Can I compare real and complex eigenvalues?












20












$begingroup$


I'm calculating the eigenvalues of the matrix $begin{pmatrix} 2 &0 &0& 1\
0 &1& 0& 1\
0 &0& 3& 1\
-1 &0 &0 &1end{pmatrix}$,



which are $1$,$3$, $frac{3}{2}+sqrt{3}i$ and $frac{3}{2}-sqrt{3}i$.



I wish to recognize the biggest and smallest of these. But how can I compare real and complex numbers?






matrices complex-numbers eigenvalues-eigenvectors







share|cite|improve this question











$endgroup$

















    20












    $begingroup$


    I'm calculating the eigenvalues of the matrix $begin{pmatrix} 2 &0 &0& 1\
    0 &1& 0& 1\
    0 &0& 3& 1\
    -1 &0 &0 &1end{pmatrix}$,



    which are $1$,$3$, $frac{3}{2}+sqrt{3}i$ and $frac{3}{2}-sqrt{3}i$.



    I wish to recognize the biggest and smallest of these. But how can I compare real and complex numbers?






    matrices complex-numbers eigenvalues-eigenvectors







    share|cite|improve this question











    $endgroup$















      20












      20








      20


      5



      $begingroup$


      I'm calculating the eigenvalues of the matrix $begin{pmatrix} 2 &0 &0& 1\
      0 &1& 0& 1\
      0 &0& 3& 1\
      -1 &0 &0 &1end{pmatrix}$,



      which are $1$,$3$, $frac{3}{2}+sqrt{3}i$ and $frac{3}{2}-sqrt{3}i$.



      I wish to recognize the biggest and smallest of these. But how can I compare real and complex numbers?






      matrices complex-numbers eigenvalues-eigenvectors







      share|cite|improve this question











      $endgroup$




      I'm calculating the eigenvalues of the matrix $begin{pmatrix} 2 &0 &0& 1\
      0 &1& 0& 1\
      0 &0& 3& 1\
      -1 &0 &0 &1end{pmatrix}$,



      which are $1$,$3$, $frac{3}{2}+sqrt{3}i$ and $frac{3}{2}-sqrt{3}i$.



      I wish to recognize the biggest and smallest of these. But how can I compare real and complex numbers?






      matrices complex-numbers eigenvalues-eigenvectors




      matrices complex-numbers eigenvalues-eigenvectors






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 31 '18 at 9:10









      Tommi Brander

      981922




      981922










      asked Apr 19 '16 at 13:52









      mavaviljmavavilj

      2,81411137




      2,81411137






















          3 Answers
          3






          active

          oldest

          votes


















          38












          $begingroup$

          In general, when talking about "largest" eigenvalue, we are usually talking about largest in absolute value (or magnitude,) where $|a+bi|=sqrt{a^2+b^2}$.



          This means sometimes that there isn't one eigenvalue that is "largest", because two different eigenvalues can have the same absolute value.



          As mentioned by others, complex numbers are not themselves ordered.



          As mentioned in the comments below, if you know a matrix has only real eigenvalues, then the question of "largest" and "smallest" eigenvalues will depend on the context.



          The "largest" eigenvalue for a matrix $A$ is often interesting, particularly when it is unique, because then for large $n$, $A^n$ is dominated by the action on the eigenvectors for those values. This is useful for putting bounds on $A^nmathbf v$.






          share|cite|improve this answer











          $endgroup$









          • 7




            $begingroup$
            One needs to be careful with one issue: when talking about the smallest eigenvalue of a Hermitian matrix, this may either mean the “farthest down” value (e.g. with the ground-state energy of a QM Hamiltonian – you can always gauge this to be positive, but it's quite common to have all energies negative) or indeed the smallest-absolute value.
            $endgroup$
            – leftaroundabout
            Apr 19 '16 at 14:39





















          4












          $begingroup$

          Suppose that there was an order on $mathbb{C}$ compatible with the natural order on $mathbb{R}$. Then either $i>0$ or $i<0$. Assume that $i>0$, then multiplying this inequality by $i$ we find that $-1=i^2>0$, but that's a contradiction. Thus $i<0$. But then multiplying by $i$ we get $-1=i^2>0$ (the inequality flipped since we multiplied by a negative number). In both cases we arrive at a contradiction. Hence there's no order on $mathbb{C}$ compatible with $mathbb{R}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            You completely missed the point of the question.
            $endgroup$
            – Najib Idrissi
            Apr 22 '16 at 11:26










