Vector Subspace Test Hey can some help me with this textbook question I think it is a subspace I just want to be sure thanks
Let $R^{2\times2}$ denote the vector space of $2 \times 2$ matrices, and let
$$
S =\left\{
\left[\begin{matrix}
a \space b \\ 
-b \space c \\
\end{matrix}\right]\mid a,b,c \in \mathbb{R}\right\}
$$
Such that a b and c are Real numbers. Either show that S is a subspace of ${R^{2 \times 2}}$, or explain why it is not a subspace.
 A: Additivity 
$$
\left[\begin{matrix}
a \space b \\ 
-b \space c \\
\end{matrix}\right] + \left[\begin{matrix}
d \space e \\ 
-e \space f \\
\end{matrix}\right] =\left[\begin{matrix}
(a+d) \space (b+e) \\ 
(-b-e) \space (c+f) \\
\end{matrix}\right] =\left[\begin{matrix}
(a+d) \space (b+e) \\ 
-(b+e) \space (c+f) \\
\end{matrix}\right] \checkmark
$$
Scalarity:
$$
\lambda \left[\begin{matrix}
a \space b \\ 
-b \space c \\
\end{matrix}\right] = \left[\begin{matrix}
\lambda a \space \lambda b \\ 
-\lambda b \space \lambda c \\
\end{matrix}\right]\checkmark
$$
Null:
$$
\left[\begin{matrix}
0 \space 0 \\ 
-0 \space 0 \\ 
\end{matrix}\right] = \left[\begin{matrix}
0 \space 0 \\ 
0 \space 0 \\ 
\end{matrix}\right] \checkmark
$$
Here we take for granted that $\mathbb{R}$ is closed under multiplication, and addition. So indeed $S$ is a subspace.
A: Your phrasing isn't quite correct. $S$ can't be both a single matrix and a subspace; I think you mean that $S$ is the set of $2 \times 2$ matrices of a given form. 

But you are correct that $S$ is a subspace, for if $\alpha \in \Bbb{R}$ is given and we choose two elements of $S$, we find that
$$ \left(\begin{array}{cc} a & b \\ -b & c\end{array} \right) + \left(\begin{array}{cc} a' & b' \\ -b' & c'\end{array} \right) = \left(\begin{array}{cc} a + a' & b + b' \\ -(b+b') & c+c'\end{array} \right)$$
has the correct form and
$$ \alpha \left(\begin{array}{cc} a & b \\ -b & c\end{array} \right) = \left(\begin{array}{cc} \alpha a & \alpha b \\ -(\alpha b) & \alpha c\end{array} \right)$$
also has the correct form. Finally, $S \neq \emptyset$ since it contains the zero matrix.
A: Here's a really easy way using the fact that kernels are subspaces (we just have to choose our map carefully).
Let $T\colon\mathbb{R}^{2\times 2}\rightarrow\mathbb{R}$ be given by $$T\left(\left[\begin{matrix}
w \space x \\ 
y \space z \\
\end{matrix}\right]\right)=x+y$$ which you can easily show is a linear transformation. It's also easy to see that $\ker T=S$ because $T(A)=0\iff x+y=0\iff x=-y$. The kernel of a linear transformation is a vector subspace of the domain and so $S$ is a subspace.

You could also use the fact that the image of a linear transformation is a subspace of the codomain.
Let $T\colon\mathbb{R}^3\rightarrow \mathbb{R}^{2\times 2}$ be given by $$T(x,y,z)=\left[\begin{matrix}
x \space y \\ 
-y \space z \\
\end{matrix}\right]$$ which, again, is clearly linear and we see that $\mbox{Im }T=S$. Hence $S$ is a linear subspace.
