Having a homogenous system of linear equations with real coefficients with a non-trivial complex solution...












1












$begingroup$


I'm struggeling a bit with this proof.




Suppose we have a homogenous system of linear equations with real coefficients with a non-trivial complex solution than there is a real solution too.




This looks really simple and my first thought was...



In our course we proved that, if $alpha,beta in L(A,0) implies alpha + beta in L(A,0)$.



So let $z_1 = a+ib, z_2 = bar{z_1} = a-ib$ both $in L(A,0)$.



Than the sum of them $z_1+z_2 = a+ib + a-ib = 2a in L(A,0)$. Since $ainmathbb{R}$ there is a real solution too.



Additionally i figured out, that the complex conjugation is a field homorphism.



My questions are:




  1. Can somebody show me an example for a homogenous system of linear equations with real coefficients with a non-trivial complex solution?

  2. if my idea is correct, why can I assume that $bar{z_1}$ is a solution too?

  3. if my idea is not correct, where is my mistake?


Many thanks in advance






linear-algebra complex-numbers homogeneous-equation







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I'm struggeling a bit with this proof.




    Suppose we have a homogenous system of linear equations with real coefficients with a non-trivial complex solution than there is a real solution too.




    This looks really simple and my first thought was...



    In our course we proved that, if $alpha,beta in L(A,0) implies alpha + beta in L(A,0)$.



    So let $z_1 = a+ib, z_2 = bar{z_1} = a-ib$ both $in L(A,0)$.



    Than the sum of them $z_1+z_2 = a+ib + a-ib = 2a in L(A,0)$. Since $ainmathbb{R}$ there is a real solution too.



    Additionally i figured out, that the complex conjugation is a field homorphism.



    My questions are:




    1. Can somebody show me an example for a homogenous system of linear equations with real coefficients with a non-trivial complex solution?

    2. if my idea is correct, why can I assume that $bar{z_1}$ is a solution too?

    3. if my idea is not correct, where is my mistake?


    Many thanks in advance






    linear-algebra complex-numbers homogeneous-equation







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I'm struggeling a bit with this proof.




      Suppose we have a homogenous system of linear equations with real coefficients with a non-trivial complex solution than there is a real solution too.




      This looks really simple and my first thought was...



      In our course we proved that, if $alpha,beta in L(A,0) implies alpha + beta in L(A,0)$.



      So let $z_1 = a+ib, z_2 = bar{z_1} = a-ib$ both $in L(A,0)$.



      Than the sum of them $z_1+z_2 = a+ib + a-ib = 2a in L(A,0)$. Since $ainmathbb{R}$ there is a real solution too.



      Additionally i figured out, that the complex conjugation is a field homorphism.



      My questions are:




      1. Can somebody show me an example for a homogenous system of linear equations with real coefficients with a non-trivial complex solution?

      2. if my idea is correct, why can I assume that $bar{z_1}$ is a solution too?

      3. if my idea is not correct, where is my mistake?


      Many thanks in advance






      linear-algebra complex-numbers homogeneous-equation







      share|cite|improve this question











      $endgroup$




      I'm struggeling a bit with this proof.




      Suppose we have a homogenous system of linear equations with real coefficients with a non-trivial complex solution than there is a real solution too.




      This looks really simple and my first thought was...



      In our course we proved that, if $alpha,beta in L(A,0) implies alpha + beta in L(A,0)$.



      So let $z_1 = a+ib, z_2 = bar{z_1} = a-ib$ both $in L(A,0)$.



      Than the sum of them $z_1+z_2 = a+ib + a-ib = 2a in L(A,0)$. Since $ainmathbb{R}$ there is a real solution too.



      Additionally i figured out, that the complex conjugation is a field homorphism.



      My questions are:




      1. Can somebody show me an example for a homogenous system of linear equations with real coefficients with a non-trivial complex solution?

      2. if my idea is correct, why can I assume that $bar{z_1}$ is a solution too?

      3. if my idea is not correct, where is my mistake?


