Page 197 - 35Linear Algebra

9.2 Building Subspaces 197

Note that the requirements of the subspace theorem are often referred to as
“closure”.
We can use this theorem to check if a set is a vector space. That is, if we
have some set U of vectors that come from some bigger vector space V , to
check if U itself forms a smaller vector space we need check only two things:

1. If we add any two vectors in U, do we end up with a vector in U?

2. If we multiply any vector in U by any constant, do we end up with a
vector in U?

If the answer to both of these questions is yes, then U is a vector space. If
not, U is not a vector space.

Reading homework: problem 1

9.2 Building Subspaces

Consider the set
   
 1 0 
3
1
0
,
U =     ⊂ R .
0 0
 
Because U consists of only two vectors, it clear that U is not a vector space,
since any constant multiple of these vectors should also be in U. For example,
the 0-vector is not in U, nor is U closed under vector addition.
But we know that any two vectors define a plane:

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