          • $begingroup$
            Yes and no, Thomas Andrews' answer already answers the point of the question, I merely wanted to give this simple argument why no natural order on $mathbb{C}$ exists that extends the order of $mathbb{R}$.
            $endgroup$
            – Mathematician 42
            Apr 22 '16 at 12:54



















          0












          $begingroup$

          You can't, in the sense that there is no natural total order in $mathbb{C}$.






          share|cite|improve this answer









          $endgroup$









          • 7




            $begingroup$
            But I'm asked to do it. This is an assignment.
            $endgroup$
            – mavavilj
            Apr 19 '16 at 13:55






          • 3




            $begingroup$
            Well is $(0,1)$ or $(1,0)$ larger? You need to boil things down to real numbers in some way to compare complex values. Either you have miscalculated the eigenvalues, or you have misunderstood the question.
            $endgroup$
            – Matias Heikkilä
            Apr 19 '16 at 13:56






          • 4




            $begingroup$
            There are total orders on $mathbb{C}$ (e.g. the lexicographic order, for which $1 > i > -i > -1$). I don't know what you mean by "natural", though it is true that $mathbb{C}$ cannot be made into an ordered field. Anyway, in the context of eigenvalues it's the absolute value that counts, as Thomas Andrews says.
            $endgroup$
            – Najib Idrissi
            Apr 19 '16 at 14:01












          • $begingroup$
            Yes there are total orders on $mathbb{C}$, but which one you would use if you were asked to compare $a in mathbb{C}$ and $b in mathbb{C}$? Could you be sure that your friend uses the same order if given the same task? If $a, b in mathbb{R}$ instead, we tend to use the usual order relation, and there would be no problem. That's what I mean by natural. But this is not very interesting, since the original question should have been read as "compare the absolute values". Good to know that convention in the future, though.
            $endgroup$
            – Matias Heikkilä
            Apr 20 '16 at 9:02













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          3 Answers
          3






          active

          oldest

          votes








          3 Answers
          3






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          38












          $begingroup$

          In general, when talking about "largest" eigenvalue, we are usually talking about largest in absolute value (or magnitude,) where $|a+bi|=sqrt{a^2+b^2}$.



          This means sometimes that there isn't one eigenvalue that is "largest", because two different eigenvalues can have the same absolute value.



          As mentioned by others, complex numbers are not themselves ordered.



          As mentioned in the comments below, if you know a matrix has only real eigenvalues, then the question of "largest" and "smallest" eigenvalues will depend on the context.



          The "largest" eigenvalue for a matrix $A$ is often interesting, particularly when it is unique, because then for large $n$, $A^n$ is dominated by the action on the eigenvectors for those values. This is useful for putting bounds on $A^nmathbf v$.






          share|cite|improve this answer











          $endgroup$









          • 7




            $begingroup$
            One needs to be careful with one issue: when talking about the smallest eigenvalue of a Hermitian matrix, this may either mean the “farthest down” value (e.g. with the ground-state energy of a QM Hamiltonian – you can always gauge this to be positive, but it's quite common to have all energies negative) or indeed the smallest-absolute value.
            $endgroup$
            – leftaroundabout
            Apr 19 '16 at 14:39


















          38












          $begingroup$

          In general, when talking about "largest" eigenvalue, we are usually talking about largest in absolute value (or magnitude,) where $|a+bi|=sqrt{a^2+b^2}$.



          This means sometimes that there isn't one eigenvalue that is "largest", because two different eigenvalues can have the same absolute value.



          As mentioned by others, complex numbers are not themselves ordered.



          As mentioned in the comments below, if you know a matrix has only real eigenvalues, then the question of "largest" and "smallest" eigenvalues will depend on the context.



          The "largest" eigenvalue for a matrix $A$ is often interesting, particularly when it is unique, because then for large $n$, $A^n$ is dominated by the action on the eigenvectors for those values. This is useful for putting bounds on $A^nmathbf v$.






          share|cite|improve this answer











          $endgroup$









          • 7




            $begingroup$
            One needs to be careful with one issue: when talking about the smallest eigenvalue of a Hermitian matrix, this may either mean the “farthest down” value (e.g. with the ground-state energy of a QM Hamiltonian – you can always gauge this to be positive, but it's quite common to have all energies negative) or indeed the smallest-absolute value.
            $endgroup$
            – leftaroundabout
            Apr 19 '16 at 14:39
















          38












          38








          38





          $begingroup$

          In general, when talking about "largest" eigenvalue, we are usually talking about largest in absolute value (or magnitude,) where $|a+bi|=sqrt{a^2+b^2}$.



