Solving Linear System Ax=b with complex A and complex b

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I present you a very simple linear system with one unknown.



beginequationlabelatag1
Ax=b \
left ( 3+4i right )x=(6+8i)
endequation



this paper("on the numerical solving of complex linear systems") says that I can solve the linear system by transforming A to matrix form and then solving it as follows:



beginequationlabelbtag2
beginpmatrix
3 & -4\
4 & 3
endpmatrix
binomx_rx_c
=
binomb_rb_c
endequation



Where $$b_r = 6,b_c=8$$



question: The translation of the A is fairly easy to understand. What I don't get is why b is not converted to matrix form yet solving the above system yields the correct answer. In other words, the following equations



beginequationlabelctag3
left{beginmatrix
3x_r-4x_c = 6\
4x_r+3x_c = 8
endmatrixright.
endequation
make no sense to me.



Note that, I know how to solve the linear system. I'm looking for a detailed explainantion of what's happening between (1) and (2)



Thank you






linear-algebra complex-numbers matrix-equations







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    up vote
    3
    down vote

    favorite












    I present you a very simple linear system with one unknown.



    beginequationlabelatag1
    Ax=b \
    left ( 3+4i right )x=(6+8i)
    endequation



    this paper("on the numerical solving of complex linear systems") says that I can solve the linear system by transforming A to matrix form and then solving it as follows:



    beginequationlabelbtag2
    beginpmatrix
    3 & -4\
    4 & 3
    endpmatrix
    binomx_rx_c
    =
    binomb_rb_c
    endequation



    Where $$b_r = 6,b_c=8$$



    question: The translation of the A is fairly easy to understand. What I don't get is why b is not converted to matrix form yet solving the above system yields the correct answer. In other words, the following equations



    beginequationlabelctag3
    left{beginmatrix
    3x_r-4x_c = 6\
    4x_r+3x_c = 8
    endmatrixright.
    endequation
    make no sense to me.



    Note that, I know how to solve the linear system. I'm looking for a detailed explainantion of what's happening between (1) and (2)



    Thank you






    linear-algebra complex-numbers matrix-equations







    share|cite|improve this question









    New contributor




    ObjectOrientedPirate is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.





















      up vote
      3
      down vote

      favorite









      up vote
      3
      down vote

      favorite











      I present you a very simple linear system with one unknown.



      beginequationlabelatag1
      Ax=b \
      left ( 3+4i right )x=(6+8i)
      endequation



      this paper("on the numerical solving of complex linear systems") says that I can solve the linear system by transforming A to matrix form and then solving it as follows:



      beginequationlabelbtag2
      beginpmatrix
      3 & -4\
      4 & 3
      endpmatrix
      binomx_rx_c
      =
      binomb_rb_c
      endequation



      Where $$b_r = 6,b_c=8$$



      question: The translation of the A is fairly easy to understand. What I don't get is why b is not converted to matrix form yet solving the above system yields the correct answer. In other words, the following equations



      beginequationlabelctag3
      left{beginmatrix
      3x_r-4x_c = 6\
      4x_r+3x_c = 8
      endmatrixright.
      endequation
      make no sense to me.



      Note that, I know how to solve the linear system. I'm looking for a detailed explainantion of what's happening between (1) and (2)



      Thank you






      linear-algebra complex-numbers matrix-equations







      share|cite|improve this question









      New contributor




      ObjectOrientedPirate is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      I present you a very simple linear system with one unknown.



      beginequationlabelatag1
      Ax=b \
      left ( 3+4i right )x=(6+8i)
      endequation



      this paper("on the numerical solving of complex linear systems") says that I can solve the linear system by transforming A to matrix form and then solving it as follows:



      beginequationlabelbtag2
      beginpmatrix
      3 & -4\
      4 & 3
      endpmatrix
      binomx_rx_c
      =
      binomb_rb_c
      endequation



      Where $$b_r = 6,b_c=8$$



      question: The translation of the A is fairly easy to understand. What I don't get is why b is not converted to matrix form yet solving the above system yields the correct answer. In other words, the following equations



      beginequationlabelctag3
      left{beginmatrix
      3x_r-4x_c = 6\
      4x_r+3x_c = 8
      endmatrixright.
      endequation
      make no sense to me.



