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? In other words, show that a complex 2 x 2 Matrix can in a unique way be written as
$$
M = \lambda _ 0 I+\lambda _1 \sigma _ x + \lambda _2 \sigma _y + \lambda _ 3 \sigma_z 
$$ 
If$$M = \Big(\begin{matrix}
  m_{11} & m_{12} \\
  m_{21} & m_{22}
 \end{matrix}\Big)= \lambda _ 0 I+\lambda _1 \sigma _ x + \lambda _2 \sigma _y + \lambda _ 3 \sigma_z $$
I get the following equations 
$$
m_{11}=\lambda_0+\lambda_3 \\ m_{12}=\lambda_1-i\lambda_2 \\ m_{21}=\lambda_1+i\lambda_2 \\ m_{22}=\lambda_0-\lambda_3
$$

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
$c_0=\frac{1}{2}(z_{11}+z_{22}),~c_1=\frac{1}{2}(z_{12}+z_{21}),~c_2=\frac{1}{2}i(z_{12}-z_{21}),~c_3=\frac{1}{2}(z_{11}-z_{22})$.
