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I have an $r \times r$ matrix $M$ with rational entries and suppose $M$ has $\operatorname{rank} = n-k, k < n$. Then I proceed as follows : Let $m(x)$ be the minimal polynomial of $M$, and make sure it is irreducible. Then $m(x) = m'(x)x$, where $m'(x)$ has a nonzero constant term. Then, if we set $N = m'(M)$ we have that the minimal polynomial of $N$ is of the form $x^2 + ax$, so that $-N/a$ is an idempotent matrix of rank $k$. Moreover if $E =$ the Concatenation of the nullspace of $M$ and the nullspace of $N$ then all the matrices $A$ diagonally splits in blocks of $(n-k)\times (n-k)$ and $k \times k$ matrices. It's not difficult to prove this if we diagonalize $M$ using its roots in $\Bbb C$, but I would like a pure linear algebra proof that I can generalize to other rings than $\Bbb Q$.

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    Do I take it right that you want a classification of all idempotent elements? And what coefficients are you allowing, a field? – 2017-02-18
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    In fact I only need one since in the problem I'm looking at $n = 120$ and $k = 10$ . In fact $M$ lives in a commutative ring and I can construct an action of $S_5$ as auomorphism on it, so it sufficies to trace the orbit of $N$ to find all minimal idempotents (of which there are $12$). Finding all nilpmotents would yield $2^12$ of them (including 0 and 1). Fields asz coefficients would be ok,, but rings would be welcome. – 2017-02-18
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    I don't understand your question. Is your matrix ring (let us call it $\mathcal R$) a subring of $M_n(R)$ for some commutative ring $R$? If so, then the identity matrix is already an idempotent element in $\mathcal R$ and you don't need an algorithm to find it. If $\mathcal R$ has a ring structure but it is not necessarily a subring of $M_n(R)$ (i.e. the multiplicative identity element in $\mathcal R$ is not $I_n$), how can you guarantee that your $N=m'(M)$ is an element in $\mathcal R$? – 2017-02-18
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    @user1551: The ring $\mathcal R$ is not simple, so it is the direct sum of ideals isomorphic to fields. So there are matrices of the form $\begin{pmatrix} I & 0 \\ 0& 0 \end{pmatrix}$, with $I$ a (sub-)identity matrix, and it is exactly those I want to find. Do the exercize with $M = \begin{pmatrix} 0& -1& 0& 0& 1& 1 \\ 0& -3& -1& 1& 3& 4 \\ -1& 0& -3& 1& 1&3 \\ -1&-1& -3& 1& 2& 4 \\ -3& -4& -10& 4& 7&14 \\ -1& -3& -4& 2& 4,&7 \end{pmatrix}$. – 2017-02-18
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    @user1551: Since $M \in \mathcal R$ as a matrix with entries in the rationals all its powers and combinations thereof are in $\mathcal R$. – 2017-02-18

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