      Many thanks in advance






      linear-algebra complex-numbers homogeneous-equation




      linear-algebra complex-numbers homogeneous-equation






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 11 '18 at 12:13







      Matthias

















      asked Dec 10 '18 at 10:29









      MatthiasMatthias

      303




      303






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$


          1. Let $z=a+ibinBbb C^n$ with with $a,binBbb R^n$. When $Az=0$ for a real matrix $A$, you can split this into real and imaginary part to see that $Az=0$ is equivalent to $Aa=0$ and $Ab=0$. Hence, you can combine any two real solutions as $a+ib$ to obtain a (non-trivial) complex solution.

          2. Taking the complex conjugate of $Az=0$ gives $bar Abar z=0$ and for real $A$ this just gives $Abar z=0$, hence $z$ is a solution if and only if $bar z$ is a solution. (This also follows from 1. since $-b$ is a solution iff $b$ is one)

          3. It is correct!






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for this good answer. Regarding 2) you say "Taking the complex conjugates...", it is not clear to me why we can take them. Can you please explain this.
            $endgroup$
            – Matthias
            Dec 10 '18 at 10:56










          • $begingroup$
            First split into $n$ complex equations $a_{i1} z_1 + cdots + a_{in} z_n = 0$ and now those equalities in $Bbb C$ can just be conjugated on both sides as usual to get $overline{a_{i1}} ,overline{z_1} + cdots + overline{a_{in}}, overline{z_n} = 0$. So this is just the fact that $w=0$ is equivalent to $bar w=0$ in $Bbb C$.
            $endgroup$
            – Christoph
            Dec 10 '18 at 11:09












          • $begingroup$
            I think the technique is clear, but it is not clear to my why "... in $mathbb{C}$ can just be conjugated...". So I'm not insistent for pedantic reasons, but I want to know your trains of thoughts. Is it like an automatism... here are complex numbers what can I do with them (e.g. dissociate imaginary from real parts, conjugate...) And now you know that in this particular situation the conjugation ist the technique to go with?
            $endgroup$
            – Matthias
            Dec 10 '18 at 11:20












          • $begingroup$
            If $x=y$ and $f$ is any function, then $f(x)=f(y)$. Apply this to the situation where $x$ and $y$ are complex numbers and $f(x)=bar x$. You get that $x=y$ implies $bar y=bar x$. Now since $overline{bar x}= x$, you can also go the other way. It is the obvious thing to do in this situation because you know $Az=0$ and you want to somehow get an equation involving $bar z$.
            $endgroup$
            – Christoph
            Dec 10 '18 at 11:36





















          0












          $begingroup$

          The set of solutions of $Ax=0$ is a subspace (of the correspondinf vector space), so all the linear combinations of 'solution' vectors are again solutions of $Ax=0$. Fot $A $ real, we can consider the solution set/space either as a subspace (of the vector space we're working in) with scalars from $mathbb{R}$ or $mathbb{C}$. Considering it over the complex number field, again any scalar multiple of a real solution vector $u$ is a solution. (Clearly if $Au=0$ then $A((a+ib)u)=0$.



          Conversely if $A(u+iv)=0$ ($u,v $ real) then $Au=-iAv$ so $Au=Av=0$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            But you can't pick $a+bi$ such that for a given complex vector $u$ the scalar multiple $(a+bi)u$ is real, can you? (Try $u=(1,i)$)
            $endgroup$
            – Christoph
            Dec 10 '18 at 10:46












          • $begingroup$
            @Christoph That was not really what I was trying to say. I will read the question and answer and check.
            $endgroup$
            – AnyAD
            Dec 10 '18 at 10:52










          • $begingroup$
            I see, you wanted to produce a non-trivial complex solution so you assume $u$ is real in that part?
            $endgroup$
            – Christoph
            Dec 10 '18 at 10:53












          • $begingroup$
            @Christoph Thank you for pointing that out. I've added 'real' in the answer.
            $endgroup$
            – AnyAD
            Dec 10 '18 at 10:55










          • $begingroup$
            @AnyAD. Thanks for your answer. Why is it that $Au=-iAv implies Au=Av=0$?
            $endgroup$
            – Matthias
            Dec 10 '18 at 11:23













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          2 Answers
          2






          active

          oldest

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          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$


          1. Let $z=a+ibinBbb C^n$ with with $a,binBbb R^n$. When $Az=0$ for a real matrix $A$, you can split this into real and imaginary part to see that $Az=0$ is equivalent to $Aa=0$ and $Ab=0$. Hence, you can combine any two real solutions as $a+ib$ to obtain a (non-trivial) complex solution.

