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Eigenvalues of sum of a matrix and its inverse

Eigenvalues of a matrix and its squarecomplex eigenvalues and invariant spacesProperties of a matrix that shares the set of real eigenvalues with its inverseEigenvalues of an upper Hessenberg matrixEigenvalues of complex special orthogonal matrix$A in M(n,mathbb R)$ has all its eigenvalues real , then is it true that all the eigenvalues of $A^2-A$ are also real?What are the eigenvalues of the following symmetric matrix?What is a necessary and sufficient condition for all the eigenvalues of a non-symmetric matrix to be positive?Is there any matrix with this propery?Eigenvalues of a square matrix

2

$begingroup$

Let $M$ be a $ntimes n$ matrix such that $M^3=I$.Suppose that $Mv neq v$ for any non zero vector $v$.Then which of the following isare true?

1).$M+M^-1$ has real eigenvalues

The solution i tried is

Here it is given that $M^3=I$ so from here it is confirm that its minimal polynomial can be from $(x-1),(x^2+x+1) ;or; (x-1)(x^2+x+1)$, but according to given condition $(x-1); and ;(x-1);(x^2+x+1)$ can't be minimal polynomial.So the only possibility is $(x^2+x+1)$

From above it is confirmed that roots are complex roots ,but still i have no idea how to prove $M+M^-1$ has real eigenvalue

please help!

Thankyou

linear-algebra eigenvalues-eigenvectors linear-transformations

share|cite|improve this question

edited Jun 2 at 10:50

José Carlos Santos

190k24146264

asked Jun 2 at 10:33

Honey KumarHoney Kumar

274

New contributor

Honey Kumar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

2

$begingroup$

Let $M$ be a $ntimes n$ matrix such that $M^3=I$.Suppose that $Mv neq v$ for any non zero vector $v$.Then which of the following isare true?

1).$M+M^-1$ has real eigenvalues

The solution i tried is

Here it is given that $M^3=I$ so from here it is confirm that its minimal polynomial can be from $(x-1),(x^2+x+1) ;or; (x-1)(x^2+x+1)$, but according to given condition $(x-1); and ;(x-1);(x^2+x+1)$ can't be minimal polynomial.So the only possibility is $(x^2+x+1)$

From above it is confirmed that roots are complex roots ,but still i have no idea how to prove $M+M^-1$ has real eigenvalue

please help!

Thankyou

linear-algebra eigenvalues-eigenvectors linear-transformations

share|cite|improve this question

edited Jun 2 at 10:50

José Carlos Santos

190k24146264

asked Jun 2 at 10:33

Honey KumarHoney Kumar

274

New contributor

Honey Kumar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

Let $M$ be a $ntimes n$ matrix such that $M^3=I$.Suppose that $Mv neq v$ for any non zero vector $v$.Then which of the following isare true?

1).$M+M^-1$ has real eigenvalues

The solution i tried is

Here it is given that $M^3=I$ so from here it is confirm that its minimal polynomial can be from $(x-1),(x^2+x+1) ;or; (x-1)(x^2+x+1)$, but according to given condition $(x-1); and ;(x-1);(x^2+x+1)$ can't be minimal polynomial.So the only possibility is $(x^2+x+1)$

From above it is confirmed that roots are complex roots ,but still i have no idea how to prove $M+M^-1$ has real eigenvalue

please help!

Thankyou

linear-algebra eigenvalues-eigenvectors linear-transformations

share|cite|improve this question

edited Jun 2 at 10:50

José Carlos Santos

190k24146264

asked Jun 2 at 10:33

Honey KumarHoney Kumar

274

New contributor

Honey Kumar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited Jun 2 at 10:50

José Carlos Santos

190k24146264

asked Jun 2 at 10:33

Honey KumarHoney Kumar

274

New contributor

Honey Kumar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited Jun 2 at 10:50

José Carlos Santos

190k24146264

asked Jun 2 at 10:33

Honey KumarHoney Kumar

274

New contributor

Honey Kumar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited Jun 2 at 10:50

José Carlos Santos

190k24146264

edited Jun 2 at 10:50

José Carlos Santos

190k24146264

asked Jun 2 at 10:33

Honey KumarHoney Kumar

274

New contributor

Honey Kumar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Jun 2 at 10:33

Honey KumarHoney Kumar

274

2 Answers
2

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3

$begingroup$

If $x^2+x+1$ is the minimal polynomial of $M$ then $M^2+M+I=0$. Also $M^3=I$ implies $M^-1=M^2$.
Hence $$M^-1+M=M^2+M=-I$$.

share|cite|improve this answer

answered Jun 2 at 10:43

NamanNaman

3207

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Naman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$

