12—Tensors 365The two expressions ( 19 ) and ( 20 ) represent the same thing, so equate the coefficients ofˆe 1 and ofˆe 2 :
T 11 +T 12 =T 11 ′ −T 21 ′
T 21 +T 22 =T 11 ′ +T 21 ′.Solve these simultaneous equations forT 11 ′ andT 21 ′ get
T 11 ′ =
1
2
[
T 11 +T 12 +T 21 +T 22
]
=
3
2
, T 21 ′ =
1
2
[
T 21 +T 22 −T 11 −T 12
]
=
1
2
. (21)
Repeat the process for^11 T(ˆe′ 2 ):
1
1 T(ˆe′
2 ) =1
1 T(
ˆe 2 −ˆe 1
√
2)
=
1
√
2
[ 1
1 T(ˆe^2 )−1
1 T(ˆe^1 )]
=
1
√
2
[
T 12 ˆe 1 +T 22 ˆe 2 −T 11 ˆe 1 −T 21 ˆe 2]
.
The left side of this is
T 12 ′ˆe′ 1 +T 22 ′ˆe′ 2 =1
√
2
[
T 12 ′(ˆe 1 +ˆe 2 ) +T 22 ′(ˆe 2 −eˆ 1 )]
.
Comparing coefficients, you have
T 12 −T 11 =T 12 ′ −T 22 ′ , T 22 −T 21 =T 12 ′ +T 22 ′.Solve forT 12 ′ andT 22 ′:
T 12 ′ =
1
2
[
T 12 −T 11 +T 22 −T 21
]
=−
1
2
, T 22 ′ =
1
2
[
T 22 −T+ 21−T 12 +T 11
]
= +
1
2
.
So, the matrix of components in the primed basis is
(
T 11 ′ T 12 ′
T 21 ′ T 22 ′