          2. Taking the complex conjugate of $Az=0$ gives $bar Abar z=0$ and for real $A$ this just gives $Abar z=0$, hence $z$ is a solution if and only if $bar z$ is a solution. (This also follows from 1. since $-b$ is a solution iff $b$ is one)

          3. It is correct!






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for this good answer. Regarding 2) you say "Taking the complex conjugates...", it is not clear to me why we can take them. Can you please explain this.
            $endgroup$
            – Matthias
            Dec 10 '18 at 10:56










          • $begingroup$
            First split into $n$ complex equations $a_{i1} z_1 + cdots + a_{in} z_n = 0$ and now those equalities in $Bbb C$ can just be conjugated on both sides as usual to get $overline{a_{i1}} ,overline{z_1} + cdots + overline{a_{in}}, overline{z_n} = 0$. So this is just the fact that $w=0$ is equivalent to $bar w=0$ in $Bbb C$.
            $endgroup$
            – Christoph
            Dec 10 '18 at 11:09












          • $begingroup$
            I think the technique is clear, but it is not clear to my why "... in $mathbb{C}$ can just be conjugated...". So I'm not insistent for pedantic reasons, but I want to know your trains of thoughts. Is it like an automatism... here are complex numbers what can I do with them (e.g. dissociate imaginary from real parts, conjugate...) And now you know that in this particular situation the conjugation ist the technique to go with?
            $endgroup$
            – Matthias
            Dec 10 '18 at 11:20












          • $begingroup$
            If $x=y$ and $f$ is any function, then $f(x)=f(y)$. Apply this to the situation where $x$ and $y$ are complex numbers and $f(x)=bar x$. You get that $x=y$ implies $bar y=bar x$. Now since $overline{bar x}= x$, you can also go the other way. It is the obvious thing to do in this situation because you know $Az=0$ and you want to somehow get an equation involving $bar z$.
            $endgroup$
            – Christoph
            Dec 10 '18 at 11:36


















          1












          $begingroup$


          1. Let $z=a+ibinBbb C^n$ with with $a,binBbb R^n$. When $Az=0$ for a real matrix $A$, you can split this into real and imaginary part to see that $Az=0$ is equivalent to $Aa=0$ and $Ab=0$. Hence, you can combine any two real solutions as $a+ib$ to obtain a (non-trivial) complex solution.

          2. Taking the complex conjugate of $Az=0$ gives $bar Abar z=0$ and for real $A$ this just gives $Abar z=0$, hence $z$ is a solution if and only if $bar z$ is a solution. (This also follows from 1. since $-b$ is a solution iff $b$ is one)

          3. It is correct!






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for this good answer. Regarding 2) you say "Taking the complex conjugates...", it is not clear to me why we can take them. Can you please explain this.
            $endgroup$
            – Matthias
            Dec 10 '18 at 10:56










          • $begingroup$
            First split into $n$ complex equations $a_{i1} z_1 + cdots + a_{in} z_n = 0$ and now those equalities in $Bbb C$ can just be conjugated on both sides as usual to get $overline{a_{i1}} ,overline{z_1} + cdots + overline{a_{in}}, overline{z_n} = 0$. So this is just the fact that $w=0$ is equivalent to $bar w=0$ in $Bbb C$.
            $endgroup$
            – Christoph
            Dec 10 '18 at 11:09












          • $begingroup$
            I think the technique is clear, but it is not clear to my why "... in $mathbb{C}$ can just be conjugated...". So I'm not insistent for pedantic reasons, but I want to know your trains of thoughts. Is it like an automatism... here are complex numbers what can I do with them (e.g. dissociate imaginary from real parts, conjugate...) And now you know that in this particular situation the conjugation ist the technique to go with?
            $endgroup$
            – Matthias
            Dec 10 '18 at 11:20












          • $begingroup$
            If $x=y$ and $f$ is any function, then $f(x)=f(y)$. Apply this to the situation where $x$ and $y$ are complex numbers and $f(x)=bar x$. You get that $x=y$ implies $bar y=bar x$. Now since $overline{bar x}= x$, you can also go the other way. It is the obvious thing to do in this situation because you know $Az=0$ and you want to somehow get an equation involving $bar z$.
            $endgroup$
            – Christoph
            Dec 10 '18 at 11:36
















          1












          1








          1





          $begingroup$


          1. Let $z=a+ibinBbb C^n$ with with $a,binBbb R^n$. When $Az=0$ for a real matrix $A$, you can split this into real and imaginary part to see that $Az=0$ is equivalent to $Aa=0$ and $Ab=0$. Hence, you can combine any two real solutions as $a+ib$ to obtain a (non-trivial) complex solution.

