Find $det A$ and $operatorname{Tr} A$ if $det(A-sqrt[n]{3}I_n)=0$












4












$begingroup$


$A in M_{n}(mathbb{Q})$ and I have to find $det A$ and $operatorname{Tr} A$ if $det(A+sqrt[n]{3}I_n)=0$. I observed that $sqrt[n]{3}$ is an eigenvalue of $A$,but I don't know how to continue.

EDIT : My bad,the matrix has rational entries.






linear-algebra matrices







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I don't think this is possible to answer if we only know that $a in M_n(Bbb C)$. Do we have some other information? Does $A$ have integer or rational entries?
    $endgroup$
    – Omnomnomnom
    Dec 22 '18 at 20:52










  • $begingroup$
    Yes,I made an edit,sorry for the typo
    $endgroup$
    – Math Guy
    Dec 22 '18 at 20:56
















4












$begingroup$


$A in M_{n}(mathbb{Q})$ and I have to find $det A$ and $operatorname{Tr} A$ if $det(A+sqrt[n]{3}I_n)=0$. I observed that $sqrt[n]{3}$ is an eigenvalue of $A$,but I don't know how to continue.

EDIT : My bad,the matrix has rational entries.






linear-algebra matrices







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I don't think this is possible to answer if we only know that $a in M_n(Bbb C)$. Do we have some other information? Does $A$ have integer or rational entries?
    $endgroup$
    – Omnomnomnom
    Dec 22 '18 at 20:52










  • $begingroup$
    Yes,I made an edit,sorry for the typo
    $endgroup$
    – Math Guy
    Dec 22 '18 at 20:56














4












4








4





$begingroup$


$A in M_{n}(mathbb{Q})$ and I have to find $det A$ and $operatorname{Tr} A$ if $det(A+sqrt[n]{3}I_n)=0$. I observed that $sqrt[n]{3}$ is an eigenvalue of $A$,but I don't know how to continue.

EDIT : My bad,the matrix has rational entries.






linear-algebra matrices







share|cite|improve this question











$endgroup$




$A in M_{n}(mathbb{Q})$ and I have to find $det A$ and $operatorname{Tr} A$ if $det(A+sqrt[n]{3}I_n)=0$. I observed that $sqrt[n]{3}$ is an eigenvalue of $A$,but I don't know how to continue.

EDIT : My bad,the matrix has rational entries.






linear-algebra matrices




linear-algebra matrices






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 22 '18 at 20:55







Math Guy

















asked Dec 22 '18 at 20:35









Math GuyMath Guy

1267




1267








  • 1




    $begingroup$
    I don't think this is possible to answer if we only know that $a in M_n(Bbb C)$. Do we have some other information? Does $A$ have integer or rational entries?
    $endgroup$
    – Omnomnomnom
    Dec 22 '18 at 20:52










  • $begingroup$
    Yes,I made an edit,sorry for the typo
    $endgroup$
    – Math Guy
    Dec 22 '18 at 20:56














  • 1




    $begingroup$
    I don't think this is possible to answer if we only know that $a in M_n(Bbb C)$. Do we have some other information? Does $A$ have integer or rational entries?
    $endgroup$
    – Omnomnomnom
    Dec 22 '18 at 20:52










  • $begingroup$
    Yes,I made an edit,sorry for the typo
    $endgroup$
    – Math Guy
    Dec 22 '18 at 20:56








1




1




$begingroup$
I don't think this is possible to answer if we only know that $a in M_n(Bbb C)$. Do we have some other information? Does $A$ have integer or rational entries?
$endgroup$
– Omnomnomnom
Dec 22 '18 at 20:52




$begingroup$
I don't think this is possible to answer if we only know that $a in M_n(Bbb C)$. Do we have some other information? Does $A$ have integer or rational entries?
$endgroup$
– Omnomnomnom
Dec 22 '18 at 20:52












$begingroup$
Yes,I made an edit,sorry for the typo
$endgroup$
– Math Guy
Dec 22 '18 at 20:56




