Justify matrices form basis for SO(4)

[OhNoYaDidn't]

I am given the following set of 4x4 matrices. How can i justify that they form a basis for the Lie Algebra of the group SO(4)? I know that they must be real matrices, and $AA^{T}=\mathbb{I}$, and the $detA = +-1.$ Do i show that the matrices are linearly independent, verify these properties, and so they are a basis? Why are they 6 elements?

$$_{A_1}=\begin{pmatrix} 0 &0 &0 &0 \\ 0 & 0 & 1 & 0 \\ 0& -1 &0 &0 \\ 0& 0& 0 & 0 \end{pmatrix} ,\\ \ _{A_2}=\begin{pmatrix} 0 &0 &-1 &0 \\ 0 & 0 & 0 & 0 \\ 1& 0 &0 &0 \\ 0& 0& 0 & 0 \end{pmatrix} ,\\ _{A_3}=\begin{pmatrix} 0 &-1 &0 &0 \\ 1 & 0 & 0 & 0 \\ 0& 0 &0 &0 \\ 0& 0& 0 & 0 \end{pmatrix} \\ _{B_1}=\begin{pmatrix} 0 &0 &0 &-1 \\ 0 & 0 & 0 & 0 \\ 0& 0 &0 &0 \\ 1& 0& 0 & 0 \end{pmatrix} ,\\ _{B_2}=\begin{pmatrix} 0 &0 &0 &0 \\ 0 & 0 & 0 & -1 \\ 0& 0 &0 &0 \\ 0& 1& 0 & 0 \end{pmatrix} ,\\ _{B_3}=\begin{pmatrix} 0 &0 &0 &0 \\ 0 & 0 & 0 & 0 \\ 0& 0 &0 &1 \\ 0& 0& -1 & 0 \end{pmatrix}$$

 

[fresh_42]

I am given the following set of 4x4 matrices. How can i justify that they form a basis for the Lie Algebra of the group SO(4)? I know that they must be real matrices, and $AA^{T}=\mathbb{I}$, and the $detA = +-1.$ Do i show that the matrices are linearly independent, verify these properties, and so they are a basis? Why are they 6 elements?

$$_{A_1}=\begin{pmatrix} 0 &0 &0 &0 \\ 0 & 0 & 1 & 0 \\ 0& -1 &0 &0 \\ 0& 0& 0 & 0 \end{pmatrix} ,\\ \ _{A_2}=\begin{pmatrix} 0 &0 &-1 &0 \\ 0 & 0 & 0 & 0 \\ 1& 0 &0 &0 \\ 0& 0& 0 & 0 \end{pmatrix} ,\\ _{A_3}=\begin{pmatrix} 0 &-1 &0 &0 \\ 1 & 0 & 0 & 0 \\ 0& 0 &0 &0 \\ 0& 0& 0 & 0 \end{pmatrix} \\ _{B_1}=\begin{pmatrix} 0 &0 &0 &-1 \\ 0 & 0 & 0 & 0 \\ 0& 0 &0 &0 \\ 1& 0& 0 & 0 \end{pmatrix} ,\\ _{B_2}=\begin{pmatrix} 0 &0 &0 &0 \\ 0 & 0 & 0 & -1 \\ 0& 0 &0 &0 \\ 0& 1& 0 & 0 \end{pmatrix} ,\\ _{B_3}=\begin{pmatrix} 0 &0 &0 &0 \\ 0 & 0 & 0 & 0 \\ 0& 0 &0 &1 \\ 0& 0& -1 & 0 \end{pmatrix}$$

I get the impression you confuse the special orthogonal group $SO(4)$ and its Lie algebra $\mathfrak{so}(4)$. At least on the level of their defining conditions. Your matrices are elements of a Lie algebra, whereas neither is invertible $(AA^T=\mathbb{1})$ nor is of determinant $\pm 1$, which are the conditions for the group. In the Lie algebra $\mathfrak{so}(4)$ the defining conditions are $A+A^T=0$ and $tr(A)=0$.

