Mathematical Tools for Physics

12—Tensors 369

for all vectors~uand~v. You should see that using the same symbol,T, for both functions doesn’t cause any
trouble. Given the bilinear functional, what is the explicit form forT(~v)? The answer is most readily found by a
bit of trial and error until you reach the following result:

T(~v) =ˆeiT(ˆei, ~v). (27)

(Remember the summation convention.) To verify this relation, multiply by an arbitrary vector,~u=ujˆej:

~u.T(~v) = (ujˆej).ˆeiT(ˆei, ~v),

which is, by the orthonormality of theˆe’s,

ujδjiT(ˆei, ~v) =uiT(eˆi, ~v) =T(~u, ~v).

This says that the above expression is in fact the correct one. Notice also the similarity between this construction
and the one in equation ( 26 ) forA~.
Now takeT(~v)from Eq. ( 27 ) and express~vin terms of its components

~v=vjˆej, then T(~v) =ˆeiT(ˆei,vjˆej) =eˆiT(eˆi,ˆej)vj.

Theicomponent of this expression is
T(ˆei,ˆej)vj=Tijvj,

a result already obtained in Eq. ( 16 ).
There’s a curiosity involved here; why should the left hand entry inT(, )be singled out to construct

ˆeiT(ˆei,~v)?

Why not use the right hand one instead? Answer: No reason at all. It’s easy enough to find out what happens
when you do this. Examine
eˆiT(~v,ˆei)≡T ̃(~v). (28)

Put~v=vjˆej, and you get
ˆeiT(vjˆej,ˆei) =ˆeiT(ˆej,ˆei)vj.