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

Find 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 $det(A)=frac164$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

edited Mar 21 at 15:19

Juju9704

asked Mar 21 at 15:10

Juju9704Juju9704

34511

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Just take $A=beginpmatrix i & 0 & 0 cr 0 & i & 0 cr 0 & 0 & i endpmatrix$. 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
  

  
  
  
  

add a comment | 

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

edited Mar 21 at 15:19

Juju9704

asked Mar 21 at 15:10

Juju9704Juju9704

34511

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Just take $A=beginpmatrix i & 0 & 0 cr 0 & i & 0 cr 0 & 0 & i endpmatrix$. 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
  

  
  
  
  

add a comment | 

$begingroup$

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

edited Mar 21 at 15:19

Juju9704

asked Mar 21 at 15:10

Juju9704Juju9704

34511

$endgroup$

share|cite|improve this question

edited Mar 21 at 15:19

Juju9704

asked Mar 21 at 15:10

Juju9704Juju9704

34511

share|cite|improve this question

edited Mar 21 at 15:19

Juju9704

asked Mar 21 at 15:10

Juju9704Juju9704

34511

asked Mar 21 at 15:10

Juju9704Juju9704

34511

asked Mar 21 at 15:10

Juju9704Juju9704

34511

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 $mathbbR$. So there always exists an eigenvalue.

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

beginpmatrix
0 & 0 & 0&-1\
1 & 0 & 0& 0\
0 & 1 & 0 & 0\
0& 0& 1& 0
endpmatrix

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

share|cite|improve this answer

edited Mar 21 at 15:23

answered Mar 21 at 15:16

TheSilverDoeTheSilverDoe

5,324216

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  
  

  
  
  
  

add a comment | 

1

$begingroup$

The $2times 2$ case suggests that you look for a matrix with integer coefficients which does not have any real eigenvalue, like
$$
beginpmatrix 0 & 1 cr -1 & 0 endpmatrixin SL_2(Bbb Z).
$$
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.,
$$
beginpmatrix
0 & 1 & 0 & 0\
-1 & 0 & 0 & 0\
0 & 0 & 0 & 1\
0 & 0 & -1& 0
endpmatrix.
$$
Its characteristic polynomial is $(X^2+1)^2$.

share|cite|improve this answer

edited Mar 22 at 15:39

answered Mar 21 at 15:19

Dietrich BurdeDietrich Burde

81.6k648106

$endgroup$

add a comment | 

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

answered Mar 21 at 15:28

user247327user247327

11.5k1516

$endgroup$

add a comment | 

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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 $mathbbR$. So there always exists an eigenvalue.

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

beginpmatrix
0 & 0 & 0&-1\
1 & 0 & 0& 0\
0 & 1 & 0 & 0\
0& 0& 1& 0
endpmatrix

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

share|cite|improve this answer

edited Mar 21 at 15:23

answered Mar 21 at 15:16

TheSilverDoeTheSilverDoe

5,324216

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  
  

  
  
  
  

add a comment | 

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 $mathbbR$. So there always exists an eigenvalue.

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

beginpmatrix
0 & 0 & 0&-1\
1 & 0 & 0& 0\
0 & 1 & 0 & 0\
0& 0& 1& 0
endpmatrix

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

share|cite|improve this answer

edited Mar 21 at 15:23

answered Mar 21 at 15:16

TheSilverDoeTheSilverDoe

5,324216

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  
  

  
  
  
  

add a comment | 

$begingroup$

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 $mathbbR$. So there always exists an eigenvalue.

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

beginpmatrix
0 & 0 & 0&-1\
1 & 0 & 0& 0\
0 & 1 & 0 & 0\
0& 0& 1& 0
endpmatrix

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

share|cite|improve this answer

edited Mar 21 at 15:23

answered Mar 21 at 15:16

TheSilverDoeTheSilverDoe

5,324216

$endgroup$

share|cite|improve this answer

edited Mar 21 at 15:23

answered Mar 21 at 15:16

TheSilverDoeTheSilverDoe

5,324216

answered Mar 21 at 15:16

TheSilverDoeTheSilverDoe

5,324216

answered Mar 21 at 15:16

TheSilverDoeTheSilverDoe

5,324216

1

$begingroup$

The $2times 2$ case suggests that you look for a matrix with integer coefficients which does not have any real eigenvalue, like
$$
beginpmatrix 0 & 1 cr -1 & 0 endpmatrixin SL_2(Bbb Z).
$$
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.,
$$
beginpmatrix
0 & 1 & 0 & 0\
-1 & 0 & 0 & 0\
0 & 0 & 0 & 1\
0 & 0 & -1& 0
endpmatrix.
$$
Its characteristic polynomial is $(X^2+1)^2$.

share|cite|improve this answer

edited Mar 22 at 15:39

answered Mar 21 at 15:19

Dietrich BurdeDietrich Burde

81.6k648106

$endgroup$

add a comment | 

1

$begingroup$

The $2times 2$ case suggests that you look for a matrix with integer coefficients which does not have any real eigenvalue, like
$$
beginpmatrix 0 & 1 cr -1 & 0 endpmatrixin SL_2(Bbb Z).
$$
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.,
$$
beginpmatrix
0 & 1 & 0 & 0\
-1 & 0 & 0 & 0\
0 & 0 & 0 & 1\
0 & 0 & -1& 0
endpmatrix.
$$
Its characteristic polynomial is $(X^2+1)^2$.

share|cite|improve this answer

edited Mar 22 at 15:39

answered Mar 21 at 15:19

Dietrich BurdeDietrich Burde

81.6k648106

$endgroup$

add a comment | 

$begingroup$

The $2times 2$ case suggests that you look for a matrix with integer coefficients which does not have any real eigenvalue, like
$$
beginpmatrix 0 & 1 cr -1 & 0 endpmatrixin SL_2(Bbb Z).
$$
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.,
$$
beginpmatrix
0 & 1 & 0 & 0\
-1 & 0 & 0 & 0\
0 & 0 & 0 & 1\
0 & 0 & -1& 0
endpmatrix.
$$
Its characteristic polynomial is $(X^2+1)^2$.

share|cite|improve this answer

edited Mar 22 at 15:39

answered Mar 21 at 15:19

Dietrich BurdeDietrich Burde

81.6k648106

$endgroup$

share|cite|improve this answer

edited Mar 22 at 15:39

answered Mar 21 at 15:19

Dietrich BurdeDietrich Burde

81.6k648106

answered Mar 21 at 15:19

Dietrich BurdeDietrich Burde

81.6k648106

answered Mar 21 at 15:19

Dietrich BurdeDietrich Burde

81.6k648106

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

answered Mar 21 at 15:28

user247327user247327

11.5k1516

$endgroup$

add a comment | 

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

answered Mar 21 at 15:28

user247327user247327

11.5k1516

$endgroup$

add a comment | 

$begingroup$

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

answered Mar 21 at 15:28

user247327user247327

11.5k1516

$endgroup$

share|cite|improve this answer

answered Mar 21 at 15:28

user247327user247327

11.5k1516

answered Mar 21 at 15:28

user247327user247327

11.5k1516

answered Mar 21 at 15:28

user247327user247327

11.5k1516