62 3. Tensor ProductsBoth of these propositions are simple applications of universality. For
example, if cp: X -+ C, 'l/J: Y -+ C are linear maps, thenXx Y-+ C, (x, y) c---t cp(x)'l/J(y)
is obviously a bilinear map.
An important special case which deserves highlighting is that of func-
tionals.Corollary 3 .1. 6 (Tensor product functionals). If cp : X -+ re, 'l/J : Y -+ re
are linear functionals, then there is a unique linear functional_cp8'l/J: X8Y-+ re
such that cp 8 'l/J(x 0 y) = cp(x)'lf.J(y) for all x EX, y E Y.
We have mixed notation in the corollary above both to cause confusion
and to see if you're paying attention. Luckily it is justified by the fact that
the map
re 8 re -+ re, a 0 {3 f-t a{J
is an isomorphism; hence the functional case follows either from Proposition
3.1.4 or Proposition 3.1.5.
Though we only need it once, the following corollary should be mentioned
now. It follows from the previous result because if cp: X -+ re is a conjugate
linear functional (i.e., scalars come out with complex conjugates), then the
map x c---t cp(x) is linear.Corollary 3.1. 7 (Conjugate linear functionals). If cp: X-+ re and 'l/J: Y-+
re are conjugate linear functionals, there is a conjugate linear tensor product
functional cp 8 'l/J: x 8 y-+ re such that cp 8 'l/J(x 0 y) = cp(x)'l/J(y) for all
x EX, y E Y.
Unfortunately, it is not always easy to decide when a set of elementary
tensors is linearly independent. Here is a useful sufficient condition, however.
Proposition 3.1.8 (Linear independence). If {x1, ... , xn} c X are linearly
independent, {y1, ... , Yn} CY are arbitrary and
n
0 = L Xi 0 Yi E x 8 Y,
i=l
then Yl = Y2 = · · · = Yn = 0.
The proof of this becomes short and sweet if we use a little tensor product
functional trick. Namely, let {cp1, ... , cpn} C X* be a dual set of functionals