12—Tensors 363Notice how the previous choice of indices has led to the conventional result, with the first index denoting the row
and the second the column of a matrix.
Now to take an example and tear it apart. Define a tensor by the equations
1
1 T(ˆx) =ˆx+ˆy, T(yˆ) =ˆy, (17)
wherexˆandyˆare given orthogonal unit vectors. These two expressions, combined with linearity, suffice to
determine the effect of the tensor on all linear combinations ofxˆandyˆ. (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 =ˆx, and ˆe 2 =y.ˆBy comparison with Eq. ( 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. ( 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 ( 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
)
.
Change of Basis
The above exercise is fairly trivial because the definition of the tensor lent itself to a particular basis. For practice,