Inversion of a matrix equation

Question

Is there a general way to invert (solve for $u$) this? $$\sum_{ij}R_{ijk}a_iu_j = -x_k$$ With $a,u,x \in \mathbb{R}^N$. $R_{ijk}$ is symmetric in the last two indices. So really I'm trying to invert this: $$A= \begin{bmatrix} a'R_1 \\ a'R_2 \\ ...\\ a'R_N \end{bmatrix}$$ Could there be a nice formula for this inverse if all the $R_i$ are invertible? If you multiply $A$ by a matrix that has as columns $R_i^{-1}a$ you get that it's equal to $a'aI_n+ E$ where $E$ is traceless (it has $0$ in every element of the diagonal).

Answer 1

I gave the bounty to greg because of really nice matrix calculus (and helpfulness in general). What I was looking for precisely was something along the lines of this formula: $$A^{-1} = \frac{1}{|A|}\text{adj} A$$ And from this I can say that the numerator is a polynomial in $a$ of degree $n-1$ and the denominator a polynomial of degree $n$.

Answer 2

$ \def\p{\partial} \def\grad#1#2{\frac{\p #1}{\p #2}} \def\R{{\cal R}} \def\qif{\quad\iff\quad} \def\Si{S^{-1}} \def\qiq{\quad\implies\quad} \def\Sp{S^{\bf+}} \def\Sh{{\widehat S}} \def\LR#1{\left(#1\right)} \def\gradLR#1#2{\LR{\grad{#1}{#2}}} \def\c#1{\color{red}{#1}} $Contracting the vector $a$ with the first component of the third-order tensor $\R$ yields $$ S=a\cdot\R \qif S_{jk} = \sum_i\,a_i\,\R_{ijk} $$ The fact that $\R$ is symmetric in its last two $(jk)$ indices means that $S$ is a symmetric matrix.

This leaves a symmetric linear system to be solved for $u$ $$x=-S\cdot u \qiq u=-\Si x$$

Introducing the indexed matrices $\left(R_k = \R_{ijk}\right)\,$ seems like an unnecessary complication.

${\sf NB\!:\: If}\;S^{-1}$ does not exist, then there is no solution and you must settle for a least-squares approximation via the pseudoinverse $S^{\bf+}$ and an arbitrary vector $b$ $$u=-\Sp x \;+\; \LR{I-\Sp S}b$$

Update

Although not stated in the question, your comments indicate that you're also interested in the dependence of the solution $u$ on the other vectors in the problem.

For $x$ this is straightforward $$ \c{\grad ux = -\Si} $$ For $a$ it is more complicated. First we'll need the effect of changes to $S$ on $\Si$ $$\eqalign{ S\Si &= I \qiq S\,d\Si + dS\,\Si = 0 \\ d\Si &= -\Si\,dS\,\Si \\ }$$ The effect on $u$ is therefore $$\eqalign{ du &= -d\Si x \\ &= +\LR{\Si\,dS\,\Si}x \\ \grad u{a_k} &= \Si\gradLR S{a_k}\Si x \\ }$$ Next, we need to interpret the component-wise gradient of $S$.

Define the indexed matrix $\Sh_i$ with components $$\eqalign{ \LR{\Sh_i}_{jk} &= \R_{ijk} \\ }$$ Use these to construct and differentiate the $S$ matrix itself $$\eqalign{ S &= a_1\Sh_1 + a_2\Sh_2 + \ldots \qiq \grad S{a_k} &= \Sh_k \\ }$$ Substituting yields the desired gradient $$ \c{\grad u{a_k} = \Si \Sh_k \Si x} $$ The component-wise gradient of $S$ also gives us $$ d\Si = -\Si\,dS\,\Si \qiq\grad\Si{a_k} = -\Si \Sh_k \Si $$