I have an equation that involves
$\sum_{abcd} x_ax_bx_cx_d M^{abcd}+\sum_{ab}x_ax_bN^{ab}=0$ where M, N are known.
M has the property that $M^{abcd}=M^{cdab}$. N is symmetric in its indices.
I would like to do something like this (although I think it is not formal):
Define matrix $X$ with elements $X_{ab} \equiv x_a x_b$. Unpack elements in a vector $\vec{X}$ so that I have indices $\alpha \equiv (ab)$.
Now write $M^{abcd}\equiv \mathcal{M}^{\alpha\beta}$. In some sense I am defining a matrix from a tensor like quantity.
Finally I pack $N^{ab}$ in a vector $\vec{N}$ with elements $N_{\alpha}$.
Now my equation becomes:
$\vec{x}^T\mathcal{M}\vec{x}+\vec{x}^T\vec{N}=0\ (1)$
(I also have some extra condition on this, but for the sake of the question I omit it as I am interested in this part)
First question:
- Is equation (1) legitimate? I feel that if I unpack everything I should reobtain the initial equation.
Second question:
- Does it make sense to take the inverse of this matrix coming from a tensor?
Sorry for this naive questions, I've never encountered a problem like this and wanted to make sure these operation would make sense.
I think that everything here is completely legitimate. If you'd like to connect to existing techniques and terminology, what you are doing to the vectors $X$ and $M$ is called tensor reshaping (or more specifically, vectorization and matricization).
Following the notation of the linked article: suppose that $M$ has size $n \times n \times n \times n$, and $N$ has size $n \times n$. If $\mu:[n] \times [n] \to [n^2]$ is a bijection, then $$ \vec x_{\mu(a,b)} = X_{ab} = x_ax_b, \quad \mathcal M^{\mu(a,b)\mu(c,d)}= M^{abcd}, \quad \vec N_{\mu(a,b)} = N^{ab}. $$ In terms of some other notation: if the standard (lexicographical) choice of $\mu$ is taken, then the resulting $X$ can be written as $X = (x \otimes x)(x \otimes x)^T$, where $x$ is a column-vector, $T$ denotes a transpose, and $\otimes$ denotes a Kronecker product.
Regarding your second question: it makes sense to take the inverse of $\mathcal M$ (assuming this inverse exists) in the same way that it makes sense to take the inverse of any square matrix. It is unclear, however, whether there is any meaningful connection to be made between the resulting inverse $\mathcal M^{-1}$ and the original tensor $M$.