Let $T^{ij}_{i'j'}$ be a 4-tensor, $1 \leq i,i' \leq n$, $1 \leq j,j' \leq m$, $n$ and $m$ may differ. That is, the first (second) covariant coordinate is of the same dimension as the first (second) contravariant one (respectively).
We have two possible contractions:
- $\Sigma_{i=i'}T^{ij}_{i'j'}$, which is a $m$-by-$m$ matrix.
- $\Sigma_{j=j'}T^{ij}_{i'j'}$, which is a $n$-by-$n$ matrix.
Question: Is there any relation between the eigenvalues of (1) and (2)? Between their ranks? Or between any other linear-algebraic invariants?
As Qiaochu Yuan pointed out, it makes sense to consider tensors of the form $T=A\otimes B$, (i.e. $T_{i'j'}^{ij}=A_{i'}^i B_{j'}^j$). An arbitrary tensor is always given as a finite sum of such so called pure tensors. In particular you find all interesting relations by just considering this subclass.
The two contractions you've mentioned can be expressed as $C= \text{tr}A\cdot B $ and $D= A\cdot \text{tr}B$ in a coordinate free way. This way you see that $\text{tr}C =\text{tr}A\text{tr}B = \text{tr}D $.
This is also the only relation of the spectra of $C$ and $D$ in the following sense: Given any collection of complex numbers $\lambda_1,...,\lambda_n,\mu_1,...,\mu_m$ with $\lambda_1+...+\lambda_n=\mu_1+...+\mu_m$ there is a tensor $T$ such that its first contraction $C$ has eigenvalues $\lambda_i$ and its second contraction $D$ has eigenvalues $\mu_j$. To see this, denote with $s$ the common sum of the eigenvalues, then put $A$ the diagonal matrix with entries $\mu_j/s$ nd $B$ the diagonal matrix with entries $\lambda_i$.