is expressed uniquely as x ¼ x 1 + x 2 . Conversely, suppose the vector representation
(x ¼ x 1 + x 2 ) is unique and x 2 W 1 \ W 2 . Then, x ¼ x + 0 = 0 + x; x,0 2 W 1 and
x,0 2 W 2 . Uniqueness of the representation implies that x ¼ 0. Consequently,
W 1 \ W 2 ¼ {0} follows.
In case W 1 \ W 2 ¼ {0}, V ¼ W 1 + W 2 is said to be a direct sum of W 1 and W 2 or
we say that V is decomposed into a direct sum of W 1 and W 2 . We symbolically
denote this by
V ¼ W 1 ⨁W 2 :
ð11:17Þ
In this case, the following equality holds:
dim V ¼ dimW 1 þ dimW 2 ,
ð11:18Þ
where “dim” stands for dimension of the vector space considered. To prove (11.18),
we suppose that V is a n-dimensional vector space and that W 1 and W 2 are spanned
by r 1 and r 2 linearly independent vectors, respectively, such that
W 1 ¼ Span e
1
ð Þ
1 , e
1
ð Þ
2 , Á Á Á, e
1
ð Þ
r 1
n
o
and W 2 ¼ Span e
2
ð Þ
1 , e
2
ð Þ
2 , Á Á Á, e
2
ð Þ
r 2
n
o
:
ð11:19Þ
This is equivalent to that dimension of W 1 and W 2 is r 1 and r 2 , respectively. If
V ¼ W 1 + W 2 (here we do not assume that the summation is a direct sum), we have
V ¼ Span e
1
ð Þ
1 , e
1
ð Þ
2 , Á Á Á, e
1
ð Þ
r 1
, e
2
ð Þ
1 , e
2
ð Þ
2 , Á Á Á, e
2
ð Þ
r 2
n
o
:
ð11:20Þ
Then, we have n r 1 + r 2 . This is almost trivial. Suppose r 1 + r 2 < n. Then, these
(r 1 + r 2 ) vectors cannot span V, but we need additional vectors for all vectors
including the additional vectors to span V. Thus, we should have n
r 1 + r 2
accordingly. That is,
dim V dimW 1 þ dimW 2 :
ð11:21Þ
Now, let us assume that V ¼ W 1 ⨁ W 2 . Then, e
2
ð Þ
i
1 i r 2
ð
Þmust be linearly
independent of e
1
ð Þ
1 , e
1
ð Þ
2 , Á Á Á, e
1
ð Þ
r 1
. If not, e
2
ð Þ
i could be described as a linear combination of e
1
ð Þ
1 , e
1
ð Þ
2 , Á Á Á, e
1
ð Þ
r 1
. But, this would imply that e
2
ð Þ
i 2 W 1 , i.e., e
2
ð Þ
i 2 W 1 \ W 2 ,
in contradiction to that we have V ¼ W 1 ⨁ W 2 . It is because W 1 \ W 2 ¼ {0} by
assumption. Likewise, e
1
ð Þ
j
1 j r 1
ð
Þis linearly independent of e
2
ð Þ
1 , e
2
ð Þ
2 , Á Á Á, e
2
ð Þ
r 2
.
Hence, e
1
ð Þ
1 , e
1
ð Þ
2 , Á Á Á, e
1
ð Þ
r 1
, e
2
ð Þ
1 , e
2
ð Þ
2 , Á Á Á, e
2
ð Þ
r 2
must be linearly independent. Thus,
n ! r 1 + r 2 . This is because in the vector space V we may well have additional
vector(s) that are independent of the above (r 1 + r 2 ) vectors. Meanwhile, n r 1 + r 2
from the above. Consequently, we must have n ¼ r 1 + r 2 . Thus, we have proven that
11.1 Vectors
437