          This means sometimes that there isn't one eigenvalue that is "largest", because two different eigenvalues can have the same absolute value.



          As mentioned by others, complex numbers are not themselves ordered.



          As mentioned in the comments below, if you know a matrix has only real eigenvalues, then the question of "largest" and "smallest" eigenvalues will depend on the context.



          The "largest" eigenvalue for a matrix $A$ is often interesting, particularly when it is unique, because then for large $n$, $A^n$ is dominated by the action on the eigenvectors for those values. This is useful for putting bounds on $A^nmathbf v$.






          share|cite|improve this answer











          $endgroup$



          In general, when talking about "largest" eigenvalue, we are usually talking about largest in absolute value (or magnitude,) where $|a+bi|=sqrt{a^2+b^2}$.



          This means sometimes that there isn't one eigenvalue that is "largest", because two different eigenvalues can have the same absolute value.



          As mentioned by others, complex numbers are not themselves ordered.



          As mentioned in the comments below, if you know a matrix has only real eigenvalues, then the question of "largest" and "smallest" eigenvalues will depend on the context.



          The "largest" eigenvalue for a matrix $A$ is often interesting, particularly when it is unique, because then for large $n$, $A^n$ is dominated by the action on the eigenvectors for those values. This is useful for putting bounds on $A^nmathbf v$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Apr 20 '16 at 13:55

























          answered Apr 19 '16 at 14:07









          Thomas AndrewsThomas Andrews

          130k12146298




          130k12146298








          • 7




            $begingroup$
            One needs to be careful with one issue: when talking about the smallest eigenvalue of a Hermitian matrix, this may either mean the “farthest down” value (e.g. with the ground-state energy of a QM Hamiltonian – you can always gauge this to be positive, but it's quite common to have all energies negative) or indeed the smallest-absolute value.
            $endgroup$
            – leftaroundabout
            Apr 19 '16 at 14:39
















          • 7




            $begingroup$
            One needs to be careful with one issue: when talking about the smallest eigenvalue of a Hermitian matrix, this may either mean the “farthest down” value (e.g. with the ground-state energy of a QM Hamiltonian – you can always gauge this to be positive, but it's quite common to have all energies negative) or indeed the smallest-absolute value.
            $endgroup$
            – leftaroundabout
            Apr 19 '16 at 14:39










          7




          7




          $begingroup$
          One needs to be careful with one issue: when talking about the smallest eigenvalue of a Hermitian matrix, this may either mean the “farthest down” value (e.g. with the ground-state energy of a QM Hamiltonian – you can always gauge this to be positive, but it's quite common to have all energies negative) or indeed the smallest-absolute value.
          $endgroup$
          – leftaroundabout
          Apr 19 '16 at 14:39






          $begingroup$
          One needs to be careful with one issue: when talking about the smallest eigenvalue of a Hermitian matrix, this may either mean the “farthest down” value (e.g. with the ground-state energy of a QM Hamiltonian – you can always gauge this to be positive, but it's quite common to have all energies negative) or indeed the smallest-absolute value.
          $endgroup$
          – leftaroundabout
          Apr 19 '16 at 14:39













          4












          $begingroup$

          Suppose that there was an order on $mathbb{C}$ compatible with the natural order on $mathbb{R}$. Then either $i>0$ or $i<0$. Assume that $i>0$, then multiplying this inequality by $i$ we find that $-1=i^2>0$, but that's a contradiction. Thus $i<0$. But then multiplying by $i$ we get $-1=i^2>0$ (the inequality flipped since we multiplied by a negative number). In both cases we arrive at a contradiction. Hence there's no order on $mathbb{C}$ compatible with $mathbb{R}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            You completely missed the point of the question.
            $endgroup$
            – Najib Idrissi
            Apr 22 '16 at 11:26










          • $begingroup$
            Yes and no, Thomas Andrews' answer already answers the point of the question, I merely wanted to give this simple argument why no natural order on $mathbb{C}$ exists that extends the order of $mathbb{R}$.
            $endgroup$
            – Mathematician 42
            Apr 22 '16 at 12:54
