      Note that, I know how to solve the linear system. I'm looking for a detailed explainantion of what's happening between (1) and (2)



      Thank you






      linear-algebra complex-numbers matrix-equations




      linear-algebra complex-numbers matrix-equations






      share|cite|improve this question









      New contributor




      ObjectOrientedPirate is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question









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      ObjectOrientedPirate is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      edited 3 hours ago





















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      ObjectOrientedPirate

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          3 Answers
          3






          active

          oldest

          votes

















          up vote
          2
          down vote



          accepted










          Let's try to justify the equations given by (3). If we write out the product $(3+4i)x$, we have
          $$
          (3+4i)x = (3 + 4i)(x_r + x_ci) = 3x_r + 3x_c i + 4x_r i + 4x_c i^2 \
          = [3x_r - 4x_c] + [4x_r + 3x_c]i
          $$
          Now, in order for two complex numbers to be equal, their real parts must be equal and their imaginary parts must be equal. So, in order to have $(3+4i)x = b$, we must have
          $$
          3x_r - 4x_c = b_r\
          4x_r + 3x_c = b_c
          $$
          which is precisely the system of equations that you've come up with.






          share|cite|improve this answer




















          • Does your answer imply that if A and b are complex numbers then x must be element of the complex numbers? I'm sorry if this is obvious, my algebra is really dusty
            – ObjectOrientedPirate
            3 hours ago







          • 1




            Yes, my answer does imply this. Just as $frac ab$ is a real number for any real $a,b$ with $b neq 0$, so is $frac AB$ a complex number for any complex numbers $A,B$ with $B neq 0 + 0i$.
            – Omnomnomnom
            3 hours ago


















          up vote
          2
          down vote













          Strictly speaking they should be converted to matrix form and you would solve a block-matrix equation system.



          $$beginbmatrix3&-4\4&3endbmatrixbeginbmatrixx_r&-x_c\x_c&x_rendbmatrix=beginbmatrixb_r&-b_c\b_c&b_rendbmatrix$$



          It just happens that this will be the same thing for this example. For more advanced fields of numbers we will need to resort to this block-matrix embedding.






          share|cite|improve this answer




















          • Why "should" the system of equations be solved in this particular way?
            – Omnomnomnom
            3 hours ago










          • @Omnomnomnom Imagine the more general case where "scalars" are permutations, we would not be able to represent a generic permutation with anything less than a matrix of same size.
            – mathreadler
            3 hours ago


















          up vote
          0
          down vote













          $x=x_r+ix_i$. When we multiply with $3+4i$ we get $3x_r+3ix_i+4ix_r-4x_i=6+8i$. We now equate the real and imaginary parts separately:$$3x_r-4x_i=6\3x_i+4x_r=8$$






          share|cite|improve this answer




















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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            2
            down vote



            accepted










            Let's try to justify the equations given by (3). If we write out the product $(3+4i)x$, we have
            $$
            (3+4i)x = (3 + 4i)(x_r + x_ci) = 3x_r + 3x_c i + 4x_r i + 4x_c i^2 \
            = [3x_r - 4x_c] + [4x_r + 3x_c]i
            $$
            Now, in order for two complex numbers to be equal, their real parts must be equal and their imaginary parts must be equal. So, in order to have $(3+4i)x = b$, we must have
            $$
            3x_r - 4x_c = b_r\
            4x_r + 3x_c = b_c
            $$
            which is precisely the system of equations that you've come up with.






            share|cite|improve this answer




















            • Does your answer imply that if A and b are complex numbers then x must be element of the complex numbers? I'm sorry if this is obvious, my algebra is really dusty
              – ObjectOrientedPirate
              3 hours ago







            • 1




              Yes, my answer does imply this. Just as $frac ab$ is a real number for any real $a,b$ with $b neq 0$, so is $frac AB$ a complex number for any complex numbers $A,B$ with $B neq 0 + 0i$.
              – Omnomnomnom
              3 hours ago















            up vote
            2
            down vote



            accepted










            Let's try to justify the equations given by (3). If we write out the product $(3+4i)x$, we have
            $$
            (3+4i)x = (3 + 4i)(x_r + x_ci) = 3x_r + 3x_c i + 4x_r i + 4x_c i^2 \
            = [3x_r - 4x_c] + [4x_r + 3x_c]i
            $$
            Now, in order for two complex numbers to be equal, their real parts must be equal and their imaginary parts must be equal. So, in order to have $(3+4i)x = b$, we must have
            $$
            3x_r - 4x_c = b_r\
            4x_r + 3x_c = b_c
            $$
            which is precisely the system of equations that you've come up with.