          2. Taking the complex conjugate of $Az=0$ gives $bar Abar z=0$ and for real $A$ this just gives $Abar z=0$, hence $z$ is a solution if and only if $bar z$ is a solution. (This also follows from 1. since $-b$ is a solution iff $b$ is one)

          3. It is correct!






          share|cite|improve this answer









          $endgroup$




          1. Let $z=a+ibinBbb C^n$ with with $a,binBbb R^n$. When $Az=0$ for a real matrix $A$, you can split this into real and imaginary part to see that $Az=0$ is equivalent to $Aa=0$ and $Ab=0$. Hence, you can combine any two real solutions as $a+ib$ to obtain a (non-trivial) complex solution.

          2. Taking the complex conjugate of $Az=0$ gives $bar Abar z=0$ and for real $A$ this just gives $Abar z=0$, hence $z$ is a solution if and only if $bar z$ is a solution. (This also follows from 1. since $-b$ is a solution iff $b$ is one)

          3. It is correct!







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 10 '18 at 10:42









          ChristophChristoph

          12.5k1642




          12.5k1642












          • $begingroup$
            Thanks for this good answer. Regarding 2) you say "Taking the complex conjugates...", it is not clear to me why we can take them. Can you please explain this.
            $endgroup$
            – Matthias
            Dec 10 '18 at 10:56










          • $begingroup$
            First split into $n$ complex equations $a_{i1} z_1 + cdots + a_{in} z_n = 0$ and now those equalities in $Bbb C$ can just be conjugated on both sides as usual to get $overline{a_{i1}} ,overline{z_1} + cdots + overline{a_{in}}, overline{z_n} = 0$. So this is just the fact that $w=0$ is equivalent to $bar w=0$ in $Bbb C$.
            $endgroup$
            – Christoph
            Dec 10 '18 at 11:09












          • $begingroup$
            I think the technique is clear, but it is not clear to my why "... in $mathbb{C}$ can just be conjugated...". So I'm not insistent for pedantic reasons, but I want to know your trains of thoughts. Is it like an automatism... here are complex numbers what can I do with them (e.g. dissociate imaginary from real parts, conjugate...) And now you know that in this particular situation the conjugation ist the technique to go with?
            $endgroup$
            – Matthias
            Dec 10 '18 at 11:20












          • $begingroup$
            If $x=y$ and $f$ is any function, then $f(x)=f(y)$. Apply this to the situation where $x$ and $y$ are complex numbers and $f(x)=bar x$. You get that $x=y$ implies $bar y=bar x$. Now since $overline{bar x}= x$, you can also go the other way. It is the obvious thing to do in this situation because you know $Az=0$ and you want to somehow get an equation involving $bar z$.
            $endgroup$
            – Christoph
            Dec 10 '18 at 11:36




















          • $begingroup$
            Thanks for this good answer. Regarding 2) you say "Taking the complex conjugates...", it is not clear to me why we can take them. Can you please explain this.
            $endgroup$
            – Matthias
            Dec 10 '18 at 10:56










          • $begingroup$
            First split into $n$ complex equations $a_{i1} z_1 + cdots + a_{in} z_n = 0$ and now those equalities in $Bbb C$ can just be conjugated on both sides as usual to get $overline{a_{i1}} ,overline{z_1} + cdots + overline{a_{in}}, overline{z_n} = 0$. So this is just the fact that $w=0$ is equivalent to $bar w=0$ in $Bbb C$.
            $endgroup$
            – Christoph
            Dec 10 '18 at 11:09












          • $begingroup$
            I think the technique is clear, but it is not clear to my why "... in $mathbb{C}$ can just be conjugated...". So I'm not insistent for pedantic reasons, but I want to know your trains of thoughts. Is it like an automatism... here are complex numbers what can I do with them (e.g. dissociate imaginary from real parts, conjugate...) And now you know that in this particular situation the conjugation ist the technique to go with?
            $endgroup$
            – Matthias
            Dec 10 '18 at 11:20












