most recent 30 from http://mathoverflow.net 2013-05-18T22:44:51Z http://mathoverflow.net/feeds/question/95328 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/95328/when-do-the-invariant-factors-of-a-direct-sum-of-matrices-correspond-to-those-of Adam 2012-04-27T08:05:41Z 2012-04-27T11:19:21Z

<p>(Tried asking this on math stackexchange, but no takers so far.)</p> <p>I'm trying to prove something about matroids, which I have reduced to the following question:</p> <p>Suppose I have a matrix $M$ which is a direct sum of submatrices $M_1,M_2,…,M_k.$ When do the invariant factors of the ${M_i}$ partition the set of invariant factors of $M$?</p> <p>To be more explicit, let $d_1,…,d_n$ be the invariant factors of the matrix $M$ (so that $d_j|d_{j+1}$ for all $1≤j≤n−1$). Let $D$ be the set of these numbers, and similarly let $D_i$ be the set of invariant factors of the summand $M_i$ for each $i.$ Are there any known conditions under which:</p> <p>$$\bigsqcup_iD_i=D?$$</p> <p>By definition, $M$ is a block-diagonal matrix, where the blocks are the ${M_i}$. And in fact, it is not hard to see that for the purposes of this question we can assume without loss of generality that $M$ is actually diagonal (that is, each $M_i$ is a diagonal matrix). This means that I simply need conditions on the order and nature of the diagonal entries.</p> <p>However, any information related to this scenario will be welcome, even if you think it is obvious! Please feel free to generally hold forth.</p> <p>Thanks!</p>

http://mathoverflow.net/questions/95328/when-do-the-invariant-factors-of-a-direct-sum-of-matrices-correspond-to-those-of/95347#95347 Denis Serre 2012-04-27T11:19:21Z 2012-04-27T11:19:21Z

<p>By definition $M-XI_n$ is equivalent, in $M_n(k[X])$ to ${\rm diag}(d_1,\ldots,d_n)$. Likewise, $M_j-XI_{n_j}$ is equivalent, in $M_n(k[X])$ to ${\rm diag}(d_{j,1},\ldots,d_{j,n_j})$. Therefore $M-X_I$ is equivalent to the diagonal matrix with diagonal entries the polynomials $d_{j,s}$ where $1\le j\le k$ and $1\le s\le n_j$. Conclusion: the latter polynomials are the invariant factors of $M$ if and only if they form an ordered sequence. In other words, if and only if, for every pairs $(j,s)$ and $(i,t)$, one among the polynomial $d_{j,s}$ and $d_{i,t}$ divides the other.</p>