            share|cite|improve this answer




















            • Does your answer imply that if A and b are complex numbers then x must be element of the complex numbers? I'm sorry if this is obvious, my algebra is really dusty
              – ObjectOrientedPirate
              3 hours ago







            • 1




              Yes, my answer does imply this. Just as $frac ab$ is a real number for any real $a,b$ with $b neq 0$, so is $frac AB$ a complex number for any complex numbers $A,B$ with $B neq 0 + 0i$.
              – Omnomnomnom
              3 hours ago













            up vote
            2
            down vote



            accepted







            up vote
            2
            down vote



            accepted






            Let's try to justify the equations given by (3). If we write out the product $(3+4i)x$, we have
            $$
            (3+4i)x = (3 + 4i)(x_r + x_ci) = 3x_r + 3x_c i + 4x_r i + 4x_c i^2 \
            = [3x_r - 4x_c] + [4x_r + 3x_c]i
            $$
            Now, in order for two complex numbers to be equal, their real parts must be equal and their imaginary parts must be equal. So, in order to have $(3+4i)x = b$, we must have
            $$
            3x_r - 4x_c = b_r\
            4x_r + 3x_c = b_c
            $$
            which is precisely the system of equations that you've come up with.






            share|cite|improve this answer












            Let's try to justify the equations given by (3). If we write out the product $(3+4i)x$, we have
            $$
            (3+4i)x = (3 + 4i)(x_r + x_ci) = 3x_r + 3x_c i + 4x_r i + 4x_c i^2 \
            = [3x_r - 4x_c] + [4x_r + 3x_c]i
            $$
            Now, in order for two complex numbers to be equal, their real parts must be equal and their imaginary parts must be equal. So, in order to have $(3+4i)x = b$, we must have
            $$
            3x_r - 4x_c = b_r\
            4x_r + 3x_c = b_c
            $$
            which is precisely the system of equations that you've come up with.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 3 hours ago









            Omnomnomnom

            123k786173




            123k786173











            • Does your answer imply that if A and b are complex numbers then x must be element of the complex numbers? I'm sorry if this is obvious, my algebra is really dusty
              – ObjectOrientedPirate
              3 hours ago







            • 1




              Yes, my answer does imply this. Just as $frac ab$ is a real number for any real $a,b$ with $b neq 0$, so is $frac AB$ a complex number for any complex numbers $A,B$ with $B neq 0 + 0i$.
              – Omnomnomnom
              3 hours ago

















            • Does your answer imply that if A and b are complex numbers then x must be element of the complex numbers? I'm sorry if this is obvious, my algebra is really dusty
              – ObjectOrientedPirate
              3 hours ago







            • 1




              Yes, my answer does imply this. Just as $frac ab$ is a real number for any real $a,b$ with $b neq 0$, so is $frac AB$ a complex number for any complex numbers $A,B$ with $B neq 0 + 0i$.
              – Omnomnomnom
              3 hours ago
















            Does your answer imply that if A and b are complex numbers then x must be element of the complex numbers? I'm sorry if this is obvious, my algebra is really dusty
            – ObjectOrientedPirate
            3 hours ago





            Does your answer imply that if A and b are complex numbers then x must be element of the complex numbers? I'm sorry if this is obvious, my algebra is really dusty
            – ObjectOrientedPirate
            3 hours ago





            1




            1




            Yes, my answer does imply this. Just as $frac ab$ is a real number for any real $a,b$ with $b neq 0$, so is $frac AB$ a complex number for any complex numbers $A,B$ with $B neq 0 + 0i$.
            – Omnomnomnom
            3 hours ago





            Yes, my answer does imply this. Just as $frac ab$ is a real number for any real $a,b$ with $b neq 0$, so is $frac AB$ a complex number for any complex numbers $A,B$ with $B neq 0 + 0i$.
            – Omnomnomnom
            3 hours ago











            up vote
            2
            down vote













            Strictly speaking they should be converted to matrix form and you would solve a block-matrix equation system.



            $$beginbmatrix3&-4\4&3endbmatrixbeginbmatrixx_r&-x_c\x_c&x_rendbmatrix=beginbmatrixb_r&-b_c\b_c&b_rendbmatrix$$



            It just happens that this will be the same thing for this example. For more advanced fields of numbers we will need to resort to this block-matrix embedding.






            share|cite|improve this answer




















            • Why "should" the system of equations be solved in this particular way?
              – Omnomnomnom
              3 hours ago