$begingroup$
Yes,I made an edit,sorry for the typo
$endgroup$
– Math Guy
Dec 22 '18 at 20:56










1 Answer
1






active

oldest

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2












$begingroup$

$det (A+lambda I_n)$ is a monic polynomial of degree $n$ in $lambda$. Note that if $n$-th root of $3$ is a root of a degree $n$-polynomial, the polynomial must be a multiple of $lambda^n-3$, hence $det$ and Tr must be $(-1)^ncdot -3$ and $0$ respectively (by using the fact that Tr is the coefficient of $lambda^{n-1}$ in the characteristic polynomial and $det$ is the constant term times $(-1)^n$ in the characteristic polynomial).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Why must the polynomial be a multiple of $lambda ^n -3$?
    $endgroup$
    – Math Guy
    Dec 22 '18 at 20:48










  • $begingroup$
    That is because the minimal polynomial of $n$-th root of $3$ is $x^n-3$, i.e. there is no polynomial that admits $n$-th root of $3$ as a root and has degree less than $n$.
    $endgroup$
    – Levent
    Dec 22 '18 at 20:50










  • $begingroup$
    @Arthur you are most definitely right, I guess I just automatically made some assumptions. However in this case there are matrices that satisfy the condition and have different trace and determinant which makes me believe that OP may have meant rational entries.
    $endgroup$
    – Levent
    Dec 22 '18 at 20:54










  • $begingroup$
    @Levent one more question : if the characteristic polynomial is a multiple of $lambda^n -3$,why do we still get that $det$ and $Tr$ must be $(-1)^ncdot -3$ and $0$?Shouldn't they be a multiple of $(-1)^ncdot -3$ and $0$?
    $endgroup$
    – Math Guy
    Dec 22 '18 at 21:02








  • 1




    $begingroup$
    You are welcome. If you have troubles with any of the facts I pull out of the hat, I can edit the answer to clarify.
    $endgroup$
    – Levent
    Dec 22 '18 at 21:09












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1 Answer
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active

oldest

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

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2












$begingroup$

$det (A+lambda I_n)$ is a monic polynomial of degree $n$ in $lambda$. Note that if $n$-th root of $3$ is a root of a degree $n$-polynomial, the polynomial must be a multiple of $lambda^n-3$, hence $det$ and Tr must be $(-1)^ncdot -3$ and $0$ respectively (by using the fact that Tr is the coefficient of $lambda^{n-1}$ in the characteristic polynomial and $det$ is the constant term times $(-1)^n$ in the characteristic polynomial).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Why must the polynomial be a multiple of $lambda ^n -3$?
    $endgroup$
    – Math Guy
    Dec 22 '18 at 20:48










  • $begingroup$
    That is because the minimal polynomial of $n$-th root of $3$ is $x^n-3$, i.e. there is no polynomial that admits $n$-th root of $3$ as a root and has degree less than $n$.
    $endgroup$
    – Levent
    Dec 22 '18 at 20:50










  • $begingroup$
    @Arthur you are most definitely right, I guess I just automatically made some assumptions. However in this case there are matrices that satisfy the condition and have different trace and determinant which makes me believe that OP may have meant rational entries.
    $endgroup$
    – Levent
    Dec 22 '18 at 20:54










  • $begingroup$
    @Levent one more question : if the characteristic polynomial is a multiple of $lambda^n -3$,why do we still get that $det$ and $Tr$ must be $(-1)^ncdot -3$ and $0$?Shouldn't they be a multiple of $(-1)^ncdot -3$ and $0$?
    $endgroup$
    – Math Guy
    Dec 22 '18 at 21:02








  • 1




    $begingroup$
    You are welcome. If you have troubles with any of the facts I pull out of the hat, I can edit the answer to clarify.
    $endgroup$
    – Levent
    Dec 22 '18 at 21:09
















