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.

  • 6
    $\begingroup$ $A \otimes I + I \otimes B$ (sometimes called "Kronecker sum") should work. $\endgroup$ Sep 28, 2015 at 14:03
  • $\begingroup$ Or just $A\oplus B$ on the direct sum. $\endgroup$ Sep 28, 2015 at 17:35
  • 1
    $\begingroup$ @paul: that's not correct. You get the union (as a multiset) of the eigenvalues that way, not the pairwise sums. $\endgroup$ Sep 28, 2015 at 21:24
  • $\begingroup$ @QiaochuYuan, aha!, you are certainly right about that! I was not thinking! $\endgroup$ 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.) $\endgroup$ Sep 29, 2015 at 4:11

1 Answer 1


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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.