            • @Omnomnomnom Imagine the more general case where "scalars" are permutations, we would not be able to represent a generic permutation with anything less than a matrix of same size.
              – mathreadler
              3 hours ago















            up vote
            2
            down vote













            Strictly speaking they should be converted to matrix form and you would solve a block-matrix equation system.



            $$beginbmatrix3&-4\4&3endbmatrixbeginbmatrixx_r&-x_c\x_c&x_rendbmatrix=beginbmatrixb_r&-b_c\b_c&b_rendbmatrix$$



            It just happens that this will be the same thing for this example. For more advanced fields of numbers we will need to resort to this block-matrix embedding.






            share|cite|improve this answer




















            • Why "should" the system of equations be solved in this particular way?
              – Omnomnomnom
              3 hours ago










            • @Omnomnomnom Imagine the more general case where "scalars" are permutations, we would not be able to represent a generic permutation with anything less than a matrix of same size.
              – mathreadler
              3 hours ago













            up vote
            2
            down vote










            up vote
            2
            down vote









            Strictly speaking they should be converted to matrix form and you would solve a block-matrix equation system.



            $$beginbmatrix3&-4\4&3endbmatrixbeginbmatrixx_r&-x_c\x_c&x_rendbmatrix=beginbmatrixb_r&-b_c\b_c&b_rendbmatrix$$



            It just happens that this will be the same thing for this example. For more advanced fields of numbers we will need to resort to this block-matrix embedding.






            share|cite|improve this answer












            Strictly speaking they should be converted to matrix form and you would solve a block-matrix equation system.



            $$beginbmatrix3&-4\4&3endbmatrixbeginbmatrixx_r&-x_c\x_c&x_rendbmatrix=beginbmatrixb_r&-b_c\b_c&b_rendbmatrix$$



            It just happens that this will be the same thing for this example. For more advanced fields of numbers we will need to resort to this block-matrix embedding.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 3 hours ago









            mathreadler

            14.1k72059




            14.1k72059











            • Why "should" the system of equations be solved in this particular way?
              – Omnomnomnom
              3 hours ago










            • @Omnomnomnom Imagine the more general case where "scalars" are permutations, we would not be able to represent a generic permutation with anything less than a matrix of same size.
              – mathreadler
              3 hours ago

















            • Why "should" the system of equations be solved in this particular way?
              – Omnomnomnom
              3 hours ago










            • @Omnomnomnom Imagine the more general case where "scalars" are permutations, we would not be able to represent a generic permutation with anything less than a matrix of same size.
              – mathreadler
              3 hours ago
















            Why "should" the system of equations be solved in this particular way?
            – Omnomnomnom
            3 hours ago




            Why "should" the system of equations be solved in this particular way?
            – Omnomnomnom
            3 hours ago












            @Omnomnomnom Imagine the more general case where "scalars" are permutations, we would not be able to represent a generic permutation with anything less than a matrix of same size.
            – mathreadler
            3 hours ago





            @Omnomnomnom Imagine the more general case where "scalars" are permutations, we would not be able to represent a generic permutation with anything less than a matrix of same size.
            – mathreadler
            3 hours ago











            up vote
            0
            down vote













            $x=x_r+ix_i$. When we multiply with $3+4i$ we get $3x_r+3ix_i+4ix_r-4x_i=6+8i$. We now equate the real and imaginary parts separately:$$3x_r-4x_i=6\3x_i+4x_r=8$$






            share|cite|improve this answer
























              up vote
              0
              down vote













              $x=x_r+ix_i$. When we multiply with $3+4i$ we get $3x_r+3ix_i+4ix_r-4x_i=6+8i$. We now equate the real and imaginary parts separately:$$3x_r-4x_i=6\3x_i+4x_r=8$$






              share|cite|improve this answer






















                up vote
                0
                down vote










                up vote
                0
                down vote









                $x=x_r+ix_i$. When we multiply with $3+4i$ we get $3x_r+3ix_i+4ix_r-4x_i=6+8i$. We now equate the real and imaginary parts separately:$$3x_r-4x_i=6\3x_i+4x_r=8$$






                share|cite|improve this answer












                $x=x_r+ix_i$. When we multiply with $3+4i$ we get $3x_r+3ix_i+4ix_r-4x_i=6+8i$. We now equate the real and imaginary parts separately:$$3x_r-4x_i=6\3x_i+4x_r=8$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 3 hours ago









                Andrei

                8,6512923




                8,6512923




















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