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Let $A$ and $B$ be two square matrices with complex entries. Let $\lambda_1, \ldots, ,\lambda_n$ be the Eigenvalues of $A$ and $\mu_1, \ldots, ,\mu_m$ be the Eigenvalues of $B$. Then the Eigenvalues of the Kronecker product are exactly the products $\lambda_i \cdot \mu_j$. Is there an analogue for the sums of Eigenvalues? My precise question is the following:

For given natural numbers $m$ and $n$ are there polynomials $f_{rs} \in \mathbb{C}[x_{ij},y_{kl}: \, 1 \leq i,j \leq m, \, 1 \leq k,l \leq n]$ such that for every $n \times n$ matrix $A$ and every $m \times m$ matrix $B$ the Eigenvalues of the matrix $C=(f_{rs}(A,B))_{1 \leq r,s \leq mn}$ are exactly the sums of an Eigenvalue of $A$ and an Eigenvalue of $B$? Here $f_{rs}(A,B)$ stands for the complex number obtained by substituting $x_{ij}$ by the $(i,j)$th entry of $A$ and $y_{ij}$ by the $(i,j)$th entry of $B$.

I am aware of some similar construction where the matrix $C$ has the desired Eigenvalues among others. But for me it is important that they are no other Eigenvalues.

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    $\begingroup$ $A \otimes I + I \otimes B$ (sometimes called "Kronecker sum") should work. $\endgroup$
    – Federico Poloni
    Commented Sep 28, 2015 at 14:03
  • $\begingroup$ Or just $A\oplus B$ on the direct sum. $\endgroup$
    – paul garrett
    Commented Sep 28, 2015 at 17:35
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    $\begingroup$ @paul: that's not correct. You get the union (as a multiset) of the eigenvalues that way, not the pairwise sums. $\endgroup$
    – Qiaochu Yuan
    Commented Sep 28, 2015 at 21:24
  • $\begingroup$ @QiaochuYuan, aha!, you are certainly right about that! I was not thinking! $\endgroup$
    – paul garrett
    Commented Sep 28, 2015 at 23:39
  • $\begingroup$ (I must note that I have seen sometimes the notation $A\oplus B$ used for the Kronecker sum as defined in my previous comment. This is, of course, very confusing, since it is also the standard notation for direct sums.) $\endgroup$
    – Federico Poloni
    Commented Sep 29, 2015 at 4:11

1 Answer 1

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Federico already mentioned the keyword. The precise answer may be found among others as Theorem 13.16, of this book. (That theorem makes a restriction to real matrices, but that is not necessary).

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