Page 195 - DMTH502_LINEAR_ALGEBRA

Unit 16: Direct Sum Decompositions of Elementary Canonical Forms

16.2 Direct-sum Decompositions Notes

Definition: Let W ,..., W be subspaces of the vector space V. We say that W ,..., W are independent
1 k 1 k
if
 + ... +  = 0,  in W
1 k i i
implies that each  is 0.
i
For k = 2, the meaning of independence is {0} intersection, i.e., W and W are independent if and
1 2
only if W  W = {0}. If k > 2, the independence of W ,..., W says much more than W  ...  W = {0}.
1 2 1 k 1 k
It says that each W intersects the sum of the other subspaces W only in the zero vector.
j i
The significance of independence is this. Let W = W + ... + W be the subspace spanned by W ,...,
1 k 1
W . Each vector  in W can be expressed as a sum
k
 =  + ... +  ,  in W .
1 k i i
If W ,..., W are independent, then that expression for  is unique; for if
1 k
 =  + ... +  ,  in W
1 k i i
then 0 = ( –  ) + ... + ( –  ), hence  –  = 0, i = 1,..., k. Thus, when W ,..., W are independent,
1 1 k k i i 1 k
we can operate with the vectors in W as k-tuples ( ,...,  ),  in W , in the same way as we operate
1 k i i
k
with vectors in R as k-tuples of numbers.
Lemma: Let V be a finite-dimensional vector space. Let W ,..., W be subspaces of V and let
1 k
W = W + ... + W . The following are equivalent.
1 k
(a) W ,..., W are independent.
1 k
(b) For each j, 2  j  k, we have
W  (W + ... + W ) = {0}
j 1 j–1
(c) If  is an ordered basis for W , 1  i  k, then the sequence  = ( ,...,  ) is an ordered basis
i i 1 k
for W.
Proof: Assume (a). Let  be a vector in the intersection W  (W + ... + W ). Then there are vectors
j 1 j–1
 ,...,  with  in W such that  =  + ... +  . Since
1 j–1 i i 1 j–1
 + ... +  + (–) + 0 + ... + 0 = 0
1 j–1
and since W , ..., W are independent, it must be that  =  = ... =  =  = 0.
1 k 1 2 j–1
Now, let us observe that (b) implies (a). Suppose
0 =  + ... +  ,  in W
1 k i i
Let j be the largest integer i such that   0. Then
i
0 =  + ... + ,   0.
1 j j
Thus  = – – ... –  is a non-zero vector in W  (W + ... + W ).
j 1 j–1 j 1 j–1
Now that we know (a) and (b) are the same, let us see why (a) is equivalent to (c). Assume (a). Let
 be basis for W , 1  i  k, and let  = ( ,...,  ). Any linear relation between the vectors in  will
i i 1 k
have the form
 + ... +  = 0
1 k
where  is some linear combination of the vectors in  . Since W ,.., W are independent, each
i i 1 k
 is 0. Since each  is independent, the relation we have between the vectors in  is the trivial
i i
relation.

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