"Let $H \subset \Bbb{R^{2x2}}$ be the set of lower-triangular matrices. Show that $H$ is a subspace of the vector space $\Bbb{R^{2x2}}$."
Proof. Clearly $H\neq \emptyset$. For example $\begin{bmatrix}1&0\\ 1&1\end{bmatrix}\in H$
Let $A,B \in H$. Both of them are lower triangular and identical in size. Hence $A+B$ is defined. Let $[a_1 a_2]$ and $[b_1 b_2]$ denote the columns of $A$ and $B$. $$A+B=[(a_1+b_1) (a_2 + b_2)]$$
Because $A$ and $B$ are lower triangular, it is clear that the first term of column $a_1+b_2$ is $0+0=0$. This means $A+B$ is lower triangular. Thus for all $A,$ and $B$ we have that $ A+B \in H$.
Let $c\in \Bbb{R}$. Now $cA=[(ca_1) (ca_2)]$. Again, the first term of column $ca_1$ is $c\cdot 0$ and $cA \in H$ for all scalars $c$. There is a follow up question : "Is $H$ a subspace if its elements are also invertible?" I don't really get what this means. Surely the proof is just the same even if $A,B$ are invertible?