Matrices over finite fields of odd characteristic as sums of diagonalizable and square-zero matrices

1. Problem Statement

The paper addresses a specific decomposition problem in linear algebra over finite fields. The central question, originally posed by Breaz (2018), asks whether every square matrix $A$ over a finite field $\mathbb{F}_q$ of odd cardinality $q$ can be decomposed into the sum of a diagonalizable matrix ($D$) and a square-zero matrix ($M$, where $M^2=0$).

Specifically, the authors investigate:

• Whether $A = D + M$ holds for all $A \in M_n(\mathbb{F}_q)$ when $q$ is odd.
• The specific constraints when $q=3$ versus $q \geq 5$.
• The relationship between this decomposition and $q$-potent matrices (matrices where $D^q = D$), noting that over $\mathbb{F}_q$, a matrix is diagonalizable if and only if it is $q$-potent.

2. Methodology

The authors employ a combination of structural matrix theory and constructive algebraic proofs:

• Rational Canonical Form: The authors utilize the fact that any matrix is similar to a direct sum of companion matrices of its invariant factors. This allows them to reduce the general problem to analyzing non-derogatory matrices (matrices where the minimal polynomial equals the characteristic polynomial).
• Constructive Decomposition: For specific cases, the authors explicitly construct the square-zero matrix $M$ and a change-of-basis matrix to demonstrate that $A+M$ becomes diagonalizable. They define $M$ with specific non-zero entries (often involving the coefficients of the characteristic polynomial) to force the resulting matrix $A+M$ to satisfy a polynomial with distinct roots (e.g., $x(x-1)(x+1)=0$).
• Case Analysis by Field Size and Trace:
  • They distinguish between fields with $q \geq 5$ and the specific case of $q=3$.
  • They analyze cases based on the trace of the matrix ($\text{Tr}(A) = 0$ vs. $\neq 0$) and the parity of the matrix order $n$.
• Counterexample Construction: For the negative results (where decomposition is impossible), they use algebraic manipulation of the defining equations ($D^3=D$ and $M^2=0$) to derive contradictions, specifically proving that the assumed square-zero matrix $M$ must be the zero matrix, which contradicts the non-diagonalizability of the original matrix.

3. Key Contributions and Results

A. Fields with Odd Characteristic $q \geq 5$

The paper provides a complete affirmative answer for fields with at least 5 elements.

• Theorem 2.2: For any finite field $\mathbb{F}_q$ with $q \geq 5$ (odd), every matrix $A \in M_n(\mathbb{F}_q)$ admits a decomposition $A = D + M$, where $D$ is diagonalizable and $M^2 = 0$.
• Implication: Since diagonalizable matrices over $\mathbb{F}_q$ satisfy $D^q = D$, this confirms that every matrix is the sum of a $q$-potent matrix and a square-zero matrix.
• Proof Strategy: The proof relies on Proposition 2.1, which establishes that any non-derogatory matrix of order $n \geq 5$ can be decomposed. The authors construct $M$ such that $A+M$ has eigenvalues $\{0, \pm u, \pm 1\}$ (depending on the trace and parity of $n$), ensuring diagonalizability. Smaller orders ($n < 5$) were previously settled in their earlier work [6].

B. The Field $\mathbb{F}_3$ (Characteristic 3)

The case $q=3$ is more complex and yields a nuanced answer.

• Positive Results:
  • Proposition 3.1: Non-derogatory matrices of even order $n$ with zero trace over $\mathbb{F}_3$ can be decomposed.
  • Corollary 3.3 & 3.4: A matrix $A \in M_n(\mathbb{F}_3)$ can be decomposed into $D+M$ unless its first invariant factor is a non-zero trace irreducible polynomial of degree 3.
• Negative Results (Counterexamples):
  • Proposition 3.5: The authors prove that matrices consisting of a direct sum of companion matrices associated with the same irreducible polynomial of degree 3 with non-zero trace cannot be decomposed.
  • Specific Counterexample: For any $n = 3k$ ($k \geq 1$), there exist matrices over $\mathbb{F}_3$ that are not sums of a diagonalizable and a square-zero matrix.
  • Proof Logic: Assuming $A+M=D$ leads to a system of equations involving $M$ and $A$. Through iterative substitution using the minimal polynomial of the companion matrix ($x^3 - x^2 - x - 1 = 0$), the authors deduce that $M$ must be the zero matrix. Since the original matrix $A$ is not diagonalizable, this is a contradiction.

4. Significance

• Resolution of Breaz's Question: The paper completely settles the question posed by Breaz regarding the decomposition of matrices over finite fields of odd cardinality.
  • Answer: Yes, for $q \geq 5$.
  • Answer: No, for $q=3$ (specifically for orders $n=3k$ involving specific irreducible polynomials).
• Refinement of Matrix Decomposition Theory: The work extends previous results (which covered infinite fields or specific small orders) to the general case of finite fields of odd characteristic. It highlights the critical role of the field size ($q=3$ vs $q \geq 5$) and the structure of invariant factors in determining decomposability.
• Identification of Obstructions: The paper precisely identifies the "obstruction" to decomposition in $\mathbb{F}_3$: the presence of irreducible polynomials of degree 3 with non-zero trace in the invariant factor decomposition.

5. Future Directions

The authors conclude by outlining open problems for fields of characteristic 2:

• While it is known that matrices over $\mathbb{F}_2$ are sums of idempotent and nilpotent matrices, the specific decomposition into diagonalizable ($D^2=D$) and square-zero ($M^2=0$) is not always possible (counterexamples exist for order 4).
• They pose two parallel questions for fields $\mathbb{F}_{2^i}$ ($i \geq 2$):
  1. Is every matrix a sum of a diagonalizable and a square-zero matrix?
  2. If not, does the decomposition hold at least for non-derogatory matrices?

In summary, this paper provides a definitive classification for the decomposition of matrices over odd-characteristic finite fields, proving universality for $q \geq 5$ and characterizing the exact failure cases for $q=3$.