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- If it's a physical quantity, like stress,
then it's usually called a
*tensor*. If it's not a physical quantity, then it's usually called a*matrix*. - The vast majority of tensors are symmetric. One common quantity that is not symmetric, and not referred to as a tensor, is a rotation matrix.
- Tensors are in fact any physical quantity that can be represented by a scalar, vector, or matrix. Zero-order tensors, like mass, are called scalars, while 1st order tensors are called vectors. Examples of higher order tensors include stress, strain, and stiffness tensors.
- The order, or rank, of a matrix or tensor is the number of subscripts it contains. A vector is a 1st rank tensor. A 3x3 stress tensor is 2nd rank.
- Coordinate Transformations of tensors are discussed in detail here.

\[ {\bf I} = \left[ \matrix{ 1&0&0 \\ 0&1&0 \\ 0&0&1 } \right] \]

Multiplying anything by the identity matrix is like multiplying by one.

Follow this link for an entertaining discussion between someone who gets it, and someone else who doesn't.

\[ \int_{-\infty}^{\infty} f(t) \delta_{(t)} dt = f(0) \]

\[ \begin{eqnarray} | {\bf A} | & = & \left| \matrix { A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} } \right| \\ \\ & = & A_{11} ( A_{22} A_{33} - A_{23} A_{32} ) + \\ & & A_{12} ( A_{23} A_{31} - A_{21} A_{33} ) + \\ & & A_{13} ( A_{21} A_{32} - A_{22} A_{31} ) \end{eqnarray} \]

If the determinant of a tensor, or matrix, is zero, then it does not have an inverse.

\[ \text{det}( {\bf A} ) \; = \; \epsilon_{ijk} A_{i1} A_{j2} A_{k3} \; = \; {1 \over 6} \epsilon_{ijk} \epsilon_{rst} A_{ir} A_{js} A_{kt} \]

\[ \text{det}( {\bf A} \cdot {\bf B} ) = \text{det}( {\bf A} ) * \text{det}( {\bf B} ) \]

The determinant of a deformation gradient gives the ratio of initial to final volume of a differential element.

\[ {\bf A} \cdot {\bf A}^{\!-1} = {\bf A}^{\!-1} \cdot {\bf A} = {\bf I} \]

If \({\bf B}\) is the inverse of \({\bf A}\), then

\[ \begin{eqnarray} B_{11} & = & (A_{22} A_{33} - A_{23} A_{32} ) \; / \; \text{det}({\bf A}) \\ B_{12} & = & (A_{13} A_{32} - A_{12} A_{33} ) \; / \; \text{det}({\bf A}) \\ B_{13} & = & (A_{12} A_{23} - A_{13} A_{22} ) \; / \; \text{det}({\bf A}) \\ B_{21} & = & (A_{23} A_{31} - A_{21} A_{33} ) \; / \; \text{det}({\bf A}) \\ B_{22} & = & (A_{11} A_{33} - A_{13} A_{31} ) \; / \; \text{det}({\bf A}) \\ B_{23} & = & (A_{13} A_{21} - A_{11} A_{23} ) \; / \; \text{det}({\bf A}) \\ B_{31} & = & (A_{21} A_{32} - A_{22} A_{31} ) \; / \; \text{det}({\bf A}) \\ B_{32} & = & (A_{12} A_{31} - A_{11} A_{32} ) \; / \; \text{det}({\bf A}) \\ B_{33} & = & (A_{11} A_{22} - A_{12} A_{21} ) \; / \; \text{det}({\bf A}) \\ \end{eqnarray} \]

\[ \left[ \matrix { 1 & 2 & 3 \\ 4 & 2 & 2 \\ 2 & 3 & 4 } \right] \left[ \matrix { \;\;\;1 & \;\;\;0.5 & -1 \\ -6 & -1 & \;\;\;5 \\ \;\;\;4 & \;\;\;0.5 & -3 } \right] = \left[ \matrix { \;\;\;1 & \;\;\;0.5 & -1 \\ -6 & -1 & \;\;\;5 \\ \;\;\;4 & \;\;\;0.5 & -3 } \right] \left[ \matrix { 1 & 2 & 3 \\ 4 & 2 & 2 \\ 2 & 3 & 4 } \right] = \left[ \matrix { 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 } \right] \]

