Counterexamples to a nullity criterion for matrix power maps

Henry Zweiman

September 15, 2026

Abstract

For every integer and every field , we construct a diagonal idempotent of nullity two for which the map is not surjective. An explicit omitted value is the direct sum of three nilpotent Jordan blocks of size two, transposed, and zero blocks. Since , this gives a negative answer to the sufficiency question posed by Saini and Singh. The first example concerns fifth powers of matrices. The proof uses two elementary facts: in dimension , a -th power with a nontrivial zero chain has rank one, and a one-dimensional perturbation of a square-zero map of rank two forces such a chain. Over algebraically closed fields of characteristic zero, the image is dense but not closed under addition.

1. Introduction and main result

Polynomial maps with nonscalar matrix coefficients extend matrix Waring problems. The behavior of a singular coefficient can differ substantially from that of a scalar coefficient, even if another coefficient is invertible. Saini and Singh [1] study the maps

over algebraically closed fields of characteristic zero, with invertible. For , they prove that surjectivity requires

and establish sufficiency when and, in every dimension, when . Their Question 6.1 asks whether (1) is sufficient in general. Related questions about surjectivity and image linearity over division algebras are considered by Gangwal, Mandal and Verma [2], particularly in dimension two.

We give counterexamples to the proposed sufficiency criterion. They use diagonal idempotent coefficients, so nontrivial nilpotent Jordan blocks in the coefficient are not needed. The argument works over every field.

Let

We use for the zero matrix of size , with a zero-size summand omitted.

Theorem 1.1.

Let be any field and . Set and

There are no satisfying

In particular, the map is not surjective although , , and

Thus Question 6.1 of [1] has a negative answer. This does not affect the necessity of (1) or the positive results in the regimes proved there. Our first example has size six; we do not claim that six is the smallest possible size for a counterexample.

2. Two linear-algebra lemmas

The first lemma isolates the special role of the dimension . It can be read from Jordan form, but a cyclic-vector argument avoids assumptions on the field.

Lemma 2.1.

Let have dimension over a field , and let . If

then is cyclic nilpotent and .

Proof.

Choose with and , and let be the least positive integer such that . Then . The vectors

are linearly independent. Indeed, if a nonzero relation has its first nonzero coefficient at , applying gives a nonzero scalar multiple of equal to zero. Hence , so and these vectors form a basis of . In this basis is a single nilpotent Jordan block. Its -th power has rank one.

The next lemma concerns an arbitrary extension of a prescribed map. There is no assumption on its action outside the indicated subspace.

Lemma 2.2.

Let be vector spaces over a field , let , and let satisfy and . Suppose that satisfies

for a linear functional on . Then

Proof.

Put and . Since , one has , and .

First suppose that . Choose a nonzero ; a single linear functional cannot be injective on a space of dimension at least two. Choose with . Since is a nonzero functional, there is such that . With , equation (4) gives . Also and , so .

Now suppose that . There is a well-defined linear functional such that for every . Choose a nonzero , possible because , and choose with . Then , so . Since and vanishes on , one also has .

In either case, the nonzero vector belongs to .

Remark 2.3.

The rank hypothesis in Lemma 2.2 cannot be replaced by . For example, on a space with basis , put , on , and define

Then (4) holds with and , but is invertible. This explains why two square-zero blocks are used on the complementary subspace in our construction.

3. Proof and the first explicit example

Proof of Theorem 1.1.

Suppose (3) holds. Transposing, and writing and , gives

Let be the standard basis and set

The coefficient vanishes on and is the identity on . Put and . Restricting (5) to gives

Thus , and Lemma 2.1 gives . Since its image contains , there is a linear functional on such that for all .

On , equation (5) now becomes

Here , , and . Lemma 2.2 shows that . Applying Lemma 2.1 to therefore gives .

However, let be the projection along . The last displayed identity gives

so . This contradiction proves the theorem.

For , the omitted target and its coefficient are the following matrices over any field:

They satisfy , , and , but for all .

4. Image geometry and the remaining classification

Corollary 4.1.

Let be algebraically closed of characteristic zero, and let . The image of

is Zariski dense and is not closed under addition. In particular, it is not a vector subspace.

Proof.

Every invertible matrix has a -th root over , so setting shows that contains , a dense open subset. For completeness, the root assertion follows blockwise: on an invertible Jordan block , choose and use the truncated binomial polynomial . It is a finite polynomial because is nilpotent.

Nonadditivity can also be witnessed explicitly. The matrices and belong to the image, whereas their sum does not by Theorem 1.1. Indeed, ; and since , a scalar with gives

The image statement is a consequence of the counterexample and the familiar existence of roots of invertible matrices; it is not a separate classification theorem. The theorem itself does not need algebraic closure or characteristic zero, while this explicit additive witness uses both the scalar root and .

The examples lie immediately above the proposed numerical boundary: and . They show that this inequality alone cannot characterize surjectivity, even for idempotent coefficients. The mechanism constrains both summands. A nontrivial nilpotent chain on the kernel of the coefficient forces the first power to have rank one; the remaining two square-zero blocks then force the second power into the same rank-one regime, which is impossible. A construction of only the first power, satisfying the rows or columns prescribed by the kernel of the coefficient, would miss this second obstruction.

A complete criterion for surjectivity remains open beyond the cases in [1]. In particular, the present construction gives no classification for fixed small exponents, does not decide all five-dimensional cases, and does not determine the entire image for the coefficients . It also does not address the classical multilinear Kaplansky–L’vov conjecture: the maps here involve powers and fixed matrix coefficients.

References

[1] P. Saini and A. Singh, Polynomial Maps with Constants on Matrix Algebra, arXiv:2604.27592v1 (2026), especially Theorems 1.2 and 1.3, Proposition 4.2, and Question 6.1. https://arxiv.org/abs/2604.27592.

[2] A. Gangwal, A. Mandal and S. Verma, Polynomial Maps with Constants over Division Algebras and the Generalized Kaplansky–L’vov Conjecture, arXiv:2607.27226v1 (2026), Question 1.1 and Theorem 1.4. https://arxiv.org/abs/2607.27226.