1979
DOI: 10.1016/0024-3795(79)90004-1
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The matrix equation AX − YB = C

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Cited by 78 publications
(23 citation statements)
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“…The first question arising in examinating the above linear equation is about its solvability. An answer to this question can be found in [16] (see also [3], [14]); we report it in the following.…”
Section: Then the Newton Iteration For (8) Is As Followsmentioning
confidence: 87%
“…The first question arising in examinating the above linear equation is about its solvability. An answer to this question can be found in [16] (see also [3], [14]); we report it in the following.…”
Section: Then the Newton Iteration For (8) Is As Followsmentioning
confidence: 87%
“…The factorizations used in Theorem 5 allow us to give the explicit general solution of BCX + YAB = B for A ∈ C m×n , B ∈ C n×n , and C ∈ C n×q . Indeed, by using the expressions of A, B, and C given in Theorem 5 it is not hard to show that BCX+YAB = B is consistent (in the unknowns X and Y) if and only if (I − C 1 C − 1 )(I − A − 1 A 1 ) = O (see [1,3]). Here, A − denotes a {1}-generalized inverse of A (that is, AA − A = A).…”
Section: Remarkmentioning
confidence: 99%
“…for arbitrary matrices Z, W, M, and N. Notice that matrices of smaller sizes are used in our computations compared to that given in [1].…”
Section: Remarkmentioning
confidence: 99%
“…Since R(A) ∩ R(B) ⊆ R(C), we have R((AXA * ) 1 2 ) ⊆ R(C) and R((BY B * ) 1 2 ) ⊆ R(C). Noting to R(AXA * ) ⊆ R((AXA * ) 1 2 ) and R(BY B * ) ⊆ R((BY B * ) 1 2 ), we obtain R(AXA * + BY B * ) ⊆ R(AXA * ) + R(BY B * ) ⊆ R(C). (3.14)…”
Section: In This Case There Exists a Unique Operator X Satisfying R(mentioning
confidence: 99%