          4












          $begingroup$

          Suppose that there was an order on $mathbb{C}$ compatible with the natural order on $mathbb{R}$. Then either $i>0$ or $i<0$. Assume that $i>0$, then multiplying this inequality by $i$ we find that $-1=i^2>0$, but that's a contradiction. Thus $i<0$. But then multiplying by $i$ we get $-1=i^2>0$ (the inequality flipped since we multiplied by a negative number). In both cases we arrive at a contradiction. Hence there's no order on $mathbb{C}$ compatible with $mathbb{R}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            You completely missed the point of the question.
            $endgroup$
            – Najib Idrissi
            Apr 22 '16 at 11:26










          • $begingroup$
            Yes and no, Thomas Andrews' answer already answers the point of the question, I merely wanted to give this simple argument why no natural order on $mathbb{C}$ exists that extends the order of $mathbb{R}$.
            $endgroup$
            – Mathematician 42
            Apr 22 '16 at 12:54














          4












          4








          4





          $begingroup$

          Suppose that there was an order on $mathbb{C}$ compatible with the natural order on $mathbb{R}$. Then either $i>0$ or $i<0$. Assume that $i>0$, then multiplying this inequality by $i$ we find that $-1=i^2>0$, but that's a contradiction. Thus $i<0$. But then multiplying by $i$ we get $-1=i^2>0$ (the inequality flipped since we multiplied by a negative number). In both cases we arrive at a contradiction. Hence there's no order on $mathbb{C}$ compatible with $mathbb{R}$.






          share|cite|improve this answer









          $endgroup$



          Suppose that there was an order on $mathbb{C}$ compatible with the natural order on $mathbb{R}$. Then either $i>0$ or $i<0$. Assume that $i>0$, then multiplying this inequality by $i$ we find that $-1=i^2>0$, but that's a contradiction. Thus $i<0$. But then multiplying by $i$ we get $-1=i^2>0$ (the inequality flipped since we multiplied by a negative number). In both cases we arrive at a contradiction. Hence there's no order on $mathbb{C}$ compatible with $mathbb{R}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Apr 20 '16 at 6:42









          Mathematician 42Mathematician 42

          8,65411438




          8,65411438












          • $begingroup$
            You completely missed the point of the question.
            $endgroup$
            – Najib Idrissi
            Apr 22 '16 at 11:26










          • $begingroup$
            Yes and no, Thomas Andrews' answer already answers the point of the question, I merely wanted to give this simple argument why no natural order on $mathbb{C}$ exists that extends the order of $mathbb{R}$.
            $endgroup$
            – Mathematician 42
            Apr 22 '16 at 12:54


















          • $begingroup$
            You completely missed the point of the question.
            $endgroup$
            – Najib Idrissi
            Apr 22 '16 at 11:26










          • $begingroup$
            Yes and no, Thomas Andrews' answer already answers the point of the question, I merely wanted to give this simple argument why no natural order on $mathbb{C}$ exists that extends the order of $mathbb{R}$.
            $endgroup$
            – Mathematician 42
            Apr 22 '16 at 12:54
















          $begingroup$
          You completely missed the point of the question.
          $endgroup$
          – Najib Idrissi
          Apr 22 '16 at 11:26




          $begingroup$
          You completely missed the point of the question.
          $endgroup$
          – Najib Idrissi
          Apr 22 '16 at 11:26












          $begingroup$
          Yes and no, Thomas Andrews' answer already answers the point of the question, I merely wanted to give this simple argument why no natural order on $mathbb{C}$ exists that extends the order of $mathbb{R}$.
          $endgroup$
          – Mathematician 42
          Apr 22 '16 at 12:54




          $begingroup$
          Yes and no, Thomas Andrews' answer already answers the point of the question, I merely wanted to give this simple argument why no natural order on $mathbb{C}$ exists that extends the order of $mathbb{R}$.
          $endgroup$
          – Mathematician 42
          Apr 22 '16 at 12:54











          0












          $begingroup$

          You can't, in the sense that there is no natural total order in $mathbb{C}$.






          share|cite|improve this answer









          $endgroup$









          • 7




            $begingroup$
            But I'm asked to do it. This is an assignment.
            $endgroup$
            – mavavilj
            Apr 19 '16 at 13:55






          • 3




            $begingroup$
            Well is $(0,1)$ or $(1,0)$ larger? You need to boil things down to real numbers in some way to compare complex values. Either you have miscalculated the eigenvalues, or you have misunderstood the question.
            $endgroup$
            – Matias Heikkilä
            Apr 19 '16 at 13:56