The inverse can be calculated using

\[ A^{-1}_{ij} = {1 \over 2 \, \text{det} ({\bf A}) } \epsilon_{jmn} \, \epsilon_{ipq} A_{mp} A_{nq} \]

\[ {\bf{A}}^{\!-T} = \left( {\bf{A}}^{\!-1} \right)^{\!T} = \left( {\bf{A}}^T \right)^{\!-1} \]

\[ \left( {\bf{A}}^{\!-1} \right)^{T} : \qquad \left[ \matrix { 1 & 2 & 3 \\ 4 & 2 & 2 \\ 2 & 3 & 4 } \right] \qquad \overrightarrow{\text{Inverse}} \qquad \left[ \matrix { \;\;\; 1 & \;\;\;\; 0.5 & -1 \\ -6 & -1 & \;\; 5 \\ \;\; 4 & \;\;\;\; 0.5 & -3 } \right] \qquad \overrightarrow{\text{Transpose}} \qquad \left[ \matrix { \;\;\;\; 1 & -6 & \;\;\;\; 4 \\ \;\;\;\; 0.5 & -1 & \;\;\;\; 0.5 \\ \;-1 & \;\;\; 5 & \;-3 } \right] \]

\[ \left( {\bf{A}}^{T} \right)^{\!-1} : \qquad \left[ \matrix { 1 & 2 & 3 \\ 4 & 2 & 2 \\ 2 & 3 & 4 } \right] \qquad \overrightarrow{\text{Transpose}} \qquad \qquad \left[ \matrix { 1 & 4 & 2 \\ 2 & 2 & 3 \\ 3 & 2 & 4 } \right] \qquad \qquad \overrightarrow{\text{Inverse}} \qquad \left[ \matrix { \;\;\;\; 1 & -6 & \;\;\;\; 4 \\ \;\;\;\; 0.5 & -1 & \;\;\;\; 0.5 \\ \;-1 & \;\;\; 5 & \;-3 } \right] \]

So \(\left( {\bf{A}}^{\!-1} \right)^T\) does indeed equal \(\left( {\bf{A}}^T \right)^{\!-1}\).

\[ C_{ij} = A_{ij} + B_{ij} \]

\[ C_{ij} = A_{ik} B_{kj} \]

(Note that no dot is used in tensor notation.) The \(k\) in both factors automatically implies

\[ C_{ij} = A_{i1} B_{1j} + A_{i2} B_{2j} + A_{i3} B_{3j} \]

which is the i

\[ C_{23} = A_{21} B_{13} + A_{22} B_{23} + A_{23} B_{33} \]

\[ \left[ \matrix { 1 & 2 & 3 \\ 4 & 2 & 2 \\ 2 & 3 & 4 } \right] \left[ \matrix { 1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 9 } \right] = \left[ \matrix { 14 & 32 & 50 \\ 14 & 38 & 62 \\ 20 & 47 & 74 } \right] \]

subroutine aa_dot_bb(n,a,b,c) dimension a(n,n), b(n,n), c(n,n) do i = 1,n do j = 1,n c(i,j) = 0 do k = 1,n c(i,j) = c(i,j) + a(i,k) * b(k,j) end do end do end do return end

\[ {\bf A} \cdot {\bf B} \ne {\bf B} \cdot {\bf A} \]

\[ \text{then } {\bf A} \cdot {\bf B} = \left[ \matrix { 1 & 2 & 3 \\ 4 & 2 & 2 \\ 2 & 3 & 4 } \right] \left[ \matrix { 1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 9 } \right] = \left[ \matrix { 14 & 32 & 50 \\ 14 & 38 & 62 \\ 20 & 47 & 74 } \right] \]

\[ \text{but } \; {\bf B} \cdot {\bf A} = \left[ \matrix { 1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 9 } \right] \left[ \matrix { 1 & 2 & 3 \\ 4 & 2 & 2 \\ 2 & 3 & 4 } \right] = \left[ \matrix { 31 & 31 & 39 \\ 38 & 38 & 48 \\ 45 & 45 & 57 } \right] \]

So it is clear that \({\bf A} \cdot {\bf B}\) is not equal to \({\bf B} \cdot {\bf A}\).

