Page 47 - Linear Algebra Done Right

Dimension
2
Because F has dimension 2, the last proposition implies that this lin-
early independent list of length 2 is a basis of F (we do not need to
2
bother checking that it spans F ). 2 33
The next theorem gives a formula for the dimension of the sum of
two subspaces of a finite-dimensional vector space.
2.18 Theorem: If U 1 and U 2 are subspaces of a finite-dimensional This formula for the
vector space, then dimension of the sum
of two subspaces is
dim(U 1 + U 2 ) = dim U 1 + dim U 2 − dim(U 1 ∩ U 2 ). analogous to a familiar
counting formula: the
number of elements in
Proof: Let (u 1 ,...,u m ) be a basis of U 1 ∩ U 2 ; thus dim(U 1 ∩ U 2 ) =
m. Because (u 1 ,...,u m ) is a basis of U 1 ∩U 2 , it is linearly independent the union of two finite
sets equals the number
in U 1 and hence can be extended to a basis (u 1 ,...,u m ,v 1 ,...,v j ) of U 1
(by 2.12). Thus dim U 1 = m + j. Also extend (u 1 ,...,u m ) to a basis of elements in the first
(u 1 ,...,u m ,w 1 ,...,w k ) of U 2 ; thus dim U 2 = m + k. set, plus the number of
We will show that (u 1 ,...,u m ,v 1 ,...,v j ,w 1 ,...,w k ) is a basis of elements in the second
U 1 + U 2 . This will complete the proof because then we will have set, minus the number
of elements in the
dim(U 1 + U 2 ) = m + j + k intersection of the two
= (m + j) + (m + k) − m sets.
= dim U 1 + dim U 2 − dim(U 1 ∩ U 2 ).

Clearly span(u 1 ,...,u m ,v 1 ,...,v j ,w 1 ,...,w k ) contains U 1 and U 2
and hence contains U 1 + U 2 . So to show that this list is a basis of
U 1 + U 2 we need only show that it is linearly independent. To prove
this, suppose

a 1 u 1 +· · ·+ a m u m + b 1 v 1 +· · ·+ b j v j + c 1 w 1 + ··· + c k w k = 0,

where all the a’s, b’s, and c’s are scalars. We need to prove that all the
a’s, b’s, and c’s equal 0. The equation above can be rewritten as

c 1 w 1 +· · ·+ c k w k =−a 1 u 1 −· · ·− a m u m − b 1 v 1 −· · ·− b j v j ,

which shows that c 1 w 1 +···+c k w k ∈ U 1 . All the w’s are in U 2 , so this
implies that c 1 w 1 +· · ·+ c k w k ∈ U 1 ∩ U 2 . Because (u 1 ,...,u m ) is a
basis of U 1 ∩ U 2 , we can write

c 1 w 1 +· · ·+ c k w k = d 1 u 1 + ··· + d m u m