          • 4




            $begingroup$
            There are total orders on $mathbb{C}$ (e.g. the lexicographic order, for which $1 > i > -i > -1$). I don't know what you mean by "natural", though it is true that $mathbb{C}$ cannot be made into an ordered field. Anyway, in the context of eigenvalues it's the absolute value that counts, as Thomas Andrews says.
            $endgroup$
            – Najib Idrissi
            Apr 19 '16 at 14:01












          • $begingroup$
            Yes there are total orders on $mathbb{C}$, but which one you would use if you were asked to compare $a in mathbb{C}$ and $b in mathbb{C}$? Could you be sure that your friend uses the same order if given the same task? If $a, b in mathbb{R}$ instead, we tend to use the usual order relation, and there would be no problem. That's what I mean by natural. But this is not very interesting, since the original question should have been read as "compare the absolute values". Good to know that convention in the future, though.
            $endgroup$
            – Matias Heikkilä
            Apr 20 '16 at 9:02


















          0












          $begingroup$

          You can't, in the sense that there is no natural total order in $mathbb{C}$.






          share|cite|improve this answer









          $endgroup$









          • 7




            $begingroup$
            But I'm asked to do it. This is an assignment.
            $endgroup$
            – mavavilj
            Apr 19 '16 at 13:55






          • 3




            $begingroup$
            Well is $(0,1)$ or $(1,0)$ larger? You need to boil things down to real numbers in some way to compare complex values. Either you have miscalculated the eigenvalues, or you have misunderstood the question.
            $endgroup$
            – Matias Heikkilä
            Apr 19 '16 at 13:56






          • 4




            $begingroup$
            There are total orders on $mathbb{C}$ (e.g. the lexicographic order, for which $1 > i > -i > -1$). I don't know what you mean by "natural", though it is true that $mathbb{C}$ cannot be made into an ordered field. Anyway, in the context of eigenvalues it's the absolute value that counts, as Thomas Andrews says.
            $endgroup$
            – Najib Idrissi
            Apr 19 '16 at 14:01












          • $begingroup$
            Yes there are total orders on $mathbb{C}$, but which one you would use if you were asked to compare $a in mathbb{C}$ and $b in mathbb{C}$? Could you be sure that your friend uses the same order if given the same task? If $a, b in mathbb{R}$ instead, we tend to use the usual order relation, and there would be no problem. That's what I mean by natural. But this is not very interesting, since the original question should have been read as "compare the absolute values". Good to know that convention in the future, though.
            $endgroup$
            – Matias Heikkilä
            Apr 20 '16 at 9:02
















          0












          0








          0





          $begingroup$

          You can't, in the sense that there is no natural total order in $mathbb{C}$.






          share|cite|improve this answer









          $endgroup$



          You can't, in the sense that there is no natural total order in $mathbb{C}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Apr 19 '16 at 13:54









          Matias HeikkiläMatias Heikkilä

          1,778713




          1,778713








          • 7




            $begingroup$
            But I'm asked to do it. This is an assignment.
            $endgroup$
            – mavavilj
            Apr 19 '16 at 13:55






          • 3




            $begingroup$
            Well is $(0,1)$ or $(1,0)$ larger? You need to boil things down to real numbers in some way to compare complex values. Either you have miscalculated the eigenvalues, or you have misunderstood the question.
            $endgroup$
            – Matias Heikkilä
            Apr 19 '16 at 13:56






          • 4




            $begingroup$
            There are total orders on $mathbb{C}$ (e.g. the lexicographic order, for which $1 > i > -i > -1$). I don't know what you mean by "natural", though it is true that $mathbb{C}$ cannot be made into an ordered field. Anyway, in the context of eigenvalues it's the absolute value that counts, as Thomas Andrews says.
            $endgroup$
            – Najib Idrissi
            Apr 19 '16 at 14:01












          • $begingroup$
            Yes there are total orders on $mathbb{C}$, but which one you would use if you were asked to compare $a in mathbb{C}$ and $b in mathbb{C}$? Could you be sure that your friend uses the same order if given the same task? If $a, b in mathbb{R}$ instead, we tend to use the usual order relation, and there would be no problem. That's what I mean by natural. But this is not very interesting, since the original question should have been read as "compare the absolute values". Good to know that convention in the future, though.
            $endgroup$
            – Matias Heikkilä
            Apr 20 '16 at 9:02
