Note that the "in reverse order" is critical. This is used extensively in the sections on deformation gradients and Green strains.

\[ ( {\bf A} \cdot {\bf B} )^T = {\bf B}^T \cdot {\bf A}^T \qquad \text{and} \qquad ( {\bf A} \cdot {\bf B} )^{-1} = {\bf B}^{-1} \cdot {\bf A}^{-1} \]

This also applies to multiple products. For example

\[ ( {\bf A} \cdot {\bf B} \cdot {\bf C} )^T = {\bf C}^T \cdot {\bf B}^T \cdot {\bf A}^T \]

then \( {\bf A} \cdot {\bf B} = \left[ \matrix { 1 & 2 & 3 \\ 4 & 2 & 2 \\ 2 & 3 & 4 } \right] \left[ \matrix { 1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 9 } \right] = \left[ \matrix { 14 & 32 & 50 \\ 14 & 38 & 62 \\ 20 & 47 & 74 } \right] \)

and \( ( {\bf A} \cdot {\bf B} )^T = \left[ \matrix { 14 & 14 & 20 \\ 32 & 38 & 47 \\ 50 & 62 & 74 } \right] \)

For comparison, \( {\bf B}^T \cdot {\bf A}^T = \left[ \matrix { 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 } \right] \left[ \matrix { 1 & 4 & 2 \\ 2 & 2 & 3 \\ 3 & 2 & 4 } \right] = \left[ \matrix { 14 & 14 & 20 \\ 32 & 38 & 47 \\ 50 & 62 & 74 } \right] \)

then \( {\bf A} \cdot {\bf A}^T = \left[ \matrix { 1 & 2 & 3 \\ 4 & 2 & 2 \\ 2 & 3 & 4 } \right] \left[ \matrix { 1 & 4 & 2 \\ 2 & 2 & 3 \\ 3 & 2 & 4 } \right] = \left[ \matrix { 14 & 14 & 20 \\ 14 & 24 & 22 \\ 20 & 22 & 29 } \right] \)

and \( {\bf A}^T \cdot {\bf A} = \left[ \matrix { 1 & 4 & 2 \\ 2 & 2 & 3 \\ 3 & 2 & 4 } \right] \left[ \matrix { 1 & 2 & 3 \\ 4 & 2 & 2 \\ 2 & 3 & 4 } \right] = \left[ \matrix { 21 & 16 & 19 \\ 16 & 17 & 22 \\ 19 & 22 & 29 } \right] \)

Both results are indeed symmetric, although they are unrelated to each other.

\[ {\bf A} : {\bf B} = A_{ij} B_{ij} \]

Since the \(i\) and \(j\) subscripts appear in both factors, they are both summed to give

\[ \matrix { {\bf A} : {\bf B} \; = \; A_{ij} B_{ij} \; = & A_{11} * B_{11} & + & A_{12} * B_{12} & + & A_{13} * B_{13} & + \\ & A_{21} * B_{21} & + & A_{22} * B_{22} & + & A_{23} * B_{23} & + \\ & A_{31} * B_{31} & + & A_{32} * B_{32} & + & A_{33} * B_{33} & } \]

\[ \matrix { {\bf A} : {\bf B}\; & = & 1 * 1 & + & 2 * 4 & + & 3 * 7 & + \\ & & 4 * 2 & + & 2 * 5 & + & 2 * 8 & + \\ & & 2 * 3 & + & 3 * 6 & + & 4 * 9 \\ & \\ & = & 124 } \]

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