|^2$.

The previous question shows that such a matrix is an orthogonal projector with trace $1$, hence with rank $1$. Conversely, let $P$ be a rank-one orthogonal projector. It is Hermitian, so $P=a_0\mathbf{1}+\mathbf a\cdot\boldsymbol\sigma$ with real coefficients, and its trace is $1$, hence $a_0=\frac12$. Moreover, $P^2=a_0^2\mathbf{1}+2a_0\,\mathbf a\cdot\boldsymbol\sigma+|\mathbf a|^2\mathbf{1}=\bigl(\frac14+|\mathbf a|^2\bigr)\mathbf{1}+\mathbf a\cdot\boldsymbol\sigma$. The equality $P^2=P$ requires $\frac14+|\mathbf a|^2=\frac12$, hence $|\mathbf a|=\frac12$, and we set $\mathbf n=2\mathbf a$, which is a unit vector.

Every pure state $\ket\psi$ of a qubit defines a rank-one projector $\ket\psi\bra\psi$, which is independent of the global phase of $\ket\psi$. From the preceding result, $\ket\psi\bra\psi=\frac12(\mathbf{1}+\mathbf n\cdot\boldsymbol\sigma)$ for a unique unit vector $\mathbf n$. The pure states of a qubit, up to a phase, are therefore in one-to-one correspondence with points on the unit sphere: this is the Bloch sphere, and $\mathbf n$ is the Bloch vector of the state. We check that $\bra\psi\sigma_j\ket\psi=\tr(\ket\psi\bra\psi\sigma_j)=\frac12\tr(\sigma_j+\sum_kn_k\sigma_k\sigma_j)=n_j$: the components of the Bloch vector are the expectation values of the Pauli matrices.