          • 7




            $begingroup$
            But I'm asked to do it. This is an assignment.
            $endgroup$
            – mavavilj
            Apr 19 '16 at 13:55






          • 3




            $begingroup$
            Well is $(0,1)$ or $(1,0)$ larger? You need to boil things down to real numbers in some way to compare complex values. Either you have miscalculated the eigenvalues, or you have misunderstood the question.
            $endgroup$
            – Matias Heikkilä
            Apr 19 '16 at 13:56






          • 4




            $begingroup$
            There are total orders on $mathbb{C}$ (e.g. the lexicographic order, for which $1 > i > -i > -1$). I don't know what you mean by "natural", though it is true that $mathbb{C}$ cannot be made into an ordered field. Anyway, in the context of eigenvalues it's the absolute value that counts, as Thomas Andrews says.
            $endgroup$
            – Najib Idrissi
            Apr 19 '16 at 14:01












          • $begingroup$
            Yes there are total orders on $mathbb{C}$, but which one you would use if you were asked to compare $a in mathbb{C}$ and $b in mathbb{C}$? Could you be sure that your friend uses the same order if given the same task? If $a, b in mathbb{R}$ instead, we tend to use the usual order relation, and there would be no problem. That's what I mean by natural. But this is not very interesting, since the original question should have been read as "compare the absolute values". Good to know that convention in the future, though.
            $endgroup$
            – Matias Heikkilä
            Apr 20 '16 at 9:02










          7




          7




          $begingroup$
          But I'm asked to do it. This is an assignment.
          $endgroup$
          – mavavilj
          Apr 19 '16 at 13:55




          $begingroup$
          But I'm asked to do it. This is an assignment.
          $endgroup$
          – mavavilj
          Apr 19 '16 at 13:55




          3




          3




          $begingroup$
          Well is $(0,1)$ or $(1,0)$ larger? You need to boil things down to real numbers in some way to compare complex values. Either you have miscalculated the eigenvalues, or you have misunderstood the question.
          $endgroup$
          – Matias Heikkilä
          Apr 19 '16 at 13:56




          $begingroup$
          Well is $(0,1)$ or $(1,0)$ larger? You need to boil things down to real numbers in some way to compare complex values. Either you have miscalculated the eigenvalues, or you have misunderstood the question.
          $endgroup$
          – Matias Heikkilä
          Apr 19 '16 at 13:56




          4




          4




          $begingroup$
          There are total orders on $mathbb{C}$ (e.g. the lexicographic order, for which $1 > i > -i > -1$). I don't know what you mean by "natural", though it is true that $mathbb{C}$ cannot be made into an ordered field. Anyway, in the context of eigenvalues it's the absolute value that counts, as Thomas Andrews says.
          $endgroup$
          – Najib Idrissi
          Apr 19 '16 at 14:01






          $begingroup$
          There are total orders on $mathbb{C}$ (e.g. the lexicographic order, for which $1 > i > -i > -1$). I don't know what you mean by "natural", though it is true that $mathbb{C}$ cannot be made into an ordered field. Anyway, in the context of eigenvalues it's the absolute value that counts, as Thomas Andrews says.
          $endgroup$
          – Najib Idrissi
          Apr 19 '16 at 14:01














          $begingroup$
          Yes there are total orders on $mathbb{C}$, but which one you would use if you were asked to compare $a in mathbb{C}$ and $b in mathbb{C}$? Could you be sure that your friend uses the same order if given the same task? If $a, b in mathbb{R}$ instead, we tend to use the usual order relation, and there would be no problem. That's what I mean by natural. But this is not very interesting, since the original question should have been read as "compare the absolute values". Good to know that convention in the future, though.
          $endgroup$
          – Matias Heikkilä
          Apr 20 '16 at 9:02






          $begingroup$
          Yes there are total orders on $mathbb{C}$, but which one you would use if you were asked to compare $a in mathbb{C}$ and $b in mathbb{C}$? Could you be sure that your friend uses the same order if given the same task? If $a, b in mathbb{R}$ instead, we tend to use the usual order relation, and there would be no problem. That's what I mean by natural. But this is not very interesting, since the original question should have been read as "compare the absolute values". Good to know that convention in the future, though.
          $endgroup$
          – Matias Heikkilä
          Apr 20 '16 at 9:02




















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