# 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:

1. That $\{I, \sigma_i\}$ are linearly independent.
2. 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 $$$$\begin{pmatrix} z_{11} & z_{12} \\ z_{21} & z_{22} \end{pmatrix} \mapsto \begin{pmatrix} \Re z_{11} & \Im z_{11} & \Re z_{12} & \Im z_{12} & \Re z_{21} & \Im z_{21} & \Re z_{22} & \Im z_{22} \end{pmatrix}^\top$$$$ where $$\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.