Some solutions to a third-order quaternion tensor equation

: The paper deals with the third-order quaternion tensor equation. Based on the Qt multiplication operation, we derive solvability conditions and also get the general solution, the least-squares solution, the minimum-norm solution and the minimum-norm least-squares solution of the tensor equation A ∗ Q X = B . Finally, two numerical examples are presented.


Introduction
In 1843, Hamilton extended the real number field R and the complex number field C to quaternions Q.By now, quaternions and quaternion matrices have been widely used in many fields such as computer graphics, color image processing and signals [1][2][3].Tensors, as multidimensional arrays of vectors and matrices, appear widely in applications such as chemometrics [4], image and signal processing [5][6][7][8].For instance, Soto-Quiros [5] considered the inverse tensor problem for denoising data, which can be represented as a least-squares problem of a linear tensor equation of third-order.The author proposed a numerical method of estimating a least-squares solution to a complex tensor linear equation and used it for audio denoising and color image deconvolution.Based on t-product, Reichel et al. [6] also solved a penalized least-squares problem of a linear tensor equation of third-order by generalized Arnoldi-type and bidiagonalization solution methods.As applications, it is used on the color image and video restoration.Guide et al. [7] proposed a tensor iterative Krylov subspace method to solve large multi-linear tensor equations M(X) = C, like AX = C and AXB = C.While Jin et al. [9] developed an algorithm to compute the Moore-Penrose inverse of p-order tensor, and then it can deal with a linear p-order tensor equation problems.All those work are about the tensor over the real field.In [10,11], they considered the third-order tensor over the quaternions Q. Qin et al. [10] proposed a numerical method to compute the singular value decomposition of A ∈ Q n 1 ×n 2 ×n 3 and presented it to compress the color video.Inspired by [5,6,8], we realized that the third-order linear tensor equation problems over Q always exists when doing the color video deconvolution, color video denoising, color video reconstruction, and so on.Thus, we aim to solve the classic third-order quaternion tensor equation A * Q X = B, especially the least-squares solutions and the minimum-norm least-squares solutions.Up to now, there are some numerical methods to solve the quaternion linear tensor equations, like [12,13].In [13], Zhang et al. solved the generalized Sylvester quaternion p-order quaternion tensor equations by tensor form of GPBiCG algorithm.But an effective way to find the least-squares solution to the tensor equation A * Q X = B is to be developed.Thus, this paper will explore this problem in theoretical way.

Preliminary results
The set of quaternions Q is a linear space over R.An element q of Q is of the form q = a + bi + cj + dk, a, b, c, d ∈ R.
Here i, j and k are three imaginary units with the following multiplication laws: For a quaternion, the conjugate quaternion of ) is a multidimensional array with n 1 n 2 n 3 entries.In this paper, we use the notations a, a, A and A to denote the scalar, vector, matrix and thirdorder quaternion tensor, respectively.In [10], the horizontal, lateral and frontal slices of a third-order tensor are denoted by A(i, :, :), A(:, i, :), A(:, :, i), respectively, and for simplicity we denote the frontal slice of a third-order tensor by A (i) = A(:, :, i).In the paper, we use A * to represent the conjugate transpose of matrix A. O represents zero tensor with all the entries being zero.I denotes the identity tensor, in which the first frontal slice is an identity matrix and the other slice matrices are zero.For a positive integer n, [n] stands for {1, 2, . . ., n}.Let A ∈ C n 1 ×n 2 ×n 3 , the block circulant matrix circ(A) ∈ C n 1 n 3 ×n 2 n 3 generated by a third-order tensor A's frontal slices A (1) , A (2) , ..., A (n 3 ) is defined as A (1)  ... A (3) . . . . . . . . . . . .
see [7].For the quaternion A ∈ Q n 1 ×n 2 ×n 3 , the block circulant matrix circ(A) ∈ Q n 1 n 3 ×n 2 n 3 can be defined in the same way.
The inverse operation of unfold(•), denoted as fold(•), turns a block tensor with the size of n 1 n 3 × n 2 into a tensor with the size of In [10], the authors defined the Qt-product of two third-order quaternion tensors.
Lemma 1. (See [10]) Let A ∈ Q n 1 ×n 2 ×n 3 , B ∈ Q n 2 ×n 4 ×n 3 and C ∈ Q n 1 ×n 4 ×n 3 , A, B, C after DFT to obtain A, B, C, respectively.Then It follows from Lemma 1 that The conjugate transpose A * of third-order complex tensor A ∈ C n 1 ×n 2 ×n 3 is defined as follows: first conjugately transpose each frontal slice of A, and then reverse the order of conjugately transposed frontal slices 2 through n 3 , see [10].For the third-order quaternion tensor is defined in a more generalized way.[10] defined the third-order quaternion tensor A * through unfold(A * ), which should satisfies For example, for the third-order quaternion tensor , we should first derive unfold(A * ) by (2.3): From the definition of third-order quaternion tensor A * , we can see that it still satisfies For the third-order quaternion tensor A ∈ Q n 1 ×n 2 ×n 3 , it should be noted that the definition of A * generalizes the definition of A * when A is a third-order complex tensor. If Next, we will show some properties for A * .

