How would I prove this is a subspace? "Determine if the set $H$ of all matrices in the form
$
\left[
\begin{array}{cc}
a & b \\
0 & d \\
\end{array}
\right]
$
is a subspace of $M_{2\times2}$."
And I'm given,

A subspace of a vector space is a subset $H$ of $V$ that has three
  properties:
a. The zero vector is in $H$.
b. $H$ is closed under vector addition. That is, for each $u$ and $v$
  in $H$, the sum $ u + v$ is in $H$.
c. $H$ is closed under multiplication by scalars. That is, for each
  $u$ in $H$ and each scalar $c$, the vector $cu$ is in $H$.

How can I show these operations on this matrix? Am I supposed to split this matrix into three vectors? I am unsure how to.
This was the solution given:

The set H is a subspace of M2×2. The zero matrix is in H, the sum of
  two upper triangular matrices is upper triangular, and a scalar
  multiple of an upper triangular matrix is upper triangular.

 A: Well, let's check it out:  
a. $$0\left[
\begin{array}{cc}
a & b \\
0 & d \\
\end{array}
\right] = \left[
\begin{array}{cc}
0 & 0 \\
0 & 0 \\
\end{array}
\right]$$  Yep $\checkmark$
b.  
$$\left[
\begin{array}{cc}
a_0 & b_0 \\
0 & d_0 \\
\end{array}
\right]+\left[
\begin{array}{cc}
a_1 & b_1 \\
0 & d_1 \\
\end{array}
\right]= \left[
\begin{array}{cc}
a_0 + a_1 & b_0 + b_1 \\
0 & d_0 + d_1 \\
\end{array}
\right]$$
Roger dodger $\checkmark$
c.  $$c\left[
\begin{array}{cc}
a & b \\
0 & d \\
\end{array}
\right]=\left[
\begin{array}{cc}
ca & cb \\
0 & cd \\
\end{array}
\right]$$  
Lookin' good $\checkmark$ 
The set of matrices of this form qualifies as a subspace under the definition given.
A: The underlying vector space ($V$) here is the space of $2\times 2$ matrices $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$. The $0$ vector according to this (a vector $x$ such that $\forall v \in V, x + v = v$) is simply the zero matrix $\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$. 
The set $H$ above is the set of all $2\times 2$ matrices with the $(2,1)$ entry ( or $c$) set to zero. Now simply run with the definition.
a) $0 \in H$, as $0_{21}=0$ for the zero matrix
b) $u, v \in H \implies u_{21} = v_{21} = 0 \implies (u+v)_{21} = 0 \implies u+v \in H$.
c) $u \in H \implies u_{21} = 0 \implies c\cdot u_{21} = 0 \implies (cu)_{21} = 0 \implies cu \in H$ $\forall c \in \mathbb{C}$
Thus, $H$ is a vector space that is a subset of $V$, therefore, a subspace of $V$.
