# 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 $$m_{11}=\lambda_0+\lambda_3 \\ m_{12}=\lambda_1-i\lambda_2 \\ m_{21}=\lambda_1+i\lambda_2 \\ m_{22}=\lambda_0-\lambda_3$$

• Take a matrix with all entries populated and write it as each single entry multiplying a matrix having $1$ in that place and $0$ elsewhere. Then group the similar ones and you will end up with 4 linearly independent matrices, exactly the above. – gented Nov 11 '16 at 9:43
• Sorry I don't quite follow. Could you explain what you mean in terms of equations? – Turbotanten Nov 11 '16 at 9:55
• Since the Pauli matrices and the identity are hermitian, you can only obtain an hermitian 2x2matrix by combining them. Taking the coefficients $\lambda_i$ as real numbers, you indeed get real diagonal elements $m_{11}$ and $m_{12}$, and complex conjugate off-diagonal elements $m_{12}$ and $m_{21}$, as expected for an hermitian matrix. – Christophe Nov 11 '16 at 10:06
• Well, in order to be a basis they would need to 1) be linearly independent (and they are) 2) any matrix can always be expressed as a linear combination thereof (and the arguments that we mentioned above show that). – gented Nov 11 '16 at 10:26

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?

• To show number 2 I need to show that they span the space of complex 2x2 matrices? – Turbotanten Nov 11 '16 at 11:11
• @user2120387 Yes, that is another way of saying the same thing. – valerio Nov 11 '16 at 11:14
• @user2120387 What I'm trying to say is that your equations are not wrong, but maybe you need to write them in a different form to prove what you need to prove... – valerio Nov 11 '16 at 11:17
• Since evidently the space of $2\times 2$ complex matrices has dimension $4$, 1 implies 2. You do not have to prove 2. – Valter Moretti Nov 11 '16 at 15:23
• @ValterMoretti Shouldn't we show before that $\{I,\sigma_i\}$ is a basis and then deduce from this that the dimension of the space is $4$? After all, the dimension of a vector space is defined as the cardinality of its bases. – valerio Nov 11 '16 at 15:43

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.

Pauli matrices $$\sigma_1,\sigma_2$$ and $$\sigma_3$$ evidently form a base of the 3-dimensional real vector space of the 2 by 2 traceless Hermitian matrices. Since every Hermitian matrix is the sum of a traceless Hermitian matrix and the real multiple of the identity matrix, $$\sigma_1\sigma_2,\sigma_3$$ an $$I$$ together form a base of the 4-dimensional real vector space of the 2 by 2 Hermitian matrices.
Since every complex 2 by 2 matrix can be decomposed to the sum of a Hermitian and an anti-Hermitian matrix, regarding that, $$M$$ is hermitian if and only if $$iM$$ is anti-Hermitian, every complex 2 by 2 matrix $$M$$ can be written as $$M=A+iB$$ where both $$A$$ and $$B$$ is Hermitian. So, there are some unique real numbers $$a_0,a_1,a_2,a_3,b_0,b_1,b_2, b_3$$ so that $$A=a_0I+\sum_i{a_i\sigma_i}$$ and $$B=b_0I+\sum_i{b_i\sigma_i}$$ hence $$M=(a_0+ib_0)I+\sum_i(a_i+ib_i)\sigma_i$$ that is, the Pauli matrices and $$I$$ together expand the complex vector space of the 2 by 2 complex matrices. Since the (complex) dimension of this vector space is 4, they form a base.