          • $begingroup$
            If $x=y$ and $f$ is any function, then $f(x)=f(y)$. Apply this to the situation where $x$ and $y$ are complex numbers and $f(x)=bar x$. You get that $x=y$ implies $bar y=bar x$. Now since $overline{bar x}= x$, you can also go the other way. It is the obvious thing to do in this situation because you know $Az=0$ and you want to somehow get an equation involving $bar z$.
            $endgroup$
            – Christoph
            Dec 10 '18 at 11:36


















          $begingroup$
          Thanks for this good answer. Regarding 2) you say "Taking the complex conjugates...", it is not clear to me why we can take them. Can you please explain this.
          $endgroup$
          – Matthias
          Dec 10 '18 at 10:56




          $begingroup$
          Thanks for this good answer. Regarding 2) you say "Taking the complex conjugates...", it is not clear to me why we can take them. Can you please explain this.
          $endgroup$
          – Matthias
          Dec 10 '18 at 10:56












          $begingroup$
          First split into $n$ complex equations $a_{i1} z_1 + cdots + a_{in} z_n = 0$ and now those equalities in $Bbb C$ can just be conjugated on both sides as usual to get $overline{a_{i1}} ,overline{z_1} + cdots + overline{a_{in}}, overline{z_n} = 0$. So this is just the fact that $w=0$ is equivalent to $bar w=0$ in $Bbb C$.
          $endgroup$
          – Christoph
          Dec 10 '18 at 11:09






          $begingroup$
          First split into $n$ complex equations $a_{i1} z_1 + cdots + a_{in} z_n = 0$ and now those equalities in $Bbb C$ can just be conjugated on both sides as usual to get $overline{a_{i1}} ,overline{z_1} + cdots + overline{a_{in}}, overline{z_n} = 0$. So this is just the fact that $w=0$ is equivalent to $bar w=0$ in $Bbb C$.
          $endgroup$
          – Christoph
          Dec 10 '18 at 11:09














          $begingroup$
          I think the technique is clear, but it is not clear to my why "... in $mathbb{C}$ can just be conjugated...". So I'm not insistent for pedantic reasons, but I want to know your trains of thoughts. Is it like an automatism... here are complex numbers what can I do with them (e.g. dissociate imaginary from real parts, conjugate...) And now you know that in this particular situation the conjugation ist the technique to go with?
          $endgroup$
          – Matthias
          Dec 10 '18 at 11:20






          $begingroup$
          I think the technique is clear, but it is not clear to my why "... in $mathbb{C}$ can just be conjugated...". So I'm not insistent for pedantic reasons, but I want to know your trains of thoughts. Is it like an automatism... here are complex numbers what can I do with them (e.g. dissociate imaginary from real parts, conjugate...) And now you know that in this particular situation the conjugation ist the technique to go with?
          $endgroup$
          – Matthias
          Dec 10 '18 at 11:20














          $begingroup$
          If $x=y$ and $f$ is any function, then $f(x)=f(y)$. Apply this to the situation where $x$ and $y$ are complex numbers and $f(x)=bar x$. You get that $x=y$ implies $bar y=bar x$. Now since $overline{bar x}= x$, you can also go the other way. It is the obvious thing to do in this situation because you know $Az=0$ and you want to somehow get an equation involving $bar z$.
          $endgroup$
          – Christoph
          Dec 10 '18 at 11:36






          $begingroup$
          If $x=y$ and $f$ is any function, then $f(x)=f(y)$. Apply this to the situation where $x$ and $y$ are complex numbers and $f(x)=bar x$. You get that $x=y$ implies $bar y=bar x$. Now since $overline{bar x}= x$, you can also go the other way. It is the obvious thing to do in this situation because you know $Az=0$ and you want to somehow get an equation involving $bar z$.
          $endgroup$
          – Christoph
          Dec 10 '18 at 11:36













          0












          $begingroup$

          The set of solutions of $Ax=0$ is a subspace (of the correspondinf vector space), so all the linear combinations of 'solution' vectors are again solutions of $Ax=0$. Fot $A $ real, we can consider the solution set/space either as a subspace (of the vector space we're working in) with scalars from $mathbb{R}$ or $mathbb{C}$. Considering it over the complex number field, again any scalar multiple of a real solution vector $u$ is a solution. (Clearly if $Au=0$ then $A((a+ib)u)=0$.



