I'm reading a paper about tensors and I come across a form like this. A and B are second order tensors, and k is a scalar. 1 is a 2nd order symmetric unit tensor, while I is a 4th order symmetric unit tensor.
$A:B = (1)Trace(B) + kB \implies A = 1\otimes1 + kI$
I am not sure how this implication works. Here is what I tried so far.
$A:B = (1)Trace(B) + kB = Trace(1\;B) + kB = B:1 + kB$
How would I obtain $A = 1\otimes1 + kI$ here?
Similar to "Inverse" of tensor product but with more details. The answer over there couldn't help me.
If $A:B=(kB+\ldots)$ then $A$ must be a 4th order tensor.
Given that, there are 3 extremely common and useful 4th order isotropic tensors $$\eqalign{ {\mathcal S}_{ijkl} &= \delta_{ik}\,\delta_{jl} \cr {\mathcal T}_{ijkl} &= \delta_{il}\,\delta_{jk} \cr {\mathcal P}_{ijkl} &= \delta_{ij}\,\delta_{kl} \cr }$$ Contracting an arbitrary matrix $B$ with each yields $$\eqalign{ B:{\mathcal S}&={\mathcal S}:B &= B \cr B:{\mathcal T}&={\mathcal T}:B &= B^T \cr B:{\mathcal P}&={\mathcal P}:B &= I\,\,{\rm tr}(B) \cr }$$ For your problem, we obtain $$\eqalign{ A = {\mathcal P} + k{\mathcal S} \cr }$$