Dyadic operations, fourth order tensors and Tensor algebra

Question

I am trying to understand the dyadic operation for a while since I am interested in Elasticity problems. I believe an intuitive understanding (rather than assuming) will give me good problem solving abilities.

1. The following equation is shown in almost all lecture notes on elasticity. However, I don't understand how this result is arrived. Can you actually derive this result? Or is this result rather intuitive? $$ \bf{A}-\rm{\dfrac{1}{3}}\bf(I:A)I=(\mathbb I-\rm\dfrac{1}{3}\bf(I \otimes I)):A $$
2. Also how does a fourth order Identity tensor look like? I see the following notation everywhere. $$ \bf \mathbb I= \rm \delta_{ij}\delta_{kl}\bf e_i \otimes e_j \otimes e_k \otimes e_l $$ Not being able to understand this could be attributed to my lack of understanding of the dyadic operator. Every lecture note elaborates on scalars. But not on tensors. Please show the actual matrix. It could help in understanding the operation better.
3. Also how does the dyad works between vectors and tensors? How do you numerically compute $\bf I \otimes I$ in the following $$ \bf \mathbb{E}=\rm \lambda \bf I \otimes I + \rm 2 \mu \bf\mathbb{I}^{\rm sym} $$ Ive taken the above equations from Stanford Notes

I believe I have detailed my issue. Also let me know if there are any notes or books that could detail on these trivia.

Answer 1

A dyadic product takes as input two vectors and outputs a second order tensor. This is what I know as a dyadic product, and a dyad is the term $\mathbf{a}\mathbf{b}$. A general second order tensor can be written as a linear combination of dyads.

Commonly the symbol $\otimes$ is referred as the tensor product and it outputs higher-order tensors. I think that this is easier to understand in index notation

$$A = B \otimes C\, ,$$

with $B$ and $C$ second order tensors, is a fourth order one. In index notation this is

$$A_{ijkl} = B_{ij} C_{kl}\, .$$

If you want to compute this computationally you need a 4th dimensional array. The following snippet does this

for i in range(3):
    for j in range(3):
        for k in range(3):
            for l in range(3):
                A[i, j, k, l] = B[i, j] * C[k, l]

That depends on the programming language that you are using. There might be capabilities for tensor products already built-in.

Regarding, how to visualize a fourth order tensor. This is more difficult, because it could be thought as a 4th dimensional array (for some coordinate system), and for that you need 4 dimensions. One thing that is commonly done is to represent it as 9 by 9 matrix, or as a matrix of matrices.

For example, the following is $\mathbf{I}\otimes\mathbf{I}$.

⎡⎡1  0  0⎤  ⎡0  0  0⎤  ⎡0  0  0⎤⎤
⎢⎢       ⎥  ⎢       ⎥  ⎢       ⎥⎥
⎢⎢0  1  0⎥  ⎢0  0  0⎥  ⎢0  0  0⎥⎥
⎢⎢       ⎥  ⎢       ⎥  ⎢       ⎥⎥
⎢⎣0  0  1⎦  ⎣0  0  0⎦  ⎣0  0  0⎦⎥
⎢                               ⎥
⎢⎡0  0  0⎤  ⎡1  0  0⎤  ⎡0  0  0⎤⎥
⎢⎢       ⎥  ⎢       ⎥  ⎢       ⎥⎥
⎢⎢0  0  0⎥  ⎢0  1  0⎥  ⎢0  0  0⎥⎥
⎢⎢       ⎥  ⎢       ⎥  ⎢       ⎥⎥
⎢⎣0  0  0⎦  ⎣0  0  1⎦  ⎣0  0  0⎦⎥
⎢                               ⎥
⎢⎡0  0  0⎤  ⎡0  0  0⎤  ⎡1  0  0⎤⎥
⎢⎢       ⎥  ⎢       ⎥  ⎢       ⎥⎥
⎢⎢0  0  0⎥  ⎢0  0  0⎥  ⎢0  1  0⎥⎥
⎢⎢       ⎥  ⎢       ⎥  ⎢       ⎥⎥
⎣⎣0  0  0⎦  ⎣0  0  0⎦  ⎣0  0  1⎦⎦

Answer 2

The first equality you wrote is not correct, as noted by other users. However, what I think you want to know is why $(I:A)I = (I \otimes I)A$. You can show this just by using dyads properties.

$$(B \otimes B) : A =(B_{ij}B_{kl} e_i \otimes e_j \otimes e_k \otimes e_l):(A_{mn} e_m \otimes e_n) = B_{ij}B_{kl} A_{mn} \delta_{km} \delta_{ln} e_i \otimes e_j = B_{ml} A_{ml} B_{ij}e_i \otimes e_j = (B:A) B$$