Page 31 - Linear Algebra Done Right

Sums and Direct Sums
First suppose that V = U 1 ⊕· · · U n . Clearly (a) holds
Proof:
(because of how sum and direct sum are defined). To prove (b), suppose
that u 1 ∈ U 1 ,...,u n ∈ U n and 17
0 = u 1 +· · ·+ u n .
Then each u j must be 0 (this follows from the uniqueness part of the
definition of direct sum because 0 = 0+· · ·+0 and 0 ∈ U 1 ,..., 0 ∈ U n ),
proving (b).
Now suppose that (a) and (b) hold. Let v ∈ V. By (a), we can write

v = u 1 +· · ·+ u n

for some u 1 ∈ U 1 ,...,u n ∈ U n . To show that this representation is
unique, suppose that we also have

v = v 1 + ··· + v n ,

where v 1 ∈ U 1 ,...,v n ∈ U n . Subtracting these two equations, we have

0 = (u 1 − v 1 ) +· · ·+ (u n − v n ).
Clearly u 1 − v 1 ∈ U 1 ,...,u n − v n ∈ U n , so the equation above and (b)
imply that each u j − v j = 0. Thus u 1 = v 1 ,...,u n = v n , as desired.

The next proposition gives a simple condition for testing which pairs Sums of subspaces are
of subspaces give a direct sum. Note that this proposition deals only analogous to unions of
with the case of two subspaces. When asking about a possible direct subsets. Similarly,
sum with more than two subspaces, it is not enough to test that any direct sums of
two of the subspaces intersect only at 0. To see this, consider the subspaces are
nonexample presented just before 1.8. In that nonexample, we had analogous to disjoint
3
3
F = U 1 + U 2 + U 3 , but F did not equal the direct sum of U 1 ,U 2 ,U 3 . unions of subsets. No
two subspaces of a
However, in that nonexample, we have U 1 ∩U 2 = U 1 ∩U 3 = U 2 ∩U 3 ={0}
(as you should verify). The next proposition shows that with just two vector space can be
subspaces we get a nice necessary and sufficient condition for a direct disjoint because both
sum. must contain 0.So
disjointness is
1.9 Proposition: Suppose that U and W are subspaces of V. Then replaced, at least in the
V = U ⊕ W if and only if V = U + W and U ∩ W ={0}. case of two subspaces,
with the requirement
Proof: First suppose that V = U ⊕ W. Then V = U + W (by the that the intersection
definition of direct sum). Also, if v ∈ U ∩ W, then 0 = v + (−v), where equals {0}.