Calculate $\sigma$ from the tensor equation $ \sigma + S \, [S:\sigma] = C$ I have a tensor equation:
$${\boldsymbol \sigma} + {\bf S} \, [{\bf S}:{\boldsymbol \sigma}] = {\bf C} \tag{1}$$ 
I want to calculate ${\boldsymbol \sigma}$, whereas $\bf S$ and $\bf C$ are known constant second-order tensors. The easiest try would be to :
$${\boldsymbol \sigma} + {\bf S} \, [{\bf S}:({\bf I \cdot \boldsymbol \sigma})] = {\bf C} $$
$$[\bf I + {\bf S} \, \otimes {\bf S}: \bf I ]\cdot{\boldsymbol \sigma} = {\bf C} $$ 
$${\boldsymbol \sigma} = {[\bf I + {\bf S} \, \otimes {\bf S}: \bf I ]}^{-1}\cdot {\bf C} $$ 
but when I calculated the result manually with indices, the result was different, so this approach seems definitely wrong.  
My question is: is it possible to factor out $\boldsymbol \sigma$ from the left hand side of equation $(1) $? 
Is there any alternate way to accomplish this task of calculating a tensor from such expression (which contains double contraction)?.
Edit: using fourth order identity tensor, defined by:
$$\mathbb I = {\mathbb I}^{-1}$$
and 
$$\mathbb I : \boldsymbol \sigma = \boldsymbol \sigma$$
is it correct to rewrite $(1)$ as:
$$\mathbb I :{\boldsymbol \sigma} + {\bf S} \, [{\bf S}:{\boldsymbol \sigma}] = {\bf C} $$ 
$$[\mathbb I + {\bf S} \otimes {\bf S}]:{\boldsymbol \sigma} = {\bf C} $$ 
and finally 
$${\boldsymbol \sigma} = [\mathbb I + {\bf S} \otimes {\bf S}]^{-1}:{\bf C} $$ 
?
 A: $
\def\s{\sigma}\def\o{{\tt1}}
\def\L{\left}\def\R{\right}\def\LR#1{\L(#1\R)}
\def\BR#1{\Big(#1\Big)}\def\br#1{\big(#1\big)}
$Fully contract each term of the equation with $S,\,$ then solve for the scalar $\br{S:\s}$
$$\eqalign{
C &= \s+S\BR{S:\s} \\
{S:C} &= \BR{S:\s}\BR{\o+S:S} \\
{S:\s} &= \LR{\frac{S:C}{\o+S:S}} \\\\
}$$
Substitute this result into the original equation to recover $\s$
$$\eqalign{
\s &= C-\LR{\frac{S:C}{\o+S:S}}S \\\\
}$$
Note that
$$\eqalign{
&\BR{S:S} = \big\|S\big\|_F^2\;\ge\;0 \\
&\BR{\o+S:S}\;\ge\;\o \\
}$$
so the denominator is always $\ge\o\,$ and can never equal zero.
A: If the following computations of coordinates on an orthonormal basis are correct, then
$$
\left[(\boldsymbol{\sigma} : {\bf S}){\bf S}\right]_{ij} = (\sigma_{ab}S_{ab})S_{ij} .
$$
In counterpart,
$$
\left[(({\bf S} \otimes {\bf S}) : {\bf I})\cdot \boldsymbol{\sigma}\right]_{ij} = (S_{aa})S_{ib}\sigma_{bj},
$$
which suggests that the factorization through $\boldsymbol\sigma$ is incorrect.
Let us assume that the vector space is $\Bbb R^3$. In the particular case where $\bf{S} = \alpha I$, the equation $(1)$ rewrites as
$$
\boldsymbol{\sigma} + \alpha^2 \text{tr}\boldsymbol{\sigma}\, {\bf I} = {\bf C} .
$$
Taking the trace, one has $(1+3\alpha^2)\,\text{tr}\boldsymbol{\sigma} = \text{tr}{\bf C}$, i.e.
$$
\boldsymbol{\sigma} = {\bf C} - \frac{\alpha^2\text{tr}{\bf C}}{1+3\alpha^2}\, {\bf I} ,
$$
but this works only for $\bf{S} = \alpha I$.
In the general case, I don't know if $(1)$ can be inverted under some conditions, and if there is a method to do so.
Edit: The proposed factorization looks fine. However, it may be not very useful in practice (how to compute the inverse of a fourth-order tensor with respect to the double-dot product?). One may have a look to related posts [1,2].
