Groups and Subgroups -- Abstract Algebra Let H be the subset of $M_2(R)$ consisting of all matrices of the form $[(a, -b),(b, a)]$ for $a, b \in R$. Is H closed under (a) matrix addition?  (b) matrix multiplication?
I know that the answers are yes to both, but I'm having a really hard time explaining this in words. Can someone help? Thanks.
 A: Let's just try it! Take two elements $x,y \in H\subset M_2(\mathbb{R})$. Then we know that $x$ and $y$ 'look like'
$$
x=\pmatrix{a_1&-b_1\\ b_1&a_1},y=\pmatrix{a_2&-b_2\\b_2&a_2}
$$
so that 
$$
x+y=\pmatrix{a_1&-b_1\\ b_1&a_1}+\pmatrix{a_2&-b_2\\b_2&a_2}=\pmatrix{a_1+a_2&-(b_1+b_2)\\b_1+b_2&a_1+a_2}
$$
which is exactly the form of a matrix in $H\subset M_2(\mathbb{R})$, i.e. $x+y \in H \subset M_2(\mathbb{R})$ when $x,y \in H\subset M_2(\mathbb{R})$. So $H$ is closed under addition.
Now let's check multiplication
$$
xy=\pmatrix{a_1&-b_1\\ b_1&a_1}\pmatrix{a_2&-b_2\\b_2&a_2}=\pmatrix{a_1a_2-b_1b_2&-(a_1b_2+b_1a_2)\\a_2b_1+a_1b_2&-b_1b_2+a_1a_2}
$$
for $xy \in H$, we need the corresponding diagonal elements to be the same. Are they here? Yes! Let's just move things around and notice that the above matrix is the same as
$$
\pmatrix{a_1a_2-b_1b_2&-(a_2b_1+a_1b_2)\\a_2b_1+a_1b_2&a_1a_2-b_1b_2}
$$
so $xy \in H \subset M_2(\mathbb{R})$ whenever $x,y \in H \subset M_2(\mathbb{R})$. So $H$ is closed under multiplication.
