What is the eigenvalue/eigenvector relationship between matrix A,B and AB?
closed as too broad by Will Jagy, Yemon Choi, Suvrit, Andrey Rekalo, Carlo Beenakker Jul 25 '13 at 10:21
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I assume this is over $\mathbb C$. If they don't commute, about all you can say is that the determinant (which is the product of the eigenvalues, counted by algebraic multiplicity) of $AB$ is the product of the determinants of $A$ and $B$. 


I do not know about eigenvectors, but for the eigenvalues this is a special case of DeligneSimpson Problem. It was completely solved by CrawleyBoevey about 10 years ago using quivers. For details, see my answer here and references therein. 


See the following paper. The review describes only results for $A+B$, but this readily transforms to $AB$, as described in the paper.
Review: The text provides a review of the amazing development resulting from the groundbreaking discovery of the author [Selecta Math. (N.S.) 4 (1998), no. 3, 419–445; MR1654578 (2000b:14054)] of the fundamental connections between inequalities for eigenvalues of Hermitian operators $A$, $B$ and $A+B$, LittlewoodRichardson coefficients $c^\gamma_{\alpha\beta}$ for the decomposition of the tensor product of irreducible $GL_n$representations $$V_\alpha⊗V_β=∑c^γ_{αβ}V_γ,$$ the decomposition of the product of two Schubert cycles in the cohomology of the Grassmannian and toric stable vector bundles on P2. The topic has further ramifications for the groups $SL_n$, $SO_n$, $Sp_{2n}$ and other symmetric spaces. The paper contains no proofs but has an extensive list of references. 

