12—Tensors 300Now to take an example and tear it apart. Define a tensor by the equationsT(xˆ) =ˆx+ˆy, T(ˆy) =ˆy, (12.17)
wherexˆandyˆare given orthogonal unit vectors. These two expressions, combined with linearity, suffice
to determine the effect of the tensor on all linear combinations ofˆxandyˆ. (This is a two dimensional
problem.)
To compute the components of the tensor pick a set of basis vectors. The obvious ones in this
instance are
ˆe 1 =x,ˆ and ˆe 2 =yˆ
By comparison with Eq. (12.13), you can read off the components ofT.
T 11 = 1 T 21 = 1
T 12 = 0 T 22 = 1.
Write these in the form of a matrix as in Eq. (12.15)
(Trow,column
)
=
(
T 11 T 12
T 21 T 22
)
=
(
1 0
1 1
)
and writing the vector components in the same way, the components of the vectorsˆxandˆyare
respectively (
1
0
)
and(
0
1
)
The original equations (12.17), that defined the tensor become the components
(
1 0
1 1)(
1
0
)
=
(
1
1
)
and(
1 0
1 1
)(
0
1
)
=
(
0
1
)
12.3 Relations between Tensors
Go back to the fundamental representation theorem for linear functionals and see what it looks like in
component form. Evaluatef(~v), where~v=vieˆi. (The linear functional has one vector argument and
a scalar output.)
f(~v) =f(viˆei) =vif(ˆei) (12.18)
Denote the set of numbersf(ˆei)(i= 1, 2 , 3 ) byAi=f(eˆi), in which case,
f(~v) =Aivi=A 1 v 1 +A 2 v 2 +A 3 v 3
Now it is clear that the vectorA~of the theorem is just
A~=A 1 eˆ 1 +A 2 ˆe 2 +A 3 ˆe 3 (12.19)
Again, examine the problem of starting from a bilinear functional and splitting off one of the two
arguments in order to obtain a vector valued function of a vector. I want to say