How do one show that the Pauli Matrices together with the Unit matrix form a basis in the space of complex 2 x 2 matrices?
Let $M_2(\mathbb{C})$ denote the set of all $2\times2$ complex matrices. We also note that dim$(M_2(\mathbb{C}))=4$, because if $M\in M_2(\mathbb{C})$ and
$M=\left( \begin{array}{cc} z_{11} & z_{12}\\ z_{21} & z_{22} \\ \end{array} \right)$, where $z_{ij}\in \mathbb{C}$,
then
$M=\left( \begin{array}{cc} z_{11} & z_{12}\\ z_{21} & z_{22} \\ \end{array} \right)=z_{11}\left( \begin{array}{cc} 1 & 0\\ 0 & 0 \\ \end{array} \right)+z_{12}\left( \begin{array}{cc} 0 & 1\\ 0 & 0 \\ \end{array} \right)+z_{21}\left( \begin{array}{cc} 0 & 0\\ 1 & 0 \\ \end{array} \right)+z_{22}\left( \begin{array}{cc} 0 & 0\\ 0 & 1 \\ \end{array} \right)$.
The standard four Pauli matrices are:
$I=\left( \begin{array}{cc} 1 & 0\\ 0 & 1 \\ \end{array} \right),~~ \sigma_1=\left( \begin{array}{cc} 0 & 1\\ 1 & 0 \\ \end{array} \right),~~ \sigma_2=\left( \begin{array}{cc} 0 & -i\\ i & 0 \\ \end{array} \right),~~ \sigma_3=\left( \begin{array}{cc} 1 & 0\\ 0 & -1 \\ \end{array} \right)$.
It is straightforward to show that these four matrices are linearly independent. This can be done as follows.
Let $c_\mu\in \mathbb{C}$ such that
$c_0I+c_1\sigma_1+c_2\sigma_2+c_3\sigma_3=$ O (zero matrix).
This gives
$\left( \begin{array}{cc} c_0+c_3 & c_1-ic_2\\ c_1+ic_2 & c_0-c_3 \\ \end{array} \right)=\left( \begin{array}{cc} 0 & 0\\ 0 & 0 \\ \end{array} \right)$
which further gives the following solution: $c_0=c_1=c_1=c_3=0$.
It is left to show that $\{I,\sigma_i\}$ where $i = 1,2,3$ spans $M_2(\mathbb{C})$. And this can accomplished in the following way:
$M=c_0I+c_1\sigma_1+c_2\sigma_2+c_3\sigma_3$ gives
$\left( \begin{array}{cc} c_0+c_3 & c_1-ic_2\\ c_1+ic_2 & c_0-c_3 \\ \end{array} \right)=\left( \begin{array}{cc} z_{11} & z_{12}\\ z_{21} & z_{22} \\ \end{array} \right)$
which further gives the following equations:
$c_0+c_3=z_{11},~c_0-c_3=z_{22},~c_1-ic_2=z_{12},~c_1+ic_2=z_{21}$.
Solving these equations, one obtains
$c_0=\frac{1}{2}(z_{11}+z_{22}),~c_1=\frac{1}{2}(z_{12}+z_{21}),~c_2=\frac{1}{2}i(z_{12}-z_{21}),~c_3=\frac{1}{2}(z_{11}-z_{22})$.
To show that $\{I, \sigma_i\}$ is a base of the complex vector space of all $2 \times 2$ matrices, you need to prove two things:
- That $\{I, \sigma_i\}$ are linearly independent.
- That every complex $2 \times 2$ matrix can be written as a combination of $\{I, \sigma_i\}$.
To prove point 1, you need to show that the only four complex numbers $a_0,a_1,a_2,a_3$ such that
$$a_0 I + a_1 \sigma_1 + a_2 \sigma_2 + a_3 \sigma_3 = 0$$
where $0$ is the zero matrix, are $a_0=a_1=a_2=a_3=0$.
To prove point 2, you need to show that every complex $2 \times 2$ matrix $M$ can be written as
$$M = c_0 I + c_1 \sigma_1 + c_2 \sigma_2 + c_3 \sigma_3 $$
where $c_0,c_1,c_2,c_3$ are complex numbers. Your equations are correct, but what do you need to show in order to prove 2?
Even though the question has already been sufficiently answered, I would like to offer the sketch of another "elegant" proof:
The space of complex $2\times 2$ matrices, denoted $M_{2\times 2}(\mathbb{C})$, is isomorphic to $\mathbb{R}^8$ via \begin{equation} \begin{pmatrix} z_{11} & z_{12} \\ z_{21} & z_{22} \end{pmatrix} \mapsto \begin{pmatrix} \Re z_{11} & \Im z_{11} & \Re z_{12} & \Im z_{12} & \Re z_{21} & \Im z_{21} & \Re z_{22} & \Im z_{22} \end{pmatrix}^\top \end{equation} where $\Re$ and $\Im$ denote real and imaginary parts.
Now you want to show that $(I,\sigma_i)$ is a basis of $M_{2\times 2}(\mathbb{C})$ as complex vector space, which is equivalent to $(I,\sigma_i, iI,i\sigma_i)$ being a basis of $M_{2\times 2}(\mathbb{C})$ as real vector space.
The above isomorphism maps the identity and Pauli matrices like: \begin{align*} I &\mapsto \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \end{pmatrix}^\top\\ \sigma_1 &\mapsto \begin{pmatrix} 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 \end{pmatrix}^\top\\ \sigma_2 &\mapsto \begin{pmatrix} 0 & 0 & 0 & -1 & 0 & 1 & 0 & 0 \end{pmatrix}^\top\\ \sigma_3 &\mapsto \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & -1 & 0 \end{pmatrix}^\top\\ iI &\mapsto \begin{pmatrix} 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix}^\top\\ i\sigma_1 &\mapsto \begin{pmatrix} 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 \end{pmatrix}^\top\\ i\sigma_2 &\mapsto \begin{pmatrix} 0 & 0 & 1 & 0 & -1 & 0 & 0 & 0 \end{pmatrix}^\top\\ i\sigma_3 &\mapsto \begin{pmatrix} 0 & 1 & 0 & 0 & 0 & 0 & 0 & -1 \end{pmatrix}^\top \end{align*} which can trivially be seen to be a basis of $\mathbb{R}^8$ as a real vector space.
Therefore, by the property of isomorphisms to map basis to basis vice versa, we are done.