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Determinant and trace of a matrix

Show that $det(A+A^t)neq p$ where $p$ is a prime number and $A in M_{p times p}(mathbb{Z})$I need hints on showing a matrix with certain properties defines a special transformationEigenvalues of complex special orthogonal matrix$f(A)$ is invertible iff $A,B$ have no common eigenvaluesIf $f=a_{0}+a_{1}x+…+a_{n}x^{n}in Z[x]$ has a rational root $r/s$, how to prove $r|a_0$ and $s|a_n$?Is there a $5 times 5$ matrix with entries in $mathbb{Q}$ such that its characteristic polynomial is not solvable by radicalsEigenvalues and corresponding eigenspaceShow that $A$ is a scalar matrix $kI$ if and only if the minimum polynomial of $A$ is $m(t) = t - k$.Prove that an integer matrix cannot have $frac{1}{4}(-3+isqrt{5})$ as eigenvaluefind a recursive relation for the characteristic polynomial of the $k times k $ matrix?

2

$begingroup$

Find $det A$ and $text{Tr} A$ for the matrix $Ain M_n(mathbb{Q})$ such that $sqrt[n]{p}$ is an eigenvalue of $A$, where $p$ is a prime number or a positive integer such that the square root is irrational number.

My attempt is described below. From hypothesis we know that
$$ det(A-sqrt[n]{p}I)=0,
$$ hence $ det(A+sqrt[n]{p}I)=0$ because the characteristic polynomial of matrix $A$ has rational coefficients. Multiplying these two relation we get $ det(A^2-sqrt[n]{p^2}I)=0$ and so on.

From here, I don't find something. How to proceed? Thanks.

linear-algebra matrices

share|cite|improve this question

edited Mar 11 at 11:20

stefano

asked Mar 11 at 10:30

stefanostefano

1638

$endgroup$

add a comment | 

2

$begingroup$

Find $det A$ and $text{Tr} A$ for the matrix $Ain M_n(mathbb{Q})$ such that $sqrt[n]{p}$ is an eigenvalue of $A$, where $p$ is a prime number or a positive integer such that the square root is irrational number.

My attempt is described below. From hypothesis we know that
$$ det(A-sqrt[n]{p}I)=0,
$$ hence $ det(A+sqrt[n]{p}I)=0$ because the characteristic polynomial of matrix $A$ has rational coefficients. Multiplying these two relation we get $ det(A^2-sqrt[n]{p^2}I)=0$ and so on.

From here, I don't find something. How to proceed? Thanks.

linear-algebra matrices

share|cite|improve this question

edited Mar 11 at 11:20

stefano

asked Mar 11 at 10:30

stefanostefano

1638

$endgroup$

add a comment | 

$begingroup$

Find $det A$ and $text{Tr} A$ for the matrix $Ain M_n(mathbb{Q})$ such that $sqrt[n]{p}$ is an eigenvalue of $A$, where $p$ is a prime number or a positive integer such that the square root is irrational number.

My attempt is described below. From hypothesis we know that
$$ det(A-sqrt[n]{p}I)=0,
$$ hence $ det(A+sqrt[n]{p}I)=0$ because the characteristic polynomial of matrix $A$ has rational coefficients. Multiplying these two relation we get $ det(A^2-sqrt[n]{p^2}I)=0$ and so on.

From here, I don't find something. How to proceed? Thanks.

linear-algebra matrices

share|cite|improve this question

edited Mar 11 at 11:20

stefano

asked Mar 11 at 10:30

stefanostefano

1638

$endgroup$

share|cite|improve this question

edited Mar 11 at 11:20

stefano

asked Mar 11 at 10:30

stefanostefano

1638

share|cite|improve this question

edited Mar 11 at 11:20

stefano

asked Mar 11 at 10:30

stefanostefano

1638

asked Mar 11 at 10:30

stefanostefano

1638

asked Mar 11 at 10:30

stefanostefano

1638

2 Answers
2

active

oldest

votes

1

$begingroup$

It seems that your question misses details, otherwise there are many possible solutions.

