Page 197 - DMTH502_LINEAR_ALGEBRA

Unit 16: Direct Sum Decompositions of Elementary Canonical Forms

Projections can be used to describe direct-sum decompositions of the space V. For, suppose Notes
V = W    W . For each j we shall define an operator E on V. Let  be in V, say  =  + 
1 k j 1
+  with  in W . Define E  = . Then E is a well-defined rule. It is easy to see that E is linear,
k i i j j j j
2
that the range of E is W , and that E = E . The null space of E is the subspace
j j j j j
(W +  + W + W +  + W )
1 j–1 j+1 k
for, the statement that E  = 0 simply means  = 0, i.e., that  is actually a sum of vectors from the
j j
spaces W with i  j. In terms of the projection E we have
i j
 = E  +  + E 
1 k
for each  in V. What (1) says is that
I = E +  + E
1 k
Note also that if i  j, then E E = 0, because the range of E is the subspace W which is contained
i j j j
in the null space of E . We shall now summarize our findings and state and prove a converse.
i
Theorem 1: If V = W    W , then there exist k linear operators E ,..., E on V such that
1 k 1 k
2
(i) each E is a projection (E  E );
i 1 i
(ii) E E = 0, if i  j;
i j
(iii) I = E +  + E ;
1 k
(iv) the range of E is W .
i i
Conversely, if E ,..., E are k linear operators on V which satisfy conditions (i), (ii) and (iii), and
1 k
if we let W be the range of E , then V = W    W .
i i i k
Proof: We have only to prove the converse statement. Suppose E ,..., E are linear operators on V
1 k
which satisfy the first three conditions, and let W be the range of E . Then certainly
i i
V = W +  + W ;
1 k
for, by condition (iii) we have
 = E  +  + E 
1 k
for each  in V, and E  is in W . This expression for  is unique, because if
i i
 =  +  + 
1 k
with  in W , say  = E  , then using (i) and (ii) we have
i i i i i
k
E  =  E 
j j i
i 1
k
=  E E i
i
j
i 1
2
= E  j
j
= E 
j j
= 
j
This shows that V is the direct sum of the W .
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