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Eigenvalues of M

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1

1

$begingroup$

Let M be square matrix of order n with real entries Satisfying $M^3=I$ and $Mvneq v$ for any non-zero vector $v$.Then Which of the followings are true?

$1.$ M has real eigen-values?

$2.$ $M+M^-1$ has real eigen-values?

$3.n$ is divisible by 2.

$4.n$ is divisible by 3.

$Mvneq v$ for any non-zero vector $v$ means 1 is not an eigenvalue of M.Hence $x^2+x+1=0$ is minimal polynomial of M.Hence 2 is correct option.But I am not able to understand why 1st and 4th options are correct?

linear-algebra eigenvalues-eigenvectors

share|cite|improve this question

asked Jun 15 '18 at 5:35

ASHWINI SANKHEASHWINI SANKHE

12710

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The first statement definitely cannot be true, since if $lambda$ is an eigenvalue of $M$ then $Mv= lambda v implies v = lambda^3 v$ so $lambda^3 = 1$ but $lambda neq 1$ so $lambda$ is necessarily complex. The second is right since $M + M^-1 = -I$.
  $endgroup$
  – астон вілла олоф мэллбэрг
  Jun 15 '18 at 6:08
  
  

  
  
  
  

add a comment | 

1

1

$begingroup$

Let M be square matrix of order n with real entries Satisfying $M^3=I$ and $Mvneq v$ for any non-zero vector $v$.Then Which of the followings are true?

$1.$ M has real eigen-values?

$2.$ $M+M^-1$ has real eigen-values?

$3.n$ is divisible by 2.

$4.n$ is divisible by 3.

$Mvneq v$ for any non-zero vector $v$ means 1 is not an eigenvalue of M.Hence $x^2+x+1=0$ is minimal polynomial of M.Hence 2 is correct option.But I am not able to understand why 1st and 4th options are correct?

linear-algebra eigenvalues-eigenvectors

share|cite|improve this question

asked Jun 15 '18 at 5:35

ASHWINI SANKHEASHWINI SANKHE

12710

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The first statement definitely cannot be true, since if $lambda$ is an eigenvalue of $M$ then $Mv= lambda v implies v = lambda^3 v$ so $lambda^3 = 1$ but $lambda neq 1$ so $lambda$ is necessarily complex. The second is right since $M + M^-1 = -I$.
  $endgroup$
  – астон вілла олоф мэллбэрг
  Jun 15 '18 at 6:08
  
  

  
  
  
  

add a comment | 

$begingroup$

Let M be square matrix of order n with real entries Satisfying $M^3=I$ and $Mvneq v$ for any non-zero vector $v$.Then Which of the followings are true?

$1.$ M has real eigen-values?

$2.$ $M+M^-1$ has real eigen-values?

$3.n$ is divisible by 2.

$4.n$ is divisible by 3.

$Mvneq v$ for any non-zero vector $v$ means 1 is not an eigenvalue of M.Hence $x^2+x+1=0$ is minimal polynomial of M.Hence 2 is correct option.But I am not able to understand why 1st and 4th options are correct?

linear-algebra eigenvalues-eigenvectors

share|cite|improve this question

asked Jun 15 '18 at 5:35

ASHWINI SANKHEASHWINI SANKHE

12710

$endgroup$

share|cite|improve this question

asked Jun 15 '18 at 5:35

ASHWINI SANKHEASHWINI SANKHE

12710

share|cite|improve this question

asked Jun 15 '18 at 5:35

ASHWINI SANKHEASHWINI SANKHE

12710

asked Jun 15 '18 at 5:35

ASHWINI SANKHEASHWINI SANKHE

12710

asked Jun 15 '18 at 5:35

ASHWINI SANKHEASHWINI SANKHE

12710

1 Answer
1

active

oldest

votes

2

$begingroup$

1. $M^3=I$ so $M^3-I=(M-I)(M^2+M+I)=0$ but for hypothesis you have that $M-I $ is invertibile and so:
   $M^2+M+I=0$
   By contraddiction if there exists a real eighenvalue $lambda$ of M then , if $vneq0$ is his eighenvector, you have that
   $(M^2+M+I)v=(lambda^2+lambda+1)v=0$
   and so
   $lambda^2+lambda+1=0$
   But this polinomial equation not have solutions in $mathbbR$.




2. $M^2+M+I=0$ and you can multiply both members for $M^-1$ :

$M+I+M^-1=0$

So

$M+M^-1=-I $
that is diagonalizzable oviously.