Indeed, suppose that $p$ is an integer at the power $n$, let's say $p=a^n$, with $a in mathbb{N}$. Then any diagonal matrix, with rational coefficients and $a$ on the diagonal, has $sqrt[n]{p}$ as an eigenvalue. Then you can have any trace and determinant you want...

share|cite|improve this answer

answered Mar 11 at 11:02

TheSilverDoeTheSilverDoe

3,866112

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for the answear. I made a mistake, actualy $p$ is a prime number or an integer such that the square root is irrational.
  $endgroup$
  – stefano
  Mar 11 at 11:22
  

  
  
  
  

add a comment | 

0

$begingroup$

I see that $n$ appears in $sqrt[n]{p}$ and in $M_n(mathbb{Q})$, where in the last one I take it as the dimension (= number of columns) of the square matrix $A$. So the characteristic polynomial is a polynomial of degree $n$, with rational coefficients and with $sqrt[n]{p}$ as a root:

begin{equation}
p(lambda)= lambda^n + c_{1} lambda^{n-1} +dots + c_{n}
end{equation}

Now each coefficient $c_{i}$ will be a function involving sums of monomials of degree $i$ constructed with $sqrt[n]{p}$ and the other (possibly complex) roots. For example one can show that $c_{n}=det(A)$ and $c_{1} = text{Tr}(A)$. Now the key argument is that to get a rational number involving products of $sqrt[n]{p}$, you must at least raise it to the power $n$! This shuold imply that every coefficient is $0$ except from $c_{n}=p$. So you have $c_{n}=det(A)=p$ and $c_{1} = text{Tr}(A)=0$.

Now one should make this kind of argument sound by filling the gaps and proving every step. I hope that I've not made any mistakes and that this is helpful!

EDIT: Note that this argument depends on the fact that $sqrt[n]{p}^k$ is irrational for all $k<n$

share|cite|improve this answer

answered Mar 11 at 11:38

G.CarugnoG.Carugno

164

New contributor

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

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Since $a$ defined by the square root of order $n$ of $p$ is an eigenvalue, we have that $p(a)=0$. Hence, $0=p+det(A)Leftrightarrow det(A)=-p$. Am I wrong? Why is $c_n=p$?
  $endgroup$
  – stefano
  2 days ago
  

  
  
  
  

add a comment | 

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1

$begingroup$

It seems that your question misses details, otherwise there are many possible solutions.

Indeed, suppose that $p$ is an integer at the power $n$, let's say $p=a^n$, with $a in mathbb{N}$. Then any diagonal matrix, with rational coefficients and $a$ on the diagonal, has $sqrt[n]{p}$ as an eigenvalue. Then you can have any trace and determinant you want...

share|cite|improve this answer

answered Mar 11 at 11:02

TheSilverDoeTheSilverDoe

3,866112

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for the answear. I made a mistake, actualy $p$ is a prime number or an integer such that the square root is irrational.
  $endgroup$
  – stefano
  Mar 11 at 11:22
  

  
  
  
  

add a comment | 

1

$begingroup$

It seems that your question misses details, otherwise there are many possible solutions.

Indeed, suppose that $p$ is an integer at the power $n$, let's say $p=a^n$, with $a in mathbb{N}$. Then any diagonal matrix, with rational coefficients and $a$ on the diagonal, has $sqrt[n]{p}$ as an eigenvalue. Then you can have any trace and determinant you want...

share|cite|improve this answer

answered Mar 11 at 11:02

TheSilverDoeTheSilverDoe

3,866112

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for the answear. I made a mistake, actualy $p$ is a prime number or an integer such that the square root is irrational.
  $endgroup$
  – stefano
  Mar 11 at 11:22
  

  
  
  
  

add a comment | 

$begingroup$

It seems that your question misses details, otherwise there are many possible solutions.

Indeed, suppose that $p$ is an integer at the power $n$, let's say $p=a^n$, with $a in mathbb{N}$. Then any diagonal matrix, with rational coefficients and $a$ on the diagonal, has $sqrt[n]{p}$ as an eigenvalue. Then you can have any trace and determinant you want...

share|cite|improve this answer

answered Mar 11 at 11:02

TheSilverDoeTheSilverDoe

3,866112

$endgroup$

share|cite|improve this answer

answered Mar 11 at 11:02

TheSilverDoeTheSilverDoe

3,866112

answered Mar 11 at 11:02

TheSilverDoeTheSilverDoe

3,866112

answered Mar 11 at 11:02

TheSilverDoeTheSilverDoe

3,866112

0

$begingroup$

I see that $n$ appears in $sqrt[n]{p}$ and in $M_n(mathbb{Q})$, where in the last one I take it as the dimension (= number of columns) of the square matrix $A$. So the characteristic polynomial is a polynomial of degree $n$, with rational coefficients and with $sqrt[n]{p}$ as a root:

begin{equation}
p(lambda)= lambda^n + c_{1} lambda^{n-1} +dots + c_{n}
end{equation}

Now each coefficient $c_{i}$ will be a function involving sums of monomials of degree $i$ constructed with $sqrt[n]{p}$ and the other (possibly complex) roots. For example one can show that $c_{n}=det(A)$ and $c_{1} = text{Tr}(A)$. Now the key argument is that to get a rational number involving products of $sqrt[n]{p}$, you must at least raise it to the power $n$! This shuold imply that every coefficient is $0$ except from $c_{n}=p$. So you have $c_{n}=det(A)=p$ and $c_{1} = text{Tr}(A)=0$.

