iterated characteristic polynomials

If I have $N$ $M\times M$ symmetric positive definite matrices $A_i$ and an $N\times N$ positive semi-definite symmetric matrix B, let the $N\times N$ matrix $C_{ij}(\lambda)=B_{ij}$ for $i\ne j$ and $C_{ii}(\lambda)=B_{ii}+\lvert A_i-\lambda 1_M\rvert$, so we are adding the characteristic polynomials of $A_i$ to the diagonal of $B$. Is there a simple construction – like a Kronecker product – that produces an $(NM)\times(NM)$ matrix with characteristic polynomial $P(\lambda)=\lvert C(\lambda)\rvert$?

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This is a really strange construction. Could you provide some motivation for it? – Qiaochu Yuan Jan 1 '10 at 22:12
I agree it is quite strange. It arose in studying a large linear system. $B$ represents the coupling between degrees of freedom. In the $B=0$ case, each degree of freedom has a response that can be described as a sum of $M$ modes, with the moments given by the roots of a polynomial. Finding modes of the coupled system entails finding $\lambda$ such that $C(\lambda)$ is singular. But it is easy to write the polynomials on the diagonal as characteristic polynomials, so I thought I might be able to transform the whole thing into a big eigenvalue problem. – mifune Jan 2 '10 at 0:15
What you've described seems to me like a very strange way to couple your "degrees of freedom" together. Could you explain the meaning of the coupling coefficients B? It seems strange to me that say the ith degree of freedom is controlled by an entire matrix A_i and yet coupled to the jth degree of freedom by a single number B_{ij}. – j.c. Jan 2 '10 at 3:08
For each $i$, I have an $x_i(t)$ that obeys an equation $T_ix_i(t)+\sum_jB_{ij}x_j(t)=0$, where $T_i$ is some linear operator. There are $NM$ independent solutions in general. For the uncoupled case $B=0$, each $x_i(t)$ is a sum of $M$ modes $y(t;\lambda_j)$, each of which is specified by a parameter $\lambda_j$ which is the root of a polynomial $P_i(lambda)$. In the $B\ne 0$ case, my solutions are of the form $x_i(t)=z_iy(t;lambda')$, where $\sum_jC_{ij}(lambda')z_j=0$ for all $j$. – mifune Jan 2 '10 at 3:55
The matrices $A_i$ are just something I constructed that would have the right characteristic polynomial &ndash; the simplest choice would be $diag(\lambda_1,\dots,\lambda_n)$. Instead of trying to find the values of where some matrix with crazy polynomials on the diagonal is singular, I am hoping to transform it to a regular eigenvalue problem. – mifune Jan 2 '10 at 3:58