3. What you said is correct.

By contraddiction if $ n=deg(det(M-lambda I) $ is not divisibile for 2 then the polinomial have necessary a real root because any polinomial of degree odd have al least a root in $mathbbR$

4. It is false because there exists a $2x2$- real matrix $M$ such that $M^3=I$ and $M-I$ is invertibile. An example can be

$beginbmatrix-frac12&fracsqrt32\ -fracsqrt32&-frac12endbmatrix$

share|cite|improve this answer

edited Mar 16 at 20:42

answered Jun 15 '18 at 10:38

Federico FalluccaFederico Fallucca

2,280210

$endgroup$

add a comment | 

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2

$begingroup$

1. $M^3=I$ so $M^3-I=(M-I)(M^2+M+I)=0$ but for hypothesis you have that $M-I $ is invertibile and so:
   $M^2+M+I=0$
   By contraddiction if there exists a real eighenvalue $lambda$ of M then , if $vneq0$ is his eighenvector, you have that
   $(M^2+M+I)v=(lambda^2+lambda+1)v=0$
   and so
   $lambda^2+lambda+1=0$
   But this polinomial equation not have solutions in $mathbbR$.




2. $M^2+M+I=0$ and you can multiply both members for $M^-1$ :

$M+I+M^-1=0$

So

$M+M^-1=-I $
that is diagonalizzable oviously.

3. What you said is correct.

By contraddiction if $ n=deg(det(M-lambda I) $ is not divisibile for 2 then the polinomial have necessary a real root because any polinomial of degree odd have al least a root in $mathbbR$

4. It is false because there exists a $2x2$- real matrix $M$ such that $M^3=I$ and $M-I$ is invertibile. An example can be

$beginbmatrix-frac12&fracsqrt32\ -fracsqrt32&-frac12endbmatrix$

share|cite|improve this answer

edited Mar 16 at 20:42

answered Jun 15 '18 at 10:38

Federico FalluccaFederico Fallucca

2,280210

$endgroup$

add a comment | 

2

$begingroup$

1. $M^3=I$ so $M^3-I=(M-I)(M^2+M+I)=0$ but for hypothesis you have that $M-I $ is invertibile and so:
   $M^2+M+I=0$
   By contraddiction if there exists a real eighenvalue $lambda$ of M then , if $vneq0$ is his eighenvector, you have that
   $(M^2+M+I)v=(lambda^2+lambda+1)v=0$
   and so
   $lambda^2+lambda+1=0$
   But this polinomial equation not have solutions in $mathbbR$.




2. $M^2+M+I=0$ and you can multiply both members for $M^-1$ :

$M+I+M^-1=0$

So

$M+M^-1=-I $
that is diagonalizzable oviously.

3. What you said is correct.

By contraddiction if $ n=deg(det(M-lambda I) $ is not divisibile for 2 then the polinomial have necessary a real root because any polinomial of degree odd have al least a root in $mathbbR$

4. It is false because there exists a $2x2$- real matrix $M$ such that $M^3=I$ and $M-I$ is invertibile. An example can be

$beginbmatrix-frac12&fracsqrt32\ -fracsqrt32&-frac12endbmatrix$

share|cite|improve this answer

edited Mar 16 at 20:42

answered Jun 15 '18 at 10:38

Federico FalluccaFederico Fallucca

2,280210

$endgroup$

add a comment | 

$begingroup$

1. $M^3=I$ so $M^3-I=(M-I)(M^2+M+I)=0$ but for hypothesis you have that $M-I $ is invertibile and so:
   $M^2+M+I=0$
   By contraddiction if there exists a real eighenvalue $lambda$ of M then , if $vneq0$ is his eighenvector, you have that
   $(M^2+M+I)v=(lambda^2+lambda+1)v=0$
   and so
   $lambda^2+lambda+1=0$
   But this polinomial equation not have solutions in $mathbbR$.




2. $M^2+M+I=0$ and you can multiply both members for $M^-1$ :

$M+I+M^-1=0$

So

$M+M^-1=-I $
that is diagonalizzable oviously.

3. What you said is correct.

By contraddiction if $ n=deg(det(M-lambda I) $ is not divisibile for 2 then the polinomial have necessary a real root because any polinomial of degree odd have al least a root in $mathbbR$

4. It is false because there exists a $2x2$- real matrix $M$ such that $M^3=I$ and $M-I$ is invertibile. An example can be

$beginbmatrix-frac12&fracsqrt32\ -fracsqrt32&-frac12endbmatrix$

share|cite|improve this answer

edited Mar 16 at 20:42

answered Jun 15 '18 at 10:38

Federico FalluccaFederico Fallucca

2,280210

$endgroup$

share|cite|improve this answer

edited Mar 16 at 20:42

answered Jun 15 '18 at 10:38

Federico FalluccaFederico Fallucca

2,280210

answered Jun 15 '18 at 10:38

Federico FalluccaFederico Fallucca

2,280210

answered Jun 15 '18 at 10:38

Federico FalluccaFederico Fallucca

2,280210