Find a matrix 4x4 that has no eigenvalues on $R^4$ The 2019 Stack Overflow Developer Survey...

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Find a matrix 4x4 that has no eigenvalues on $R^4$



The 2019 Stack Overflow Developer Survey Results Are InFind a 2x2 matrix with positive eigenvalues, but a negative quadratic form for some x in $R^{2}$Suppose $A$ is a 4x4 matrix such that $det(A)=frac{1}{64}$Proof that an involutory matrix has eigenvalues 1,-1Are there explicit formulas for the eigenvalues and eigenvectors of a generic 4x4 density matrix?Find “REAL” cannonical form of 4x4 matrixEigenvalues for 4x4 matrixIs there a 4x4 unitary matrix with 0 diagonal and non-zero off-diagonal?Is there an easy way to find the (analytical form of) remaining eigenvalues of a 4x4 matrix?Given this 4x4 matrix, is there an easier way to calculate the eigenvalues?Matrix with complex eigenvalues but real entries












0












$begingroup$


I had to find a matrix 2x2 and 4x4 that have no eigenvalues, for the 2x2 it was not that hard to do $a_{11}= 0$ $a_{12}= 1$ $a_{21}= -1$ $a_{22}= 0$ so that the possible eigenvalues are $det(xId-A)=x^2+1=x={-i,i}$, where A is the matrix with the $a_{ij}$ entries.



But I was not able to get to a 4x4 matrix






linear-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    Just take $A=begin{pmatrix} i & 0 & 0 cr 0 & i & 0 cr 0 & 0 & i end{pmatrix}$. It has no real eigenvalues.
    $endgroup$
    – Dietrich Burde
    Mar 21 at 15:13








  • 2




    $begingroup$
    You mean to ask no real eigenvalue I guess...
    $endgroup$
    – Prakhar Neema
    Mar 21 at 15:14










  • $begingroup$
    Presumably the matrix must have real entries. Is this correct?
    $endgroup$
    – amd
    Mar 21 at 18:33
















0












$begingroup$


I had to find a matrix 2x2 and 4x4 that have no eigenvalues, for the 2x2 it was not that hard to do $a_{11}= 0$ $a_{12}= 1$ $a_{21}= -1$ $a_{22}= 0$ so that the possible eigenvalues are $det(xId-A)=x^2+1=x={-i,i}$, where A is the matrix with the $a_{ij}$ entries.



But I was not able to get to a 4x4 matrix






linear-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    Just take $A=begin{pmatrix} i & 0 & 0 cr 0 & i & 0 cr 0 & 0 & i end{pmatrix}$. It has no real eigenvalues.
    $endgroup$
    – Dietrich Burde
    Mar 21 at 15:13








  • 2




    $begingroup$
    You mean to ask no real eigenvalue I guess...
    $endgroup$
    – Prakhar Neema
    Mar 21 at 15:14










  • $begingroup$
    Presumably the matrix must have real entries. Is this correct?
    $endgroup$
    – amd
    Mar 21 at 18:33














0












0








0





$begingroup$


I had to find a matrix 2x2 and 4x4 that have no eigenvalues, for the 2x2 it was not that hard to do $a_{11}= 0$ $a_{12}= 1$ $a_{21}= -1$ $a_{22}= 0$ so that the possible eigenvalues are $det(xId-A)=x^2+1=x={-i,i}$, where A is the matrix with the $a_{ij}$ entries.



But I was not able to get to a 4x4 matrix






linear-algebra







share|cite|improve this question











$endgroup$




I had to find a matrix 2x2 and 4x4 that have no eigenvalues, for the 2x2 it was not that hard to do $a_{11}= 0$ $a_{12}= 1$ $a_{21}= -1$ $a_{22}= 0$ so that the possible eigenvalues are $det(xId-A)=x^2+1=x={-i,i}$, where A is the matrix with the $a_{ij}$ entries.



