Page 71 - DMTH502_LINEAR_ALGEBRA

Unit 2: Vector Subspaces

Similarly one can show that Notes
aB C A a BA CA , if the matrix sums and products are defined.

Thus A aX Y a AX AY a 0 0 0
Theorem 3: Let V be a vector space over the field F. The intersection of any collection of subspaces
of V is a subspace of V.

Proof: As shown in theorem 1, here let W be a collection of subspaces of V, and let W W
a a
a
be their intersection. Remember that W is defined as the set of all elements belonging to every
W . Also since each W is a subspace, each contains the zero vector. Thus W is a non-empty set. Let
a a
u, v be vectors in W and , . F Then
u W ,v W

So , u v W and , F

Therefore u v W since u v is in all W ’s. Thus W W is a subspace of V(F).
i a
a
Definition: Let S , S , ..., S are subsets of a vector space V, the set of all sums
1 2 n
...
1 2 k
of vectors S is called the sum of the subsets ,S S ,...S and is denoted by
i i 1 2 k
k
S S ... S or by S .
1 2 k i
i=1
If W ,W ,W ...W are subspaces of V, then the sum
1 2 3 k
W W W ... W
1 2 k
is easily seen to be a subspace of V which contains each of the subspaces W . From this it follows,
i
that W is a subspace spanned by the union of W ,W ,W ...W .
1 2 3 k

Example 8: Let F be a subfield of the field c of complex numbers. Suppose
1,2,0,3,0
1
0,0,1,4,0
2
0,0,0,0,1
3
5
Now a vector is in the subspace W of F spanned by , , if and only if there exist scalars
1 2 3
c 1 , ,c 3 in F such that
c
2
c 1 1 c 2 2 c 3 3
Thus W consists of all vectors of the form

c
c 1 ,2 , ,3c 1 4 ,c 3
c
c
2
2
1

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