3.10. Ex 25.
Let L(S) denotes the subspace spanned by a subset S of a linear space V. Prove each
of the statements (a) through (g).
(a) S  L(S)
Proof
Since L(S) contains all linear combinations of the elements in S, every element in S
must also be in L(S). Hence S  L( S ) .
(b) If S  T  V and if T is a subspace of V, then L(S)  T
This property is described by saying that L(S) is the smallest subspace of V that
contains S.
Proof
Since T is a linear space, it contains the zero element, and is closed under vector
addition and scalar multiplication. Since S  T , all linear combinations of
elements in S are contained in T, i.e., L  S   T .
(c) A subset S of V is a subspace of V if and only if
L(S) = S
Proof of “if”
Given S  L(S), every linear combination of elements in S, including the zero element,
are in S. Hence, by theorem 3.4, S is a subspace of V.
Proof of “only if”
Since S is a linear space, it is closed under both vector additions and scalar
multiplications. Hence, all linear combinations of the elements of S are contained in
S, i.e., L(S)  S. On the other hand, any set must be contained in it linear span, i.e., S
 L(S) . Hence, L(S)  S.
(d) If S  T  V then L(S)  L(T)
Proof
Given S  T, every element in S is in T. Hence, every linear combination of
elements in S can be expressed as a linear combination of elements in T. Which
means L(S)  L(T) .
(e) If S and T are subspaces of V, then so is S ∩T
Proof
Since S and T are subspaces, they must both contain the zero element O. Hence,
O  S T . Consider now any x, y  S T so that x, y  S and x, y  T . By
the closure axioms of S and T, we have ax  by  S and ax  by  T for all scalars
a and b. This means ax  by  S T for all scalars a and b, i.e.,
L  S T   S T . Since S T  L  S T  , we have L  S T   S T .
According to (c), S ∩T must be then a subspace of V.
(f) If S and T are subsets of V, then
L(S∩T)  L(S)∩L(T)
Proof
Any element of L  S
x, y  S
T  can be written as ax  by where a, b are scalars and
Since x, y  S , we have ax  by  L  S  .
T.
At the same time,
x, y  T , so that ax  by  L T  . Therefore, every ax  by is in L  S 
Hence, L  S
T   LS 
L T  .
(g) Give an example in which
L(S∩T)  L(S)∩L(T)
Answer
Let V  R 3 , S   i, j  and T   i  j  . Hence,
S
T 
L S
T    empty set.
L T  .
L  S   ai  bj  x-y plane.
L T    a  i  j   line running through origin and point 1,1,0 .
LS
L  T   L T 