# Basis of 2x2 matrices with real entries

• I
• LagrangeEuler
In summary, the basis of 2x2 matrices with real entries is similar to selecting a basis for ##\mathbf{R}^4##. It consists of four linearly independent matrices that span the space, such as the standard basis consisting of the unit matrix and three Pauli matrices. However, these matrices do not commute with all other matrices in the space. In contrast, the Pauli matrices plus the identity are a basis for the Hermitian 2x2 complex matrices, but not for all 2x2 complex matrices. The number of matrices in the basis also differs depending on whether the field of scalars is the real or complex numbers.
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
LagrangeEuler said:
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
ergospherical said:
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})##.

Mark44 said:
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})##.

LagrangeEuler said:
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 Bill said:
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.
LagrangeEuler said:
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

## 1. What is a 2x2 matrix with real entries?

A 2x2 matrix with real entries is a mathematical object that represents a set of numbers arranged in a rectangular grid of 2 rows and 2 columns. The entries in the matrix are real numbers, meaning they can be positive, negative, or zero.

## 2. What is the basis of a 2x2 matrix with real entries?

The basis of a 2x2 matrix with real entries is a set of two linearly independent vectors that can be used to generate all other vectors in the matrix. These vectors form the columns of the matrix and are typically denoted as a and b.

## 3. How do you find the basis of a 2x2 matrix with real entries?

To find the basis of a 2x2 matrix with real entries, you can use the Gaussian elimination method. This involves performing row operations on the matrix until it is in its reduced row echelon form. The resulting columns will be the basis vectors a and b.

## 4. Why is the basis important in 2x2 matrices with real entries?

The basis is important in 2x2 matrices with real entries because it allows us to represent any vector in the matrix as a linear combination of the basis vectors. This makes it easier to perform operations on the matrix and solve equations involving the matrix.

## 5. Can a 2x2 matrix with real entries have more than one basis?

No, a 2x2 matrix with real entries can only have one basis. This is because the basis vectors must be linearly independent, meaning they cannot be written as a multiple of each other. In a 2x2 matrix, there are only two linearly independent vectors, so there can only be one basis.

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