12—Tensors 366As a check to be sure nothing has gone wrong, it’s easy to compute the effect of^11 Ton the original vectors
ˆxandˆyin this basis. Refer to the diagram accompanying equation ( 18 ),
ˆx=ˆe′ 1 −ˆe′ 2
√
2, and ˆy=ˆe′ 1 +ˆe′ 2
√
2.
or in component form, respectively
1
√
2(
1
− 1
)
, and1
√
2
(
1
1
)
.
Operate on these column matrices with the above matrix.
(
3 / 2 − 1 / 2
1 /2 1/ 2
)
1
√
2
(
1
− 1
)
=
1
√
2
(
2
0
)
=
1
√
2
(
1
− 1
)
+
1
√
2
(
1
1
)
and(
3 / 2 − 1 / 2
1 /2 1/ 2
)
1
√
2
(
1
1
)
=
1
√
2
(
1
1
)
,
so that both of these agree with the starting point, Eq. ( 17 )
Change of Basis (more efficient)
Next, in order to illustrate the second (and much easier) method for changing bases it’s necessary to note some
further relationships between vector valued functions and bilinear functionals. Namely, I shall prove that theTij
defined by Eq. ( 13 ) satisfies
Tij=^02 T(ˆei,ˆej), (22)
where this bilinear functional is the one associated with the vector valued function by
0
2 T(~u,~v) =~u.1
1 T(~v).In order to prove this relationship, just write~uand~vin terms of their components:
~u=uiˆei, ~v=viˆei, and^11 T(~v) =vi^11 T(ˆei)
then^02 T(~u, ~v) =^02 T(uiˆei, vjˆej) =uivj^02 T(ˆei,ˆej)