Math and science::Algebra::Aluffi

Subgroups of the general linear group

Below are three subgroups of ${GL}_{n} (R)$ , the group of invertible $n \times n$ matrices with real entries.

${SL}_{n} (R)$: The subset of ${GL}_{n} (R)$ consisting of matrices with determinant 1. "S" is short for "special".
$O_{n} (R)$: The subset of ${GL}_{n} (R)$ consisting of matrices $M$ such that $M^{T} M = M M^{T} = I_{n}$ . "O" here is short for "orthogonal".
${SO}_{n} (R)$: The subset of $O_{n} (R)$ consisting of matrices with determinant 1.

Can you proof that these 3 subsets form subgroups?

To prove that a subset of a group is a subgroup, we need to show that:

It contains the identity element.
It is closed under the group operation.
It is closed under inverses.

Orthogonal group vs. special orthogonal group

The orthogonal group has the requirement that $M^{T} M = M M^{T} = I_{n}$ . This means that the columns of $M$ are orthogonal to each other and that they all have length 1. This is quite a strong requirement and you may wonder what is left to distinguish this group from the special orthogonal group. The answer is that the special orthogonal group has the additional requirement that the determinant of $M$ is positive 1. What is not in the special orthogonal group are all the matrices with determinant -1.

${SL}_{n} (R)$

${SL}_{n} (R)$ is a subgroup of ${GL}_{n} (R)$ .

Proof. The identity matrix for ${GL}_{n} (R)$ has determinant 1, so it is in ${SL}_{n} (R)$ . For any matrices $A$ and $B$ in $G_{n} (R$ , $det (A B) = det (A) det (B)$ . An so if $det (A) = det (B) = 1$ , then $det (A B) = 1$ . So $A B$ is in ${SL}_{n} (R)$ . As $A A^{- 1} = I_{n}$ , if $A$ has determinant 1, then so too must $A^{- 1}$ . So we have closure under matrix multiplication and inverses, and the identity is present.

$O_{n} (R)$

${SO}_{n} (R)$

${SO}_{n} (R)$ is a subgroup of $O_{n} (R)$ .

Proof idea. An orthogonal matrix is one whose columns are orthogonal to each other. The main task is to show that the matrix multiplication of two orthogonal matrices is also orthogonal. Let $A$ and $B$ be orthogonal matrices and consider $A B = C$ . Let's be concrete and fix the matrix dimensions to be $3 \times 3$ . Each column of $C$ is a linear combination of the columns of $A$ with coefficients given by the columns of $B$ . Consider testing the dot product of two columns $c_{1}$ and $c_{2}$ of $C$ . $c_{1}$ will be $B_{11} a_{1} + B_{12} a_{2} + B_{13} a_{3}$ where $a_{x}$ is the $x$ th column of $A$ . Similarly, $c_{2}$ will be $B_{21} a_{1} + B_{22} a_{2} + B_{23} a_{3}$ . When we take the dot product of $c_{1}$ and $c_{2}$ , out of the 9 sub-products that make up the dot product, the only terms that will survive are those like $B_{11} a_{1} \cdot B_{21} a_{1}$ , as the other terms will be orthogonal. But $B_{11} a_{1} \cdot B_{21} a_{1}$ is just $B_{11} B_{21} ‖ a_{1} ‖$ , and so the dot product of $c_{1}$ and $c_{2}$ is just $[\begin{matrix} B_{11} B_{21} ‖ a_{1} ‖ \\ B_{12} B_{22} ‖ a_{2} ‖ \\ B_{13} B_{23} ‖ a_{3} ‖ \end{matrix}]$ .

Now remember that $A A^{T} = I_{3}$ which is another way of saying that the dot product of any column with itself is 1, and the dot product of any two different columns is 0. So $‖ a_{1} ‖ = ‖ a_{2} ‖ = ‖ a_{3} ‖ = 1$ and so the dot product of $c_{1}$ and $c_{2}$ is $[\begin{matrix} B_{11} B_{21} \\ B_{12} B_{22} \\ B_{13} B_{23} \end{matrix}]$ .

But this is just the dot product of the first column of $B$ with the second column of $B$ , which is 0. For the same two columns of $C$ , the dot product will be 1. This is what we wanted in order for $C$ to be orthogonal.

Proof.

If $G_{1}$ and $G_{2}$ are subgroups of a group $G$ , then their intersection $G_{1} \cap G_{2}$ is also a subgroup of $G$ . Applying this rule to ${SL}_{n} (R)$ and $O_{n} (R)$ as subgroups of ${GL}_{n} (R)$ , we see that ${SO}_{n} (R)$ is a subgroup of $O_{n} (R)$ also.

Subgroups of the general linear group

Orthogonal group vs. special orthogonal group

SLn(R)

On(R)

SOn(R)

${SL}_{n} (R)$

$O_{n} (R)$

${SO}_{n} (R)$