□
Here we show that Proposition 1 is also true over quaternion skew field.
It is shown by [10] that From Proposition 1 and Eq (2.5), we can get diag( A * ) = (diag( A)) * .□ We correct two equations in [10] (page 3, line-4 and line-6 ) as follows: It should be noted that, for A = A 1,i + jA j,k , A = A 1,i + j A j,k , the following equations still hold

And in general,
Based on Qt-product, we can find that the multiplication operation between tensors obeys an excellent law, similar to matrix multiplication.
Proof.For the simplicity, we only prove (f).Denote C = A * Q B. By Proposition 2 and Lemma 1,

□
The Frobenius norm of a quaternion tensor A is the sum of all norms of its entries, i.e.
For a tensor A = A 1,i + jA j,k , A 1,i , A j,k ∈ C n 1 ×n 2 ×n 3 , its Frobenius norm can also be expressed as According to equality (2.2) and (2.6), it is easy to show that (2.7)

Generalized inverses
In this section, we will define some generalized inverses and explore their properties.
Definition 3.For an A ∈ Q n 1 ×n 2 ×n 3 , if there exists a quaternion tensor X ∈ Q n 2 ×n 1 ×n 3 satisfying: then we call X the Moore-Penrose inverse of the tensor A. Also, denote it as A † .
In [10], based on the Qt-product between the two third-order quaternion tensors, the authors derived the SVD decomposition of A ∈ Q n 1 ×n 2 ×n 3 .
e is an n 3 -dimensional column vector with all elements being 1.
From Lemma 2, we can derive the SVD decomposition of A † .
Theorem 1.Let the Qt-SVD of quaternion tensor A ∈ Q n 1 ×n 2 ×n 3 be A = U * Q S * Q V * .Then tensor A has a unique Moore-Penrose inverse Proof.First of all, it can be verified that fold( ) is the Moore-Penrose inverse of S by substituting it into the four equations in Definition 3.Then, obviously, V * Q S † * Q U * also satisfies the four equations in Definition 3 as U, V are unitrary tensors.Next, by Proposition 3, using the exactly same method as proving the uniqueness of Moore-Penrose inverse of a matrix, we can show that the Moore-Penrose inverse of a tensor A ∈ Q n 1 ×n 2 ×n 3 is unique.
□ Next, we list some properties of the Moore-Penrose inverse of a quaternion tensor A ∈ Q n 1 ×n 2 ×n 3 .
(f) A † always exists and is unique.
Proof.We only prove (b) as all of those are using the same approach with the proof of matrix.By Proposition 3, (A which means (A † ) * satisfying the definition of the Moore-Penrose inverse of A * .□ Definition 4. Given a tensor A ∈ Q n 1 ×n 2 ×n 3 , let A {1, 3} denote the set of tensor X ∈ Q n 2 ×n 1 ×n 3 , which satisfies Eqs (1), ( 3) of (3.1).In this case, X ∈ A {1, 3} is called a {1, 3}-inverse of A. It can also be written as A (1,3) .A (1,4) and A (1) can be defined in the same way.

