When powers of a matrix coincide with its Hadamard powers

We characterize matrices whose powers coincide with their Hadamard powers.

The latter question has been recently posed in [5] for the case of real matrices, and two characterizations of such matrices have been given in [2] and [4]. In this note we give another description of such matrices. (c) There exist k ∈ N, distinct non-zero elements λ 1 , . . ., λ k ∈ F, and idempotent Proof. The implication (a) ⇒ (b) is trivial. We begin the proof of the implication (b) ⇒ (c) by letting p(λ) = c m λ m + c m−1 λ m−1 + · · · + c 1 λ be a polynomial of degree m ≤ n + 1. If A = (a ij ) n i,j=1 , then our assumptions give that (1) This implies that . Then the degree of p(λ) is at most n + 1 and p(0) = 0, so that the equivalence (2) implies that p(a ij ) = 0 for all i, j. Let q(λ) be the minimal polynomial annihilating the element 0 and all entries of A. Then the polynomial q(λ) divides the polynomial p(λ), so that its degree is at most n + 1. Therefore, the equivalence (2) gives that q(A) = 0, and so m(λ) divides q(λ), as m(λ) is the minimal polynomial of A. This means that m(λ) factors into distinct linear factors, so that the matrix A is diagonalizable over F and the set {λ 1 , . . . , λ k } of all non-zero eigenvalues of A coincides with the set of all non-zero entries of A.
on the spectrum of the diagonalizable matrix A. Since each entry of A belongs to the set For the proof of the implication (c) ⇒ (a), we just compute the powers: We now give the canonical form of an idempotent (0, 1)-matrix. When the field F is the field R of all real numbers, this can be obtained easily from the canonical form of a nonnegative idempotent matrix (see e.g. [1, Theorem 3.1 on page 65]).
Suppose that the characteristic of the field F is either zero or larger than n. Then there exists a permutation matrix P such that where I is the identity matrix of size m, and U, V are (0, 1)-matrices such that U has no zero columns, V has no zero rows, and V U is also a (0, 1)-matrix. (It is possible that U or V act on zero-dimensional spaces.) Proof. Suppose first that E has no zero rows and no zero columns. We must show that m = n and E = I. Assume on the contrary that m < n. Since tr (E) = m and E is a (0, 1)-matrix, there exists a permutation matrix P such that where the diagonal entries of A ∈ M m (F) are equal to 1, while the diagonal entries of D ∈ M n−m (F) are equal to 0. Since E is an idempotent, we have A 2 + BC = A, so that, in view of our characteristic assumption, A 2 is also a (0, 1)-matrix. It follows that A must be the identity matrix. As P EP T is an idempotent, we obtain that BC = 0, By the first part of the proof, we obtain that T = I which gives the desired form.
In Theorem 2 we cannot omit the assumption on the characteristic of the field F.
Namely, if the field F has prime characteristic p < n, then, for example, take the (p + 1) × (p + 1) matrix of all ones and enlarge it by adding zeros to get an idempotent (0, 1)-matrix in M n (F) which is not of the above form.
If we apply Theorem 2 for idempotent (0, 1)-matrices in the assertion (c) of Theorem 1, we obtain the following decription of a matrix whose powers coincide with its Hadamard powers. (d) There exist a permutation matrix P , non-zero elements µ 1 , . . ., µ m ∈ F, and (0, 1)matrices U, V such that U has no zero columns, V has no zero rows, V U is also a (0, 1)-matrix, and where u T 1 , . . ., u T m are the rows of the m × n matrix I U 0 0 , and v 1 , . . ., v m are the columns of the n × m matrix I 0 V 0 T .
Proof. We must explain only how to obtain the assertion (d) from the assertion (c) of Theorem 1. We first observe that the matrix E = E 1 + · · · + E k is an idempotent (0, 1)matrix of rank m. By Theorem 2, there is a permutation matrix P such that where I is the identity matrix of size m, and U, V are (0, 1)-matrices such that U has no zero columns, V has no zero rows, and V U is also a (0, 1)-matrix. Let u T 1 , . . ., u T m be the rows of the matrix I U 0 0 , and let v 1 , . . ., v m be the columns of the matrix . ., E k and E = E 1 + · · · + E k are (0, 1)-matrices, all the ones of a matrix E j (j = 1, . . . , k) are at positions where also E has ones. Thus, we have where U j is a matrix obtained from U by replacing some ones with zeros, and likewise for I j , V j , and (V U) j . Now, it follows from E j = E j E = EE j that U j = I j U and V j = V I j , so that each of the first m rows (resp. columns) of P E j P T is either equal to 0 or to a corresponding row (resp. column) of P EP T . Thus, the matrix P E j P T is a sum of some of rank-one matrices v i u T i ; for these indices i, put µ i = λ j . Then we have It is worth mentioning that we can eliminate the permutation matrix P in the assertion (d) of Theorem 3 by setting that gives the form This way, of course, we lose information about the first m coordinates of these vectors.
As one of the referees noticed, the same form of A can be derived from the theorem in [2], since an "anchored" submatrix with index set I × J (introduced there) can be represented by µE, where E is an idempotent (0, 1)-matrix of the form vu T and u, v are (0, 1)-vectors associated with the indices in I, J, respectively.
Finally, we give a simple example showing that the assertion (c) of Theorem 1 does not imply that, up to a permutation similarity, A has a block diagonal form with k blocks.
Given any non-zero real numbers α and β, define the matrix A ∈ M 4 (R) by