How is the subspace identified? In Linear Algebra class, I was learning about vector spaces and subspaces but can't quite grasp the concepts. We were given the following:


*

*$w = \{(a,b,c) \in R^3 : a = 1\}$

*$w = \{(a,b,c) \in R^3 : a = 0\}$
We were told that (2) is a subspace, while (1) is not because (1) is "not closed under addition". What does that mean?

Set 1 is "not closed under addition" because vectors in this set are of the form $\left [ \begin{array}{c}
1 \\
y\\
z\\
\end{array} \right ]$.  If I add any two of these together I get the result
$$
\left [ \begin{array}{c}
1 \\
a\\
b\\
\end{array} \right ] + \left [ \begin{array}{c}
1 \\
c\\
d\\
\end{array} \right ]\;\; =\;\; \left [ \begin{array}{c}
2 \\
a+c\\
b+d\\
\end{array} \right ]
$$
and this doesn't belong to the set since our first entry isn't 1.  The second set is "closed under addition" since vectors in this set are of the form $\left [ \begin{array}{c}
0 \\
y\\
z\\
\end{array} \right ]$ and adding any two of them yields:
$$
\left [ \begin{array}{c}
0 \\
a\\
b\\
\end{array} \right ] + \left [ \begin{array}{c}
0 \\
c\\
d\\
\end{array} \right ] \;\; =\;\; \left [ \begin{array}{c}
0 \\
a+c\\
b+d\\
\end{array} \right ].
$$
We see that addition in this set preserves the first entry remaining zero, guaranteeing that adding two vectors in this set keeps you in the set.