Solutions to third-order quaternion tensor equation
In this section, we will derive general solutions, least-squares solutions, minimum-norm solutions and minimum-norm least-squares solution of the third-order quaternion tensor equation by some generalized inverses.We first introduce a well-known result of matrix equation: Lemma 3. (See [16]) For the complex matrix equation then: (a) If (4.2) is consistent, then X = A (1) B is the solution of (4.2), moreover, X = A (1,4) B is the least-norm solution of (4.2), where A (1,4) is the (1, 4)− inverse of matrix A. (b) The matrix equation (4.2) don't have to be consistent.X = A (1,3) B is the least-squares solution of (4.2), where A (1,3) is the (1, 3)− inverse of matrix A.
(c) The matrix equation (4.2) don't have to be consistent.X = A † B is the minimum-norm least-squares solution of (4.2), where A † is the Moore-Penrose inverse of matrix A.
Remark 1.The statement also holds when the matrix equation (4.2) is over quaternion skew field.
Next, we will describe our required solutions by some generalized inverses of tensor.
* Q B is the solution of the quaternion tensor equation (4.1), when it is consistent.
Proof.By Lemma 1 and (2.4), . By (a) in Lemma 3 and its Remark 1, for the consistent quaternion matrix equation Thus, X = T * Q B is the solution of consistent tensor equation A * Q X = B, where T ∈ A{1}.□ Corollary 2. Let A ∈ Q n 1 ×n 2 ×n 3 , B ∈ Q n 1 ×n 4 ×n 3 .The tensor equation (4.1) is consistent if and only if In this case, the general solution is given by where Y ∈ Q n 2 ×n 4 ×n 3 is arbitrary. Proof.
Considering the definition of A (1) , we have Next, we show that (4.4) is the general expression of the solution to (4.1).Firstly, we can verify that X is the solution to (4.1).Secondly, we aim to show that any solution of (4.1) is in the form of (4.4).Assume that X 0 is an arbitrary solution to (4.1).Setting Y = X 0 , then (1) is replaced by A † , the following result is also true.Since the proof is almost the same with the proof of Corollary 2, thus we omit for simplicity.
When the equation is consistent, then the general solution is where Y is an arbitrary quaternion tensor with appropriate size.
Next, we aim to derive the least-squares solution of the tensor equation (4.1).
We now provide a result for the Frobenius norm of the sum of two third-order tensors.
Proof.By (2.7) and ∥ K ∥ 2 F = tr(K * K), tr(L) = tr(L * ), where K, L are any quaternion matrices and tr(•) represents the trace of a matrix, we have Proof.It follows from Theorem 3 where The property of generalized inverse A (1,3) gives us
Then we can get which means that X = A (1,3)  * Q B is the least-squares solution of Eq (4.1).□ In solving practical applications, we sometimes need to find solutions for which the norm is minimal.The next theorem provides the mimimum-norm solution of tensor equation (4.1). (1,4)* Q B is the minimum-norm solution of the tensor equation (4.1), when it is consistent.
Proof.Since A (1,4) ∈ A{1}, by Theorem 2, A (1,4)  * Q B is a solution of (4.1), if the tensor equation (4.1) is consistent.Next, we prove X = A (1,4)  * Q B is the minimum-norm solution.By Corollary 2, for any solution X 0 to (4.1), X 0 can be written in the form of X 0 = A (1,4) where By the property of generalized inverse A (1,4) , we have We conclude that In the assumption, the tensor equation is consistent means it has a general solution.The solution don't have to be a minimum-norm solution.So, According to Corollary 3, Theorem 5 can be rewritten as follows: It is well known that the least-squares solution of an equation is not unique, neither is the minimumnorm solution.Then we consider the minimum-norm least-squares solution to this problem.Theorem 6.Let A ∈ Q n 1 ×n 2 ×n 3 , B ∈ Q n 1 ×n 4 ×n 3 , and X 0 ∈ Q n 2 ×n 4 ×n 3 .The tensor X 0 is the least-squares solutions of (4.1) if and only if X 0 is the solution of the consistent tensor equation (4.8) Proof."⇒" Assuming X 0 is a least-squares solution of the tensor equation (4.1), from Theorem 4, we have From Theorem 3, we have where Based on (4.7) and (4.9), ∥ A * Q (X 0 − A (1,3) which means that X 0 is a solution of the tensor equation (4.8).
"⇐" If X 0 is a solution of the tensor equation (4.8), note that By Lemma 1 and Proposition 2, which is equivalent to In other words, X (i) 0 is the least-squares solution to Or, diag( X 0 ) is the least-squares solution to Then, we can see that X 0 is the least-squares solution of (4.1).□ Proof."⇒" If X 0 = T * Q B is the minimum-norm least-squares solution of the tensor equation (4.1), by Theorem 6, X 0 is the minimum-norm solution of Eq (4.8).Then, by Corollary 1 and Theorem 5, we have which means that T = A † ."⇐" If T = A † , since A † ∈ A{1, 2, 3, 4}, it satisfies both the properties of a least-squares solution and minimum-norm solution to the Eq (4.1) by Theorem 4 and Theorem 5. □

Numerical examples
In this section, we give two numerical examples.
We can see from the Table 1 that our restored color video achieves a good accuracy and has a satisfied result.
the least-squares solution of the tensor equation (4.1).

Remark 3 .
The tensor equation (4.1) always has a least-squares solution, thus by Theorem 6, the tensor equation (4.8) is always consistent.Theorem 7. Let A ∈ Q n 1 ×n 2 ×n 3 , B ∈ Q n 1 ×n 4 ×n 3 .The tensor X 0 = T * Q B isthe minimum-norm leastsquares solution of the tensor equation (4.1) if and only if T = A † .