But I was not able to get to a 4x4 matrix






linear-algebra




linear-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 21 at 15:19







Juju9704

















asked Mar 21 at 15:10









Juju9704Juju9704

34511




34511












  • $begingroup$
    Just take $A=begin{pmatrix} i & 0 & 0 cr 0 & i & 0 cr 0 & 0 & i end{pmatrix}$. It has no real eigenvalues.
    $endgroup$
    – Dietrich Burde
    Mar 21 at 15:13








  • 2




    $begingroup$
    You mean to ask no real eigenvalue I guess...
    $endgroup$
    – Prakhar Neema
    Mar 21 at 15:14










  • $begingroup$
    Presumably the matrix must have real entries. Is this correct?
    $endgroup$
    – amd
    Mar 21 at 18:33


















  • $begingroup$
    Just take $A=begin{pmatrix} i & 0 & 0 cr 0 & i & 0 cr 0 & 0 & i end{pmatrix}$. It has no real eigenvalues.
    $endgroup$
    – Dietrich Burde
    Mar 21 at 15:13








  • 2




    $begingroup$
    You mean to ask no real eigenvalue I guess...
    $endgroup$
    – Prakhar Neema
    Mar 21 at 15:14










  • $begingroup$
    Presumably the matrix must have real entries. Is this correct?
    $endgroup$
    – amd
    Mar 21 at 18:33
















$begingroup$
Just take $A=begin{pmatrix} i & 0 & 0 cr 0 & i & 0 cr 0 & 0 & i end{pmatrix}$. It has no real eigenvalues.
$endgroup$
– Dietrich Burde
Mar 21 at 15:13






$begingroup$
Just take $A=begin{pmatrix} i & 0 & 0 cr 0 & i & 0 cr 0 & 0 & i end{pmatrix}$. It has no real eigenvalues.
$endgroup$
– Dietrich Burde
Mar 21 at 15:13






2




2




$begingroup$
You mean to ask no real eigenvalue I guess...
$endgroup$
– Prakhar Neema
Mar 21 at 15:14




$begingroup$
You mean to ask no real eigenvalue I guess...
$endgroup$
– Prakhar Neema
Mar 21 at 15:14












$begingroup$
Presumably the matrix must have real entries. Is this correct?
$endgroup$
– amd
Mar 21 at 18:33




$begingroup$
Presumably the matrix must have real entries. Is this correct?
$endgroup$
– amd
Mar 21 at 18:33










3 Answers
3






active

oldest

votes


















6












$begingroup$

For a $3 times 3$ real matrix, this does not exist.



Indeed, the characteristic polynomial of a $3 times 3$ matrix is a polynomial of degree $3$, which has to vanish in $mathbb{R}$. So there always exists an eigenvalue.



For a $4 times 4$ matrix, consider the matrix



begin{pmatrix}
0 & 0 & 0&-1\
1 & 0 & 0& 0\
0 & 1 & 0 & 0\
0& 0& 1& 0
end{pmatrix}



Its characteritic polynomial is $X^4 +1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I am so dumb, I double checked the question and it was on 4x4
    $endgroup$
    – Juju9704
    Mar 21 at 15:19










  • $begingroup$
    The coefficients of the characteristic polynomial don't have to be real numbers. A general complex polynomial of degree 3 can in fact have no real roots.
    $endgroup$
    – PierreCarre
    Mar 21 at 15:19





















1












$begingroup$

The $2times 2$ case suggests that you look for a matrix with integer coefficients which does not have any real eigenvalue, like
$$
begin{pmatrix} 0 & 1 cr -1 & 0 end{pmatrix}in SL_2(Bbb Z).
$$
Try to generalize this for $n=4$ working with the group $SL_4(Bbb Z)$. You can take block matrices with the matrix from above, i.e.,
$$
begin{pmatrix}
0 & 1 & 0 & 0\
-1 & 0 & 0 & 0\
0 & 0 & 0 & 1\
0 & 0 & -1& 0
end{pmatrix}.
$$
Its characteristic polynomial is $(X^2+1)^2$.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    Every matrix has eigenvalues so you must mean "no real eigenvalues". Non-real roots of a polynomial equation with real coefficients must come in complex conjugate pairs so there is NO 3 by 3 matrix, with real entries, that has 3 non-real eigenvalues. Dietrich Burde gave an example of a 3 by 3 matrix with imaginary entries that has only imaginary entries.






    share|cite|improve this answer









    $endgroup$














      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      6












      $begingroup$

      For a $3 times 3$ real matrix, this does not exist.



      Indeed, the characteristic polynomial of a $3 times 3$ matrix is a polynomial of degree $3$, which has to vanish in $mathbb{R}$. So there always exists an eigenvalue.