2












$begingroup$

$det (A+lambda I_n)$ is a monic polynomial of degree $n$ in $lambda$. Note that if $n$-th root of $3$ is a root of a degree $n$-polynomial, the polynomial must be a multiple of $lambda^n-3$, hence $det$ and Tr must be $(-1)^ncdot -3$ and $0$ respectively (by using the fact that Tr is the coefficient of $lambda^{n-1}$ in the characteristic polynomial and $det$ is the constant term times $(-1)^n$ in the characteristic polynomial).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Why must the polynomial be a multiple of $lambda ^n -3$?
    $endgroup$
    – Math Guy
    Dec 22 '18 at 20:48










  • $begingroup$
    That is because the minimal polynomial of $n$-th root of $3$ is $x^n-3$, i.e. there is no polynomial that admits $n$-th root of $3$ as a root and has degree less than $n$.
    $endgroup$
    – Levent
    Dec 22 '18 at 20:50










  • $begingroup$
    @Arthur you are most definitely right, I guess I just automatically made some assumptions. However in this case there are matrices that satisfy the condition and have different trace and determinant which makes me believe that OP may have meant rational entries.
    $endgroup$
    – Levent
    Dec 22 '18 at 20:54










  • $begingroup$
    @Levent one more question : if the characteristic polynomial is a multiple of $lambda^n -3$,why do we still get that $det$ and $Tr$ must be $(-1)^ncdot -3$ and $0$?Shouldn't they be a multiple of $(-1)^ncdot -3$ and $0$?
    $endgroup$
    – Math Guy
    Dec 22 '18 at 21:02








  • 1




    $begingroup$
    You are welcome. If you have troubles with any of the facts I pull out of the hat, I can edit the answer to clarify.
    $endgroup$
    – Levent
    Dec 22 '18 at 21:09














2












2








2





$begingroup$

$det (A+lambda I_n)$ is a monic polynomial of degree $n$ in $lambda$. Note that if $n$-th root of $3$ is a root of a degree $n$-polynomial, the polynomial must be a multiple of $lambda^n-3$, hence $det$ and Tr must be $(-1)^ncdot -3$ and $0$ respectively (by using the fact that Tr is the coefficient of $lambda^{n-1}$ in the characteristic polynomial and $det$ is the constant term times $(-1)^n$ in the characteristic polynomial).






share|cite|improve this answer









$endgroup$



$det (A+lambda I_n)$ is a monic polynomial of degree $n$ in $lambda$. Note that if $n$-th root of $3$ is a root of a degree $n$-polynomial, the polynomial must be a multiple of $lambda^n-3$, hence $det$ and Tr must be $(-1)^ncdot -3$ and $0$ respectively (by using the fact that Tr is the coefficient of $lambda^{n-1}$ in the characteristic polynomial and $det$ is the constant term times $(-1)^n$ in the characteristic polynomial).







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 22 '18 at 20:43









LeventLevent

2,741925




2,741925












  • $begingroup$
    Why must the polynomial be a multiple of $lambda ^n -3$?
    $endgroup$
    – Math Guy
    Dec 22 '18 at 20:48










  • $begingroup$
    That is because the minimal polynomial of $n$-th root of $3$ is $x^n-3$, i.e. there is no polynomial that admits $n$-th root of $3$ as a root and has degree less than $n$.
    $endgroup$
    – Levent
    Dec 22 '18 at 20:50










  • $begingroup$
    @Arthur you are most definitely right, I guess I just automatically made some assumptions. However in this case there are matrices that satisfy the condition and have different trace and determinant which makes me believe that OP may have meant rational entries.
    $endgroup$
    – Levent
    Dec 22 '18 at 20:54










  • $begingroup$
    @Levent one more question : if the characteristic polynomial is a multiple of $lambda^n -3$,why do we still get that $det$ and $Tr$ must be $(-1)^ncdot -3$ and $0$?Shouldn't they be a multiple of $(-1)^ncdot -3$ and $0$?
    $endgroup$
    – Math Guy
    Dec 22 '18 at 21:02








