Page 210 - Matrix Analysis & Applied Linear Algebra

4.4 Basis and Dimension 205

If X and Y are subspaces of a vector space V, then the sum of X and
Y was defined in §4.1 to be

X + Y = {x + y | x ∈X and y ∈Y},
and it was demonstrated in (4.1.1) that X + Y is again a subspace of V. You
were asked in Exercise 4.1.8 to prove that the intersection X∩ Y is also a
subspace of V. We are now in a position to exhibit an important relationship
between dim (X + Y) and dim (X∩ Y) .

Dimension of a Sum

If X and Y are subspaces of a vector space V, then
dim (X + Y) = dim X + dim Y− dim (X∩ Y) . (4.4.19)

Proof. The strategy is to construct a basis for X + Y and count the number
of vectors it contains. Let S = {z 1 , z 2 ,..., z t } be a basis for X∩ Y. Since
S⊆ X and S⊆ Y, there must exist extension vectors {x 1 , x 2 ,..., x m } and
{y 1 , y 2 ,..., y n } such that

B X = {z 1 ,..., z t , x 1 ,..., x m } = a basis for X
and
B Y = {z 1 ,..., z t , y 1 ,..., y n } = a basis for Y.
We know from (4.1.2) that B = B X ∪B Y spans X + Y, and we wish show that
B is linearly independent. If
t m n

α i z i + β j x j + γ k y k = 0, (4.4.20)
i=1 j=1 k=1
then
 
n t m

γ k y k = −  α i z i + β j x j  ∈X.
k=1 i=1 j=1
Since it is also true that γ k y k ∈Y, we have that γ k y k ∈X ∩ Y, and
k k
hence there must exist scalars δ i such that

n t n t

γ k y k = δ i z i or, equivalently, γ k y k − δ i z i = 0.
k=1 i=1 k=1 i=1