Now one should make this kind of argument sound by filling the gaps and proving every step. I hope that I've not made any mistakes and that this is helpful!

EDIT: Note that this argument depends on the fact that $sqrt[n]{p}^k$ is irrational for all $k<n$

share|cite|improve this answer

answered Mar 11 at 11:38

G.CarugnoG.Carugno

164

New contributor

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

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Since $a$ defined by the square root of order $n$ of $p$ is an eigenvalue, we have that $p(a)=0$. Hence, $0=p+det(A)Leftrightarrow det(A)=-p$. Am I wrong? Why is $c_n=p$?
  $endgroup$
  – stefano
  2 days ago
  

  
  
  
  

add a comment | 

0

$begingroup$

I see that $n$ appears in $sqrt[n]{p}$ and in $M_n(mathbb{Q})$, where in the last one I take it as the dimension (= number of columns) of the square matrix $A$. So the characteristic polynomial is a polynomial of degree $n$, with rational coefficients and with $sqrt[n]{p}$ as a root:

begin{equation}
p(lambda)= lambda^n + c_{1} lambda^{n-1} +dots + c_{n}
end{equation}

Now each coefficient $c_{i}$ will be a function involving sums of monomials of degree $i$ constructed with $sqrt[n]{p}$ and the other (possibly complex) roots. For example one can show that $c_{n}=det(A)$ and $c_{1} = text{Tr}(A)$. Now the key argument is that to get a rational number involving products of $sqrt[n]{p}$, you must at least raise it to the power $n$! This shuold imply that every coefficient is $0$ except from $c_{n}=p$. So you have $c_{n}=det(A)=p$ and $c_{1} = text{Tr}(A)=0$.

Now one should make this kind of argument sound by filling the gaps and proving every step. I hope that I've not made any mistakes and that this is helpful!

EDIT: Note that this argument depends on the fact that $sqrt[n]{p}^k$ is irrational for all $k<n$

share|cite|improve this answer

answered Mar 11 at 11:38

G.CarugnoG.Carugno

164

New contributor

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

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Since $a$ defined by the square root of order $n$ of $p$ is an eigenvalue, we have that $p(a)=0$. Hence, $0=p+det(A)Leftrightarrow det(A)=-p$. Am I wrong? Why is $c_n=p$?
  $endgroup$
  – stefano
  2 days ago
  

  
  
  
  

add a comment | 

$begingroup$

I see that $n$ appears in $sqrt[n]{p}$ and in $M_n(mathbb{Q})$, where in the last one I take it as the dimension (= number of columns) of the square matrix $A$. So the characteristic polynomial is a polynomial of degree $n$, with rational coefficients and with $sqrt[n]{p}$ as a root:

begin{equation}
p(lambda)= lambda^n + c_{1} lambda^{n-1} +dots + c_{n}
end{equation}

Now each coefficient $c_{i}$ will be a function involving sums of monomials of degree $i$ constructed with $sqrt[n]{p}$ and the other (possibly complex) roots. For example one can show that $c_{n}=det(A)$ and $c_{1} = text{Tr}(A)$. Now the key argument is that to get a rational number involving products of $sqrt[n]{p}$, you must at least raise it to the power $n$! This shuold imply that every coefficient is $0$ except from $c_{n}=p$. So you have $c_{n}=det(A)=p$ and $c_{1} = text{Tr}(A)=0$.

Now one should make this kind of argument sound by filling the gaps and proving every step. I hope that I've not made any mistakes and that this is helpful!

EDIT: Note that this argument depends on the fact that $sqrt[n]{p}^k$ is irrational for all $k<n$

share|cite|improve this answer

answered Mar 11 at 11:38

G.CarugnoG.Carugno

164

New contributor

G.Carugno 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 Mar 11 at 11:38

G.CarugnoG.Carugno

164

New contributor

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

answered Mar 11 at 11:38

G.CarugnoG.Carugno

164

New contributor

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

answered Mar 11 at 11:38

G.CarugnoG.Carugno

164