      For a $4 times 4$ matrix, consider the matrix



      begin{pmatrix}
      0 & 0 & 0&-1\
      1 & 0 & 0& 0\
      0 & 1 & 0 & 0\
      0& 0& 1& 0
      end{pmatrix}



      Its characteritic polynomial is $X^4 +1$.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        I am so dumb, I double checked the question and it was on 4x4
        $endgroup$
        – Juju9704
        Mar 21 at 15:19










      • $begingroup$
        The coefficients of the characteristic polynomial don't have to be real numbers. A general complex polynomial of degree 3 can in fact have no real roots.
        $endgroup$
        – PierreCarre
        Mar 21 at 15:19


















      6












      $begingroup$

      For a $3 times 3$ real matrix, this does not exist.



      Indeed, the characteristic polynomial of a $3 times 3$ matrix is a polynomial of degree $3$, which has to vanish in $mathbb{R}$. So there always exists an eigenvalue.



      For a $4 times 4$ matrix, consider the matrix



      begin{pmatrix}
      0 & 0 & 0&-1\
      1 & 0 & 0& 0\
      0 & 1 & 0 & 0\
      0& 0& 1& 0
      end{pmatrix}



      Its characteritic polynomial is $X^4 +1$.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        I am so dumb, I double checked the question and it was on 4x4
        $endgroup$
        – Juju9704
        Mar 21 at 15:19










      • $begingroup$
        The coefficients of the characteristic polynomial don't have to be real numbers. A general complex polynomial of degree 3 can in fact have no real roots.
        $endgroup$
        – PierreCarre
        Mar 21 at 15:19
















      6












      6








      6





      $begingroup$

      For a $3 times 3$ real matrix, this does not exist.



      Indeed, the characteristic polynomial of a $3 times 3$ matrix is a polynomial of degree $3$, which has to vanish in $mathbb{R}$. So there always exists an eigenvalue.



      For a $4 times 4$ matrix, consider the matrix



      begin{pmatrix}
      0 & 0 & 0&-1\
      1 & 0 & 0& 0\
      0 & 1 & 0 & 0\
      0& 0& 1& 0
      end{pmatrix}



      Its characteritic polynomial is $X^4 +1$.






      share|cite|improve this answer











      $endgroup$



      For a $3 times 3$ real matrix, this does not exist.



      Indeed, the characteristic polynomial of a $3 times 3$ matrix is a polynomial of degree $3$, which has to vanish in $mathbb{R}$. So there always exists an eigenvalue.



      For a $4 times 4$ matrix, consider the matrix



      begin{pmatrix}
      0 & 0 & 0&-1\
      1 & 0 & 0& 0\
      0 & 1 & 0 & 0\
      0& 0& 1& 0
      end{pmatrix}



      Its characteritic polynomial is $X^4 +1$.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Mar 21 at 15:23

























      answered Mar 21 at 15:16









      TheSilverDoeTheSilverDoe

      5,427216




      5,427216












      • $begingroup$
        I am so dumb, I double checked the question and it was on 4x4
        $endgroup$
        – Juju9704
        Mar 21 at 15:19










      • $begingroup$
        The coefficients of the characteristic polynomial don't have to be real numbers. A general complex polynomial of degree 3 can in fact have no real roots.
        $endgroup$
        – PierreCarre
        Mar 21 at 15:19




















      • $begingroup$
        I am so dumb, I double checked the question and it was on 4x4
        $endgroup$
        – Juju9704
        Mar 21 at 15:19










      • $begingroup$
        The coefficients of the characteristic polynomial don't have to be real numbers. A general complex polynomial of degree 3 can in fact have no real roots.
        $endgroup$
        – PierreCarre
        Mar 21 at 15:19


















      $begingroup$
      I am so dumb, I double checked the question and it was on 4x4
      $endgroup$
      – Juju9704
      Mar 21 at 15:19




      $begingroup$
      I am so dumb, I double checked the question and it was on 4x4
      $endgroup$
      – Juju9704
      Mar 21 at 15:19












      $begingroup$
      The coefficients of the characteristic polynomial don't have to be real numbers. A general complex polynomial of degree 3 can in fact have no real roots.
      $endgroup$
      – PierreCarre
      Mar 21 at 15:19






      $begingroup$
      The coefficients of the characteristic polynomial don't have to be real numbers. A general complex polynomial of degree 3 can in fact have no real roots.
      $endgroup$
      – PierreCarre
      Mar 21 at 15:19













      1












      $begingroup$

      The $2times 2$ case suggests that you look for a matrix with integer coefficients which does not have any real eigenvalue, like
      $$
      begin{pmatrix} 0 & 1 cr -1 & 0 end{pmatrix}in SL_2(Bbb Z).
      $$
      Try to generalize this for $n=4$ working with the group $SL_4(Bbb Z)$. You can take block matrices with the matrix from above, i.e.,
      $$
      begin{pmatrix}
      0 & 1 & 0 & 0\
      -1 & 0 & 0 & 0\
      0 & 0 & 0 & 1\
      0 & 0 & -1& 0
      end{pmatrix}.
      $$
      Its characteristic polynomial is $(X^2+1)^2$.






