# Basis of 2x2 matrices with real entries

• I
• LagrangeEuler

#### LagrangeEuler

What is the basis of 2x2 matrices with real entries? I know that the basis of 2x2 matrices with complex entries are 3 Pauli matrices and unit matrix:
$$\begin{bmatrix} 0 & 1 \\[0.3em] 1 & 0 \\[0.3em] \end{bmatrix}$$,
$$\begin{bmatrix} 0 & -i \\[0.3em] i & 0 \\[0.3em] \end{bmatrix}$$
$$\begin{bmatrix} 1 & 0 \\[0.3em] 0 & -1 \\[0.3em] \end{bmatrix}$$
and
$$\begin{bmatrix} 1 & 0 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$$
What about in the case of real 2x2 matrices? How many matrices is there in the basis?

It's the same problem as selecting a basis of ##\mathbf{R}^4## e.g. via the correspondence ##\begin{pmatrix} a_1 & a_2 \\ a_3 & a_4 \end{pmatrix} \leftrightarrow (a_1,a_2,a_3,a_4)##

LagrangeEuler
So I again need four matrices? As in a complex case? Is the basis then
$$\begin{bmatrix} 1 & 0 \\[0.3em] 0 & 0 \\[0.3em] \end{bmatrix}$$,
$$\begin{bmatrix} 0 & 1 \\[0.3em] 0 & 0 \\[0.3em] \end{bmatrix}$$,
$$\begin{bmatrix} 0 & 0 \\[0.3em] 1 & 0 \\[0.3em] \end{bmatrix}$$
and
$$\begin{bmatrix} 0 & 0 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$$?

That's fine, but it's of course not unique.

LagrangeEuler
Yes. Thank you. So I can see if some 2x2 matrix
$$\begin{bmatrix} a_{11} & a_{12} \\[0.3em] a_{21} & a_{22} \\[0.3em] \end{bmatrix}$$
commute with
$$\begin{bmatrix} 1 & 0 \\[0.3em] 0 & 0 \\[0.3em] \end{bmatrix}$$
and
$$\begin{bmatrix} 0 & 1 \\[0.3em] 0 & 0 \\[0.3em] \end{bmatrix}$$
and
$$\begin{bmatrix} 0 & 0 \\[0.3em] 1 & 0 \\[0.3em] \end{bmatrix}$$
and
$$\begin{bmatrix} 0 & 0 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$$
then the matrix commute with any 2x2 matrix?

Well yes, but I think the only matrices that commute with all four of those are of the form ##\mathrm{diag}(a,a) = aI##, i.e. multiples of the identity.

LagrangeEuler
Yes. Thank you. So I can see if some 2x2 matrix
$$\begin{bmatrix} a_{11} & a_{12} \\[0.3em] a_{21} & a_{22} \\[0.3em] \end{bmatrix}$$
commute with
$$\begin{bmatrix} 1 & 0 \\[0.3em] 0 & 0 \\[0.3em] \end{bmatrix}$$
and
$$\begin{bmatrix} 0 & 1 \\[0.3em] 0 & 0 \\[0.3em] \end{bmatrix}$$
and
$$\begin{bmatrix} 0 & 0 \\[0.3em] 1 & 0 \\[0.3em] \end{bmatrix}$$
and
$$\begin{bmatrix} 0 & 0 \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix}$$
then the matrix commute with any 2x2 matrix?
Since you're asking about a basis for ##\mathbb M_{2\times2}##, square matrices of order 2 with real entries, commutivity doesn't enter into things. For a basis, you need four linearly independent matrices ##M_1, M_2, M_3, M_4## that span ##\mathbb M_{2\times2}##. The matrices you list in post 5 are the standard basis for this space, although as already stated, there are other possibilities for a basis.

LagrangeEuler
Well yes, but I think the only matrices that commute with all four of those are of the form ##\mathrm{diag}(a,a) = aI##, i.e. multiples of the identity.
Yes of course. It is also a center of group ##GL_2(\mathbb{R})##.

Since you're asking about a basis for ##\mathbb M_{2\times2}##, square matrices of order 2 with real entries, commutivity doesn't enter into things. For a basis, you need four linearly independent matrices ##M_1, M_2, M_3, M_4## that span ##\mathbb M_{2\times2}##. The matrices you list in post 5 are the standard basis for this space, although as already stated, there are other possibilities for a basis.
Ok. But using that you can for instance to find the center of group ##GL_2(\mathbb{R})##.

So I again need four matrices? As in a complex case?

The Pauli matrices plus the identity are not a basis for all complex 2x2 matrices. They are a basis for the Hermitian 2x2 complex matrices.

To cover all 2x2 complex matrices you need a basis with 8 elements.

Really I am pretty sure that I read that those matrices are basis in ##\mathbb{C}^{2x2}##.

The Pauli matrices plus the identity are not a basis for all complex 2x2 matrices. They are a basis for the Hermitian 2x2 complex matrices.

To cover all 2x2 complex matrices you need a basis with 8 elements.
Really I am pretty sure that I read that those matrices are basis in ##\mathbb{C}^{2x2}##.
This all depends whether the field of scalars are the Real or Complex numbers. Complex 2x2 matrices can be seen as a 4D Vector Space over ##\mathbb C##, and the set of Hermitian matrices is not a subspace. Or, they can be seen as an 8D Vector Space over ##\mathbb R##, in which case the set of Hermitian matrices is a 4D subspace.

The Bill