  • 1




    $begingroup$
    You are welcome. If you have troubles with any of the facts I pull out of the hat, I can edit the answer to clarify.
    $endgroup$
    – Levent
    Dec 22 '18 at 21:09


















  • $begingroup$
    Why must the polynomial be a multiple of $lambda ^n -3$?
    $endgroup$
    – Math Guy
    Dec 22 '18 at 20:48










  • $begingroup$
    That is because the minimal polynomial of $n$-th root of $3$ is $x^n-3$, i.e. there is no polynomial that admits $n$-th root of $3$ as a root and has degree less than $n$.
    $endgroup$
    – Levent
    Dec 22 '18 at 20:50










  • $begingroup$
    @Arthur you are most definitely right, I guess I just automatically made some assumptions. However in this case there are matrices that satisfy the condition and have different trace and determinant which makes me believe that OP may have meant rational entries.
    $endgroup$
    – Levent
    Dec 22 '18 at 20:54










  • $begingroup$
    @Levent one more question : if the characteristic polynomial is a multiple of $lambda^n -3$,why do we still get that $det$ and $Tr$ must be $(-1)^ncdot -3$ and $0$?Shouldn't they be a multiple of $(-1)^ncdot -3$ and $0$?
    $endgroup$
    – Math Guy
    Dec 22 '18 at 21:02








  • 1




    $begingroup$
    You are welcome. If you have troubles with any of the facts I pull out of the hat, I can edit the answer to clarify.
    $endgroup$
    – Levent
    Dec 22 '18 at 21:09
















$begingroup$
Why must the polynomial be a multiple of $lambda ^n -3$?
$endgroup$
– Math Guy
Dec 22 '18 at 20:48




$begingroup$
Why must the polynomial be a multiple of $lambda ^n -3$?
$endgroup$
– Math Guy
Dec 22 '18 at 20:48












$begingroup$
That is because the minimal polynomial of $n$-th root of $3$ is $x^n-3$, i.e. there is no polynomial that admits $n$-th root of $3$ as a root and has degree less than $n$.
$endgroup$
– Levent
Dec 22 '18 at 20:50




$begingroup$
That is because the minimal polynomial of $n$-th root of $3$ is $x^n-3$, i.e. there is no polynomial that admits $n$-th root of $3$ as a root and has degree less than $n$.
$endgroup$
– Levent
Dec 22 '18 at 20:50












$begingroup$
@Arthur you are most definitely right, I guess I just automatically made some assumptions. However in this case there are matrices that satisfy the condition and have different trace and determinant which makes me believe that OP may have meant rational entries.
$endgroup$
– Levent
Dec 22 '18 at 20:54




$begingroup$
@Arthur you are most definitely right, I guess I just automatically made some assumptions. However in this case there are matrices that satisfy the condition and have different trace and determinant which makes me believe that OP may have meant rational entries.
$endgroup$
– Levent
Dec 22 '18 at 20:54












$begingroup$
@Levent one more question : if the characteristic polynomial is a multiple of $lambda^n -3$,why do we still get that $det$ and $Tr$ must be $(-1)^ncdot -3$ and $0$?Shouldn't they be a multiple of $(-1)^ncdot -3$ and $0$?
$endgroup$
– Math Guy
Dec 22 '18 at 21:02






$begingroup$
@Levent one more question : if the characteristic polynomial is a multiple of $lambda^n -3$,why do we still get that $det$ and $Tr$ must be $(-1)^ncdot -3$ and $0$?Shouldn't they be a multiple of $(-1)^ncdot -3$ and $0$?
$endgroup$
– Math Guy
Dec 22 '18 at 21:02






1




1




$begingroup$
You are welcome. If you have troubles with any of the facts I pull out of the hat, I can edit the answer to clarify.
$endgroup$
– Levent
Dec 22 '18 at 21:09




$begingroup$
You are welcome. If you have troubles with any of the facts I pull out of the hat, I can edit the answer to clarify.
$endgroup$
– Levent
Dec 22 '18 at 21:09


















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