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2

$begingroup$

Let $lambda$ be an eigenvalue of $M$. Then, since $M^3=operatornameId$, $lambda^3=1$. Butbeginalignlambda^3=1&iff(lambda-1)(lambda^2+lambda+1)=0\&ifflambda^2+lambda+1=0,endalignsince $lambdaneq1$. There are only two numbers $lambda$ such that $lambda^2+lambda+1=0$: $-frac12pmfracsqrt32i$. In each case, $frac1lambda=overlinelambda$. If $v$ is an eigenvector corresponding to the eigenvalue $lambda$, then$$(M+M^-1).v=lambda v+frac1lambda v=left(lambda+overlinelambdaright)v=2operatornameRe(lambda)v=-v.$$

share|cite|improve this answer

edited Jun 2 at 10:54

P. Quinton

2,2301217

answered Jun 2 at 10:45

José Carlos SantosJosé Carlos Santos

190k24146264

$endgroup$

add a comment | 

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3

$begingroup$

If $x^2+x+1$ is the minimal polynomial of $M$ then $M^2+M+I=0$. Also $M^3=I$ implies $M^-1=M^2$.
Hence $$M^-1+M=M^2+M=-I$$.

share|cite|improve this answer

answered Jun 2 at 10:43

NamanNaman

3207

New contributor

Naman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

3

$begingroup$

If $x^2+x+1$ is the minimal polynomial of $M$ then $M^2+M+I=0$. Also $M^3=I$ implies $M^-1=M^2$.
Hence $$M^-1+M=M^2+M=-I$$.

share|cite|improve this answer

answered Jun 2 at 10:43

NamanNaman

3207

New contributor

Naman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

If $x^2+x+1$ is the minimal polynomial of $M$ then $M^2+M+I=0$. Also $M^3=I$ implies $M^-1=M^2$.
Hence $$M^-1+M=M^2+M=-I$$.

share|cite|improve this answer

answered Jun 2 at 10:43

NamanNaman

3207

New contributor

Naman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this answer

answered Jun 2 at 10:43

NamanNaman

3207

New contributor

Naman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered Jun 2 at 10:43

NamanNaman

3207

New contributor

Naman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered Jun 2 at 10:43

NamanNaman

3207

2

$begingroup$

Let $lambda$ be an eigenvalue of $M$. Then, since $M^3=operatornameId$, $lambda^3=1$. Butbeginalignlambda^3=1&iff(lambda-1)(lambda^2+lambda+1)=0\&ifflambda^2+lambda+1=0,endalignsince $lambdaneq1$. There are only two numbers $lambda$ such that $lambda^2+lambda+1=0$: $-frac12pmfracsqrt32i$. In each case, $frac1lambda=overlinelambda$. If $v$ is an eigenvector corresponding to the eigenvalue $lambda$, then$$(M+M^-1).v=lambda v+frac1lambda v=left(lambda+overlinelambdaright)v=2operatornameRe(lambda)v=-v.$$

share|cite|improve this answer

edited Jun 2 at 10:54

P. Quinton

2,2301217

answered Jun 2 at 10:45

José Carlos SantosJosé Carlos Santos

190k24146264

$endgroup$

add a comment | 

2

$begingroup$

Let $lambda$ be an eigenvalue of $M$. Then, since $M^3=operatornameId$, $lambda^3=1$. Butbeginalignlambda^3=1&iff(lambda-1)(lambda^2+lambda+1)=0\&ifflambda^2+lambda+1=0,endalignsince $lambdaneq1$. There are only two numbers $lambda$ such that $lambda^2+lambda+1=0$: $-frac12pmfracsqrt32i$. In each case, $frac1lambda=overlinelambda$. If $v$ is an eigenvector corresponding to the eigenvalue $lambda$, then$$(M+M^-1).v=lambda v+frac1lambda v=left(lambda+overlinelambdaright)v=2operatornameRe(lambda)v=-v.$$

share|cite|improve this answer

edited Jun 2 at 10:54

P. Quinton

2,2301217

answered Jun 2 at 10:45

José Carlos SantosJosé Carlos Santos

190k24146264

$endgroup$

add a comment | 

$begingroup$

Let $lambda$ be an eigenvalue of $M$. Then, since $M^3=operatornameId$, $lambda^3=1$. Butbeginalignlambda^3=1&iff(lambda-1)(lambda^2+lambda+1)=0\&ifflambda^2+lambda+1=0,endalignsince $lambdaneq1$. There are only two numbers $lambda$ such that $lambda^2+lambda+1=0$: $-frac12pmfracsqrt32i$. In each case, $frac1lambda=overlinelambda$. If $v$ is an eigenvector corresponding to the eigenvalue $lambda$, then$$(M+M^-1).v=lambda v+frac1lambda v=left(lambda+overlinelambdaright)v=2operatornameRe(lambda)v=-v.$$

share|cite|improve this answer

edited Jun 2 at 10:54

P. Quinton

2,2301217

answered Jun 2 at 10:45

José Carlos SantosJosé Carlos Santos

190k24146264

$endgroup$

share|cite|improve this answer

edited Jun 2 at 10:54

P. Quinton

2,2301217

answered Jun 2 at 10:45

José Carlos SantosJosé Carlos Santos

190k24146264

edited Jun 2 at 10:54

P. Quinton

2,2301217

edited Jun 2 at 10:54

P. Quinton

2,2301217

answered Jun 2 at 10:45

José Carlos SantosJosé Carlos Santos

190k24146264

answered Jun 2 at 10:45

José Carlos SantosJosé Carlos Santos

190k24146264