I just completed a homework problem which proves the following result: a matrix (with coefficients in some field) is a product of nilpotent matrices iff its not invertible.

The proof was broken into several parts and was quite involved. I'm wondering if there's a very simple way to demonstrate this result (just the implication non-invertible then its a product of nilpotent matrices , as the other implication is trivial).

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    $\begingroup$ Are you sure you meant asking about the forward ($\Longrightarrow$) direction? Because that one is easy, using the fact that nilpotent matrices have determinant nul and $\det (AB) = \det(A) \det(B)$. $\endgroup$ – Nigel Overmars May 3 '17 at 21:30
  • $\begingroup$ Sorry I guess forward direction was not well defined here. I'm asking for a proof of the existence of a deomposition into nilpotent matrices for a noninvertible matrix. $\endgroup$ – Joshua Benabou May 4 '17 at 6:45
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    $\begingroup$ This here claims to prove a stronger result: C.J. Hattingh, A note on Products of Nilpotent Matrices, arXiv:1608.04666v3. I don't have particularly high hopes for a short/slick/simple proof; this seems more like a fact that happens to be true for no specific reason... $\endgroup$ – darij grinberg May 4 '17 at 7:12
  • $\begingroup$ Indeed it's an interesting and not easy result. The basic outline of the proof is as follows. Let the base field be K. We show first that a non invertible matrices A is the product of a nilpotent matrix N and a noninvertible matrix W which can be expressed in block form as some invertible square matrix Q surrounded by zeros. Equivalently W is a matrix represeting an endomorphism f (with respect to a ceartain basis of K^n) such that Im(f) and Ker(f) are complementary subspaces of K^n. Next we show that such a matrix W is a product of nilpotent matrices. $\endgroup$ – Joshua Benabou May 4 '17 at 23:16
  • $\begingroup$ It will suffice to express the invertible matrix Q in its LDU composition (which we first proved exists). Then by block multiplcation, W is a product of triangular matrices whose last column and last row are zero vectors. Finally, we show that an upper triangular matrix with the last column equal to the zero vector is the product of two nilpotent matrices (one of them being the transpose if the Jordan matrix), and an analagous result for lower triangular matrices. $\endgroup$ – Joshua Benabou May 4 '17 at 23:18

Following is a proof for fields that allow a Jordan-type canonical form. Write $A=P^{-1}TP$ with $T$ in Jordan form (but this can be relaxed somewhat). $A$ is not invertible if, and only if, at least one of the diagonal elements is zero. Taking it to be the last eigenvalue for simplicity, note the following example decomposition into two nilpotent matrices: $$\pmatrix{1&2&3&4&0\\0&5&6&7&0\\0&0&8&9&0\\0&0&0&a&0\\0&0&0&0&0}=\pmatrix{0&1&2&3&4\\0&0&5&6&7\\0&0&0&8&9\\0&0&0&0&a\\0&0&0&0&0}\pmatrix{0&0&0&0&0\\1&0&0&0&0\\0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0}$$

Thus $T=M_1M_2$ nilpotent, so $A=(P^{-1}M_1P)(P^{-1}M_2P)=N_1N_2$.


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