          Conversely if $A(u+iv)=0$ ($u,v $ real) then $Au=-iAv$ so $Au=Av=0$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            But you can't pick $a+bi$ such that for a given complex vector $u$ the scalar multiple $(a+bi)u$ is real, can you? (Try $u=(1,i)$)
            $endgroup$
            – Christoph
            Dec 10 '18 at 10:46












          • $begingroup$
            @Christoph That was not really what I was trying to say. I will read the question and answer and check.
            $endgroup$
            – AnyAD
            Dec 10 '18 at 10:52










          • $begingroup$
            I see, you wanted to produce a non-trivial complex solution so you assume $u$ is real in that part?
            $endgroup$
            – Christoph
            Dec 10 '18 at 10:53












          • $begingroup$
            @Christoph Thank you for pointing that out. I've added 'real' in the answer.
            $endgroup$
            – AnyAD
            Dec 10 '18 at 10:55










          • $begingroup$
            @AnyAD. Thanks for your answer. Why is it that $Au=-iAv implies Au=Av=0$?
            $endgroup$
            – Matthias
            Dec 10 '18 at 11:23


















          0












          $begingroup$

          The set of solutions of $Ax=0$ is a subspace (of the correspondinf vector space), so all the linear combinations of 'solution' vectors are again solutions of $Ax=0$. Fot $A $ real, we can consider the solution set/space either as a subspace (of the vector space we're working in) with scalars from $mathbb{R}$ or $mathbb{C}$. Considering it over the complex number field, again any scalar multiple of a real solution vector $u$ is a solution. (Clearly if $Au=0$ then $A((a+ib)u)=0$.



          Conversely if $A(u+iv)=0$ ($u,v $ real) then $Au=-iAv$ so $Au=Av=0$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            But you can't pick $a+bi$ such that for a given complex vector $u$ the scalar multiple $(a+bi)u$ is real, can you? (Try $u=(1,i)$)
            $endgroup$
            – Christoph
            Dec 10 '18 at 10:46












          • $begingroup$
            @Christoph That was not really what I was trying to say. I will read the question and answer and check.
            $endgroup$
            – AnyAD
            Dec 10 '18 at 10:52










          • $begingroup$
            I see, you wanted to produce a non-trivial complex solution so you assume $u$ is real in that part?
            $endgroup$
            – Christoph
            Dec 10 '18 at 10:53












          • $begingroup$
            @Christoph Thank you for pointing that out. I've added 'real' in the answer.
            $endgroup$
            – AnyAD
            Dec 10 '18 at 10:55










          • $begingroup$
            @AnyAD. Thanks for your answer. Why is it that $Au=-iAv implies Au=Av=0$?
            $endgroup$
            – Matthias
            Dec 10 '18 at 11:23
















          0












          0








          0





          $begingroup$

          The set of solutions of $Ax=0$ is a subspace (of the correspondinf vector space), so all the linear combinations of 'solution' vectors are again solutions of $Ax=0$. Fot $A $ real, we can consider the solution set/space either as a subspace (of the vector space we're working in) with scalars from $mathbb{R}$ or $mathbb{C}$. Considering it over the complex number field, again any scalar multiple of a real solution vector $u$ is a solution. (Clearly if $Au=0$ then $A((a+ib)u)=0$.



          Conversely if $A(u+iv)=0$ ($u,v $ real) then $Au=-iAv$ so $Au=Av=0$.






          share|cite|improve this answer











          $endgroup$



          The set of solutions of $Ax=0$ is a subspace (of the correspondinf vector space), so all the linear combinations of 'solution' vectors are again solutions of $Ax=0$. Fot $A $ real, we can consider the solution set/space either as a subspace (of the vector space we're working in) with scalars from $mathbb{R}$ or $mathbb{C}$. Considering it over the complex number field, again any scalar multiple of a real solution vector $u$ is a solution. (Clearly if $Au=0$ then $A((a+ib)u)=0$.



