Clarification on Pauli matrices

I don't quite understand the definition of the Pauli matrices, referring to the Wikipedia entry https://en.wikipedia.org/wiki/Pauli_matrices.

How can the three Pauli matrices together with the identity matrix be a basis for the real vector space $\mathbb{R}$ and for the complex vector space $\mathbb{C}$ at the same time? And what exactly is meant by "real vector space of $2 \times 2$ Hermitian matrices"? Is it a real or a complex vector space?

How could I prove these statements mathematically?

Thanks for any type of help :)

Solution 1:

Let us consider $M := M_2(\mathbb{C})$, the $\mathbb{C}$-vector space of $2\times 2$ matrices with complex coefficients. The description of the structure of $\mathbb{C}$-vector space can be found in any textbook of linear algebra, I guess.

For each $A \in M$, let us denote $M^*$ the transpose of the matrix obtained by taking the complex conjugate of the coefficients of $M$. The matrix $M$ is called the adjoint of $M$. Consider $H$ the subset of $M$ containing the Hermitian matrices, that is, the matrices that are equal to their adjoint.

Lemma: For all $A,B \in H$, for all $\lambda \in \mathbb{R}$, $A + \lambda B \in H$.

Proof: It is straightforward, using some easy properties of the adjoint.

Lemma: If $A \in H$, and if $A \neq 0$, then $iA \not \in H$.

Proof: Also straightforward: assume both $A$ and $iA$ are elements of $H$. Then $(iA)^* = -iA^* = -iA = -(iA)^*$. So $(iA)^* = -(iA)^*$, so $(iA)^* = 0$, so $iA = 0$, so $A = 0$.

Conclusion: $H$ is a real vector subspace, but not a complex vector subspace.

Theorem:

1. Each element of $M$ can be uniquely written as a linear combination of the Pauli matrices and the identity with complex coefficients.
2. Each element of $H$ can be uniquely written as a linear combination of the Pauli matrices and the identity with real coefficients, and conversely, every linear combination of the Pauli matrices and the identity with real coefficients is a Hermitian matrix.

Proof:

1. The equality $\begin{pmatrix}a&b\\c&d\\\end{pmatrix} = \frac{a+d}{2}\begin{pmatrix}1&0\\0&1\\\end{pmatrix} + \frac{a-d}{2}\begin{pmatrix}1&0\\0&-1\\\end{pmatrix} + \frac{c+b}{2}\begin{pmatrix}0&1\\1&0\\\end{pmatrix} -i\frac{c-b}{2} \begin{pmatrix}0&-i\\i&0\\\end{pmatrix}$ gives the claim.
2. I leave you this last part of the proof as an exercise. I think it is very important for you to try to guess why the coefficients found in the proof of the first part must be real numbers, assuming that the matrix we're starting from is Hermitian. The converse is even easier. And when I say "easy", I do not mean that you should feel ashamed if you get stuck: I just think you should really take a deep breath, take a walk, and try again.