This is a question more from combinatorics although its background is in linear algebra.

Let $A$ be a real matrix of dimension $n \times n$. We know that in general the sequence of matrices $A, A^2, A^3, \dots $ can have decreasing rank - in extreme case if for some $k$ we have $A^k=0$ then the matrix is (called) nilpotent and the patterns of transforming the rank from an initial value to $0$ can be very different, in general we could write such rank pattern $ m_1\rightarrow m_2\rightarrow \dots \rightarrow m_k=0 $.

Of course if a matrix is of full rank the exponentiation preserves rank: we have always $ n \rightarrow n$ (single possible pattern).

When the rank is less than $n$ it is also possible that exponentiation doesn't change the rank.
But the situation when the rank is decreasing from some value of $m_1$ to $m_k$ is possible in many different ways. I'm interested in the number of different ways how it could be done. It obviously depends on the possible Jordan Normal Forms for $n \times n$ matrix.

  • What is the number formula for these forms if the only thing which is used in classification here is the way the rank decreases?
    (what corresponds to unique possible sequences of changes in rank for n-dimensional matrix $ m_1\rightarrow m_2\rightarrow \dots \rightarrow m_k$)
  • $\begingroup$ I think the only limitation is that the rank decrease at one step cannot be larger than the decrease in the previous step, and apart from that any sequence is possible. It shouldn't be too hard to make an example for an arbitrary such sequence, but then again I might be wrong. $\endgroup$
    – Arthur
    Mar 23 '18 at 16:35
  • $\begingroup$ @Arthur What number t of different sequences (t=f(n)) could give such constraint? $\endgroup$
    – Widawensen
    Mar 23 '18 at 16:38
  • 1
    $\begingroup$ Assuming my statement in the previous comment is correct, if you, instead of the sequence of ranks, look at the sequence of rank decreases, you get a sequence of integers, which is sorted by size with the largest first, $k-2$ long, and sums up to $m_1-m_k$. This is a partition, and they are notoriously difficult to count in general. $\endgroup$
    – Arthur
    Mar 23 '18 at 16:43
  • $\begingroup$ @Arthur There is no clear method for this method of partition? Thank you for the link.. $\endgroup$
    – Widawensen
    Mar 23 '18 at 16:46
  • $\begingroup$ @Arthur The link is very interesting, I didn't expect that this combinatorial task is so hard as it seems to be.. $\endgroup$
    – Widawensen
    Mar 23 '18 at 16:51

It suffices to consider Jordan matrices. Further, it suffices to look at each Jordan block separately, since the decrease in rank of the entire Jordan matrix is the sum of the decreases in each block, upon doing the exponentiation operation.

By thinking about what happens when you raise a Jordan block to some power, you see that the rank remains the same if the eigenvalue of the Jordan block is nonzero, and otherwise decreases by $1$ each time you raise the power by one.

As an example, suppose we have a Jordan matrix, with Jordan blocks of size $5, 3, 4, 2$ corresponding to eigenvalues $0, 7, -1, 0$.

The first Jordan block's rank decreases as $4 \to 3 \to 2 \to 1 \to 0$. The last Jordan block's rank decreases as $1 \to 0$. The other Jordan blocks do not decrease rank. Thus overall the matrix rank decreases as $12 \to 10 \to 9 \to 8 \to 7 \to 7 \to \cdots$.

From the above, you can see that one restriction on the sequence of ranks is that the decreases at each step are themselves nonincreasing: the rank decrease at a particular step is the number of zero-eigenvalue Jordan blocks that have not yet become the zero matrix from the exponentiation thus far, and this number of "active" zero-eigenvalue Jordan blocks is nonincreasing. This recovers the claim in Arthur's comment.

Another restriction is that the largest decrease (the first decrease) is $\le n - r$ where $r$ is the rank of the full matrix (and the first number in the sequence), since the number of zero-eigenvalue Jordan blocks is $\le n - r$.

[Have I missed anything else?]

Actually now, I think it would make more sense to count the number of Jordan block configurations, rather than the sequence of ranks. The size configurations of the zero-eigenvalue Jordan blocks are in bijection with the sequences of ranks(?), so I believe the number of sequences is $p(0) + \cdots + p(n)$ where $p(k)$ is the number of partitions of $k$.

[Please correct me if I have made a mistake anywhere.]

  • $\begingroup$ Hmm, I'm not sure about the number, I guess I have to study more this problem of partition.. $\endgroup$
    – Widawensen
    Mar 23 '18 at 17:07
  • $\begingroup$ I was thinking about the problem and your answer seems to be a solution. Only one remark: if I wanted to count only decreasing sequences we probably would need to substract from the sum $p(0)+\dots+p(n)$ a number $n+1$ as in the sum are included cases when the rank is stable.. $\endgroup$
    – Widawensen
    Mar 24 '18 at 13:35
  • $\begingroup$ From your links we have sequence for p(n) oeis.org/A000041/b000041.txt but I haven't found there a direct sequence for the problem stated, I wonder whether the problem is valuable to the degree as to construct such sequence for oeis.org.. $\endgroup$
    – Widawensen
    Mar 24 '18 at 13:54
  • $\begingroup$ Interesting that in sequence $\sum \dots -(n+1)$ we have initially 0,0,1,3,7,13,23,37, all numbers > 1 are primes .. then the pattern breaks .. $\endgroup$
    – Widawensen
    Mar 24 '18 at 14:22
  • $\begingroup$ I've checked thoroughly possibilities using your clues and I think that the best form for the number of different decreasing rank sequences we could assume $(\sum\limits_{k=1}^{n} p(k))-n $. $\endgroup$
    – Widawensen
    Mar 27 '18 at 10:03

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