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        The $2times 2$ case suggests that you look for a matrix with integer coefficients which does not have any real eigenvalue, like
        $$
        begin{pmatrix} 0 & 1 cr -1 & 0 end{pmatrix}in SL_2(Bbb Z).
        $$
        Try to generalize this for $n=4$ working with the group $SL_4(Bbb Z)$. You can take block matrices with the matrix from above, i.e.,
        $$
        begin{pmatrix}
        0 & 1 & 0 & 0\
        -1 & 0 & 0 & 0\
        0 & 0 & 0 & 1\
        0 & 0 & -1& 0
        end{pmatrix}.
        $$
        Its characteristic polynomial is $(X^2+1)^2$.






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          The $2times 2$ case suggests that you look for a matrix with integer coefficients which does not have any real eigenvalue, like
          $$
          begin{pmatrix} 0 & 1 cr -1 & 0 end{pmatrix}in SL_2(Bbb Z).
          $$
          Try to generalize this for $n=4$ working with the group $SL_4(Bbb Z)$. You can take block matrices with the matrix from above, i.e.,
          $$
          begin{pmatrix}
          0 & 1 & 0 & 0\
          -1 & 0 & 0 & 0\
          0 & 0 & 0 & 1\
          0 & 0 & -1& 0
          end{pmatrix}.
          $$
          Its characteristic polynomial is $(X^2+1)^2$.






          share|cite|improve this answer











          $endgroup$



          The $2times 2$ case suggests that you look for a matrix with integer coefficients which does not have any real eigenvalue, like
          $$
          begin{pmatrix} 0 & 1 cr -1 & 0 end{pmatrix}in SL_2(Bbb Z).
          $$
          Try to generalize this for $n=4$ working with the group $SL_4(Bbb Z)$. You can take block matrices with the matrix from above, i.e.,
          $$
          begin{pmatrix}
          0 & 1 & 0 & 0\
          -1 & 0 & 0 & 0\
          0 & 0 & 0 & 1\
          0 & 0 & -1& 0
          end{pmatrix}.
          $$
          Its characteristic polynomial is $(X^2+1)^2$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Mar 22 at 15:39

























          answered Mar 21 at 15:19









          Dietrich BurdeDietrich Burde

          81.9k649107




          81.9k649107























              0












              $begingroup$

              Every matrix has eigenvalues so you must mean "no real eigenvalues". Non-real roots of a polynomial equation with real coefficients must come in complex conjugate pairs so there is NO 3 by 3 matrix, with real entries, that has 3 non-real eigenvalues. Dietrich Burde gave an example of a 3 by 3 matrix with imaginary entries that has only imaginary entries.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Every matrix has eigenvalues so you must mean "no real eigenvalues". Non-real roots of a polynomial equation with real coefficients must come in complex conjugate pairs so there is NO 3 by 3 matrix, with real entries, that has 3 non-real eigenvalues. Dietrich Burde gave an example of a 3 by 3 matrix with imaginary entries that has only imaginary entries.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Every matrix has eigenvalues so you must mean "no real eigenvalues". Non-real roots of a polynomial equation with real coefficients must come in complex conjugate pairs so there is NO 3 by 3 matrix, with real entries, that has 3 non-real eigenvalues. Dietrich Burde gave an example of a 3 by 3 matrix with imaginary entries that has only imaginary entries.






                  share|cite|improve this answer









                  $endgroup$



                  Every matrix has eigenvalues so you must mean "no real eigenvalues". Non-real roots of a polynomial equation with real coefficients must come in complex conjugate pairs so there is NO 3 by 3 matrix, with real entries, that has 3 non-real eigenvalues. Dietrich Burde gave an example of a 3 by 3 matrix with imaginary entries that has only imaginary entries.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Mar 21 at 15:28









                  user247327user247327

                  11.6k1516




                  11.6k1516






























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