Page 396 - Advanced Linear Algebra

380 Advanced Linear Algebra

To be absolutely clear, we have two distinct vector spaces: the - -space
defined by the tensor product and the 2-space
-
2
> ~ 2n = - > ~ 2n = -
with scalar multiplication by elements of defined as absorption into the first
2
coordinate. The spaces > and > - 2 are identical as sets and as abelian groups.
It is only the “permission to multiply by” that is different. Accordingly, we can
recover > from > - 2 simply by restricting scalar multiplication to scalars from
-.
-
Thus, we can speak of “ -linear” maps from - = into > 2 , with the expected
meaning, that is,
² " b #³ ~ " b #
for all scalars Á  - .

If the dimension of as a vector space over is , then
2
-

- ²> ³~ ²2 - n = ³~ dim - h - ²= ³
dim - dim -
As to the dimension of > , it is not hard to see that if 2 ¸ ¹ is a basis for - = ,
then ¸ n ¹ is a basis for > 2 . Hence

2 ²> ³ ~ - ²= ³
dim 2 dim -
The map - defined by # ~ n # is easily seen to be injective and
¢= ¦ > -
- >-linear and so - contains an isomorphic copy of - = . We can also think of

2
= into > , in which case is called the -extension map = .
as mapping - 2 of -
This map has a universal property of its own, as described in the next theorem.
Theorem 14.10 The - -linear 2 -extension map ¢ = ¦ 2 n - = - has the
- = into a -space,
2
universal property for the family of all -linear maps from -
as measured by -linear maps. Specifically, for any -linear map ¢ - = - ¦ @ ,
2
2
2
where is a -space, there exists a unique -linear map ¢ 2 n = ¦ @ - for
@
which the diagram in Figure 14.6 commutes, that is, for which
k ~
Proof. If such a 2 -linear map ¢ 2 n = - ¦ @ is to exist, then it must satisfy,
for any 2 ,
n #³ ~ ² n #³ ~ ²#³ ~ ²#³
²
This shows that if exists, it is uniquely determined by . As usual, when

searching for a linear map on a tensor product such as 2n = - , we look for a
bilinear map. The map ¢ ²2 d = ³ ¦ @ defined by
-

² Á #³ ~ ²#³