2.4 Some Properties of Tensors
29
transpose of the second-order tensor T = T ij e i e j can be obtained by switching the
order of either the base vectors or the component subscripts to obtain the equivalent
forms
T
T
= T ij e j e i = T ji e i e j ,
(2.18)
where superscript T denotes the transpose.
More formally, the transpose of a tensor T satisfies the equation
b · T · a = a · T
T
· b,
(2.19)
in which a and b are arbitrary vectors. This equation implies two other relations that
will prove useful. Rearranging Eq. (2.19) and using the commutative property of the
dot product of two vectors yield b · (T · a) = (a · T T ) · b = b · (a · T T ). This and a
similar manipulation show that
T · a = a · T
T
b · T = T
T
· b.
(2.20)
Using components, it is straightforward to show that Eq. (2.18) satisfies these
equations. A second-order tensor T is symmetric if T = T T and antisymmetric
(or skew symmetric) if T = −T T .
Before proceeding, we note a subtle difference between dyadic and matrix
algebra. In matrix form, a vector can be either a column vector or a row vector.
The transpose of a column vector is a row vector, and vice versa. When a vector is
written in terms of base vectors, however, no distinction is made between column
and row vectors; hence, a T is written simply as a.
Consider, for example, the dot products T · a and a · T. In two dimensions, these
relations yield
T · a = (T ij e i e j ) · (a k e k ) = T ij a k e i (e j · e k ) = T ij a k e i δ jk = T ij a j e i
= (T 11 a 1 + T 12 a 2 )e 1 + (T 21 a 1 + T 22 a 2 )e 2
a · T = (a k e k ) · (T ij e i e j ) = T ij a k (e k · e i )e j = T ij a k δ ki e j = T ij a i e j
= (T 11 a 1 + T 21 a 2 )e 1 + (T 12 a 1 + T 22 a 2 )e 2 .
The results differ only in the terms involving T 12 and T 21 . To perform the same
calculations using matrices, the vector a must be written as a column vector/row
vector when it is dotted on the right/left side of T, respectively. For the present
problem, this gives
29
transpose of the second-order tensor T = T ij e i e j can be obtained by switching the
order of either the base vectors or the component subscripts to obtain the equivalent
forms
T
T
= T ij e j e i = T ji e i e j ,
(2.18)
where superscript T denotes the transpose.
More formally, the transpose of a tensor T satisfies the equation
b · T · a = a · T
T
· b,
(2.19)
in which a and b are arbitrary vectors. This equation implies two other relations that
will prove useful. Rearranging Eq. (2.19) and using the commutative property of the
dot product of two vectors yield b · (T · a) = (a · T T ) · b = b · (a · T T ). This and a
similar manipulation show that
T · a = a · T
T
b · T = T
T
· b.
(2.20)
Using components, it is straightforward to show that Eq. (2.18) satisfies these
equations. A second-order tensor T is symmetric if T = T T and antisymmetric
(or skew symmetric) if T = −T T .
Before proceeding, we note a subtle difference between dyadic and matrix
algebra. In matrix form, a vector can be either a column vector or a row vector.
The transpose of a column vector is a row vector, and vice versa. When a vector is
written in terms of base vectors, however, no distinction is made between column
and row vectors; hence, a T is written simply as a.
Consider, for example, the dot products T · a and a · T. In two dimensions, these
relations yield
T · a = (T ij e i e j ) · (a k e k ) = T ij a k e i (e j · e k ) = T ij a k e i δ jk = T ij a j e i
= (T 11 a 1 + T 12 a 2 )e 1 + (T 21 a 1 + T 22 a 2 )e 2
a · T = (a k e k ) · (T ij e i e j ) = T ij a k (e k · e i )e j = T ij a k δ ki e j = T ij a i e j
= (T 11 a 1 + T 21 a 2 )e 1 + (T 12 a 1 + T 22 a 2 )e 2 .
The results differ only in the terms involving T 12 and T 21 . To perform the same
calculations using matrices, the vector a must be written as a column vector/row
vector when it is dotted on the right/left side of T, respectively. For the present
problem, this gives