          Conversely if $A(u+iv)=0$ ($u,v $ real) then $Au=-iAv$ so $Au=Av=0$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 10 '18 at 10:54

























          answered Dec 10 '18 at 10:41









          AnyADAnyAD

          2,113812




          2,113812












          • $begingroup$
            But you can't pick $a+bi$ such that for a given complex vector $u$ the scalar multiple $(a+bi)u$ is real, can you? (Try $u=(1,i)$)
            $endgroup$
            – Christoph
            Dec 10 '18 at 10:46












          • $begingroup$
            @Christoph That was not really what I was trying to say. I will read the question and answer and check.
            $endgroup$
            – AnyAD
            Dec 10 '18 at 10:52










          • $begingroup$
            I see, you wanted to produce a non-trivial complex solution so you assume $u$ is real in that part?
            $endgroup$
            – Christoph
            Dec 10 '18 at 10:53












          • $begingroup$
            @Christoph Thank you for pointing that out. I've added 'real' in the answer.
            $endgroup$
            – AnyAD
            Dec 10 '18 at 10:55










          • $begingroup$
            @AnyAD. Thanks for your answer. Why is it that $Au=-iAv implies Au=Av=0$?
            $endgroup$
            – Matthias
            Dec 10 '18 at 11:23




















          • $begingroup$
            But you can't pick $a+bi$ such that for a given complex vector $u$ the scalar multiple $(a+bi)u$ is real, can you? (Try $u=(1,i)$)
            $endgroup$
            – Christoph
            Dec 10 '18 at 10:46












          • $begingroup$
            @Christoph That was not really what I was trying to say. I will read the question and answer and check.
            $endgroup$
            – AnyAD
            Dec 10 '18 at 10:52










          • $begingroup$
            I see, you wanted to produce a non-trivial complex solution so you assume $u$ is real in that part?
            $endgroup$
            – Christoph
            Dec 10 '18 at 10:53












          • $begingroup$
            @Christoph Thank you for pointing that out. I've added 'real' in the answer.
            $endgroup$
            – AnyAD
            Dec 10 '18 at 10:55










          • $begingroup$
            @AnyAD. Thanks for your answer. Why is it that $Au=-iAv implies Au=Av=0$?
            $endgroup$
            – Matthias
            Dec 10 '18 at 11:23


















          $begingroup$
          But you can't pick $a+bi$ such that for a given complex vector $u$ the scalar multiple $(a+bi)u$ is real, can you? (Try $u=(1,i)$)
          $endgroup$
          – Christoph
          Dec 10 '18 at 10:46






          $begingroup$
          But you can't pick $a+bi$ such that for a given complex vector $u$ the scalar multiple $(a+bi)u$ is real, can you? (Try $u=(1,i)$)
          $endgroup$
          – Christoph
          Dec 10 '18 at 10:46














          $begingroup$
          @Christoph That was not really what I was trying to say. I will read the question and answer and check.
          $endgroup$
          – AnyAD
          Dec 10 '18 at 10:52




          $begingroup$
          @Christoph That was not really what I was trying to say. I will read the question and answer and check.
          $endgroup$
          – AnyAD
          Dec 10 '18 at 10:52












          $begingroup$
          I see, you wanted to produce a non-trivial complex solution so you assume $u$ is real in that part?
          $endgroup$
          – Christoph
          Dec 10 '18 at 10:53






          $begingroup$
          I see, you wanted to produce a non-trivial complex solution so you assume $u$ is real in that part?
          $endgroup$
          – Christoph
          Dec 10 '18 at 10:53














          $begingroup$
          @Christoph Thank you for pointing that out. I've added 'real' in the answer.
          $endgroup$
          – AnyAD
          Dec 10 '18 at 10:55




          $begingroup$
          @Christoph Thank you for pointing that out. I've added 'real' in the answer.
          $endgroup$
          – AnyAD
          Dec 10 '18 at 10:55












          $begingroup$
          @AnyAD. Thanks for your answer. Why is it that $Au=-iAv implies Au=Av=0$?
          $endgroup$
          – Matthias
          Dec 10 '18 at 11:23






          $begingroup$
          @AnyAD. Thanks for your answer. Why is it that $Au=-iAv implies Au=Av=0$?
          $endgroup$
          – Matthias
          Dec 10 '18 at 11:23




















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