And, yes, to form a basis they have to be linear independent, which therefore has to be shown. I think this will be easier if you write them as $\delta_{ij}-\delta_{ji}$. To show they form a basis, you also have to show that they span the entire vector space of $\mathfrak{so}(4)$, which could easiest be done by counting the dimension: How many matrix entries are there for matrices in $\mathbb{M}_\mathbb{R}(4)$ and how many linear equations, conditions does this system have?

 

[OhNoYaDidn't]

I get the impression you confuse the special orthogonal group $SO(4)$ and its Lie algebra $\mathfrak{so}(4)$. At least on the level of their defining conditions. Your matrices are elements of a Lie algebra, whereas neither is invertible $(AA^T=\mathbb{1})$ nor is of determinant $\pm 1$, which are the conditions for the group. In the Lie algebra $\mathfrak{so}(4)$ the defining conditions are $A+A^T=0$ and $tr(A)=0$.

And, yes, to form a basis they have to be linear independent, which therefore has to be shown. I think this will be easier if you write them as $\delta_{ij}-\delta_{ji}$. To show they form a basis, you also have to show that they span the entire vector space of $\mathfrak{so}(4)$, which could easiest be done by counting the dimension: How many matrix entries are there for matrices in $\mathbb{M}_\mathbb{R}(4)$ and how many linear equations, conditions does this system have?

Thank you, fresh_42. I'm new to the subject, things are still a little confusing.
So, we know that a $\mathbb{M}_\mathbb{R}(4)$ matrix has 16 entries, but since $A=-A^T$, we can get rid of the elements of the upper or lower triangular part of our matrix: hence subtracting $N(N+1)/2$ from $N^2$, so now we need $n=N^2-N(N+1)/2=N(N-1)/2$ elements for this basis, which is exactly what i wanted to understand.
To show that they are linearly independent, i simply write them as a linear combination, multiplying them by constants: a, b, c, d, e, f, respectively, and prove $a=b=c=d=e=f=0$.

What do you mean by writing them as $\delta_{ij}-\delta_{ji}$.?

 

[fresh_42]

What do you mean by writing them as $\delta_{ij}-\delta_{ji}$.?

The matrices $A,B$ above are all of the form $\delta_{ij}-\delta_{ji}$ where $\delta_{ij}$ is a matrix with exactly one $1$ in the $i-$th row and $j-$th column and $0$ elsewhere. They are also written as $E_{ij}, e_{ij}$ or $\mathfrak{e}_{ij}$ or similar as they are the standard basis vectors like $(0,\ldots,1,\ldots,0) \in \mathbb{R}^n$. To chose $E_{ij}$ is probably preferable over the $\delta$ notation, because it saves the $\delta$ for pairs of integers as usual; my fault, sorry.)

It is almost obvious by itself, that the matrices above are linear independent, since all non-zero entries are all in different positions so they cannot cancel each other out. The advantage of the notation as $E_{ij}-E_{ji}$ becomes important if you start multiplying them: $E_{ij}\cdot E_{mn}=\delta_{jm}E_{in}$, i.e. match the inner indices and take the outer as the new ones; zero if the inner indices don't match.

 

[Xico Sim]

A set B of vectors (in this case matrices) is a basis of a vector space V if and only if any element of V can be written as a linear combination of the elements of B (i.e. B spans V) and the elements of B are linearly independent.

In order to show that B spans V=$\mathfrak so(4)$, you can simply find the general expression for an arbitrary element A of $\mathfrak so(4)$ using the "defining conditions" that fresh_42 referred (A is traceless and $A=-A^\dagger$). If you do this, you will be able to write this matrix A as a linear combiantion of the matrices you wrote, so they span V.

In order to show that the elements of B are linearly independent, you can write a general linear combination of them, set it equal to zero, and solve the resulting equation: you will see that all coefficients must be zero, and this by definition tells you that B is linearly independent.

 

[OhNoYaDidn't]

Thank you guys, i got it :)