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Matrix algebra

Posted 5 years ago

I am trying to do matrix rearragement in mathematica, with Assumptions = DLL [Element] Matrices[{3, 3}]; $Assumptions = DLR \[Element] Matrices[{3, 3}]; $Assumptions = DLT [Element] Matrices[{3, 3}]; $Assumptions = DLB \[Element] Matrices[{3, 3}]; $Assumptions = DRL [Element] Matrices[{3, 3}]; $Assumptions = DRR \[Element] Matrices[{3, 3}]; $Assumptions = DRT [Element] Matrices[{3, 3}]; $Assumptions = DRB \[Element] Matrices[{3, 3}]; $Assumptions = DTL [Element] Matrices[{3, 3}]; $Assumptions = DTR \[Element] Matrices[{3, 3}]; $Assumptions = DTT [Element] Matrices[{3, 3}]; $Assumptions = DTB \[Element] Matrices[{3, 3}]; $Assumptions = DBL [Element] Matrices[{3, 3}]; $Assumptions = DBR \[Element] Matrices[{3, 3}]; $Assumptions = DBT [Element] Matrices[{3, 3}]; $Assumptions = DBB \[Element] Matrices[{3, 3}]; $Assumptions = QL [Element] Matrices[{3, 1}]; $Assumptions = QR \[Element] Matrices[{3, 1}]; $Assumptions = QT [Element] Matrices[{3, 1}]; $Assumptions = QB [Element] Matrices[{3, 1}];

eq[1] = DLL QL + DLR QR + DLT QT + DLB QB; eq[2] = DRL QL + DRR QR + DRT QT + DRB QB; eq[3] = DTL QL + DTR QR + DTT QT + DTB QB; eq[4] = DBL QL + DBR QR + DBT QT + DBB QB; QR = [Lambda]1 QL; QB = [Lambda]2 QT; QR = [Lambda]1 QL; QB = [Lambda]2 QT; FT = DTL QL + DTR QR + DTT QT + DTB QB eq[6] = DBL QL + DBR QR + DBT QT + DBB QB + [Lambda]2 FT eq[7] = Collect[eq[6], QL] QT = E1 (DBL + DBR [Lambda]1 + DTL [Lambda]2 + DTR [Lambda]1 [Lambda]2) QL (-Inverse[(DBT +DBB [Lambda]2+DTT [Lambda]2+DTB [Lambda]2^2)]) eq[8] = DRL QL + DRR QR + DRT QT + DRB QB + [Lambda]1 FL [Lambda]2 = 1; QL = 1 eq[9] = Collect[eq[8], [Lambda]1]

The answer I am getting is DRL + DBL DRB E1 + DBL DRT E1 + DRB DTL E1 + DRT DTL E1 + (DLL + DRR + DBL DLB E1 + DBL DLT E1 + DBR DRB E1 + DBR DRT E1 + DLB DTL E1 + DLT DTL E1 + DRB DTR E1 + DRT DTR E1) [Lambda]1 + (DLR + DBR DLB E1 + DBR DLT E1 + DLB DTR E1 + DLT DTR E1) [Lambda]1^2

which forms a quadratic eigenvalue problem in \lambda_1 but the order of multiplication of matrices are not coming correct. Can this be fixed

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POSTED BY: Rajan Prasad
2 Replies
Posted 5 years ago

Thank you for the response I tried using Dot in QT and QB since later on everything comes as X QL=0 and we get co-eficients from X.. QR = [Lambda]1 QL ; QB = QT;

FL = DLL QL + DLR QR + DLT.QT + DLB .QB; eq[5] = DRL QL + DRR QR + DRT.QT + DRB .QB + [Lambda]1 FL;

[Lambda]2 = 1 FT = DTL QL + DTR QR + DTT.QT + DTB.QB eq[6] = DBL QL + DBR QR + DBT.QT + DBB.QB + [Lambda]2 FT eq[7] = Collect[eq[6], QL] QT = E1.(DBL + DBR [Lambda]1 + DTL [Lambda]2 + DTR [Lambda]1 [Lambda]2) eq[8] = DRL QL + DRR QR + DRT.QT + DRB .QB + [Lambda]1 FL QL = 1 (QL here i treated as some number since the the equation is X QL=0, where X=eq8] eq[10] = Collect[Expand[eq[8]], [Lambda]1]

i am getting the result as DRL + DLR [Lambda]1^2 + [Lambda]1 (DLL + DRR + DLB.E1.(DBL + DTL + DBR [Lambda]1 + DTR [Lambda]1) + DLT.E1.(DBL + DTL + DBR [Lambda]1 + DTR [Lambda]1)) + DRB.E1.(DBL + DTL + DBR [Lambda]1 + DTR [Lambda]1) + DRT.E1.(DBL + DTL + DBR [Lambda]1 + DTR [Lambda]1).

This result is not expandable. Expand command gives me the same result. So how to get the co-ffiecient after that

POSTED BY: Rajan Prasad

You appear to be confusing Dot (for matrix multiplication and related) with Times, and that latter is commutative (hence can reorder). Also redefining $Assumptions will not be of much use.

POSTED BY: Daniel Lichtblau
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