Whether the set $A$ is a compact subset of $M_3(Bbb R)$.












1












$begingroup$



Consider $$X=Big{A in M_3(Bbb R): rho_A(x)=x^3-3x^2+2x-1Big}$$ where $rho_A(x)$ is the characteristic polynomial of $A$ and $M_3(Bbb R)$ is the space of all $3 times 3$ matrices over $Bbb R$.



Is $X$ compact in $M_3(Bbb R)$ ?





My try: I confused with only the $2x$ term in $rho_A(x)$. Because , if the $2x$ term not appear , then $X$ becomes $$X={A in M_3(Bbb R): text{trace}(A)=3,det A=1} $$ which is unbounded, since $$(forall n in Bbb N):begin{pmatrix} 1&0 & n\0&1&0\0&0&1 end{pmatrix} in X$$



But here the problem is, the appearance of $2$. I Know $$2=A_{11}+A_{22}+A_{33}$$



so I think in this case the set becomes bounded and closed



Any help?






general-topology







share|cite|improve this question











$endgroup$












  • $begingroup$
    I hope my earlier incorrect comment did not mislead you. The characteristic polynomial has three distinct eigenvalues, so any $A in X$ must have the same eigenvalues.
    $endgroup$
    – copper.hat
    Dec 10 '18 at 3:00










  • $begingroup$
    If you were dealing with complex elements then the set $X$ would be unbounded (take an upper triangular matrix).
    $endgroup$
    – copper.hat
    Dec 10 '18 at 3:13
















1












$begingroup$



Consider $$X=Big{A in M_3(Bbb R): rho_A(x)=x^3-3x^2+2x-1Big}$$ where $rho_A(x)$ is the characteristic polynomial of $A$ and $M_3(Bbb R)$ is the space of all $3 times 3$ matrices over $Bbb R$.



Is $X$ compact in $M_3(Bbb R)$ ?





My try: I confused with only the $2x$ term in $rho_A(x)$. Because , if the $2x$ term not appear , then $X$ becomes $$X={A in M_3(Bbb R): text{trace}(A)=3,det A=1} $$ which is unbounded, since $$(forall n in Bbb N):begin{pmatrix} 1&0 & n\0&1&0\0&0&1 end{pmatrix} in X$$



But here the problem is, the appearance of $2$. I Know $$2=A_{11}+A_{22}+A_{33}$$



so I think in this case the set becomes bounded and closed



Any help?






general-topology







share|cite|improve this question











$endgroup$












  • $begingroup$
    I hope my earlier incorrect comment did not mislead you. The characteristic polynomial has three distinct eigenvalues, so any $A in X$ must have the same eigenvalues.
    $endgroup$
    – copper.hat
    Dec 10 '18 at 3:00










  • $begingroup$
    If you were dealing with complex elements then the set $X$ would be unbounded (take an upper triangular matrix).
    $endgroup$
    – copper.hat
    Dec 10 '18 at 3:13














1












1








1


2



$begingroup$



Consider $$X=Big{A in M_3(Bbb R): rho_A(x)=x^3-3x^2+2x-1Big}$$ where $rho_A(x)$ is the characteristic polynomial of $A$ and $M_3(Bbb R)$ is the space of all $3 times 3$ matrices over $Bbb R$.



Is $X$ compact in $M_3(Bbb R)$ ?





My try: I confused with only the $2x$ term in $rho_A(x)$. Because , if the $2x$ term not appear , then $X$ becomes $$X={A in M_3(Bbb R): text{trace}(A)=3,det A=1} $$ which is unbounded, since $$(forall n in Bbb N):begin{pmatrix} 1&0 & n\0&1&0\0&0&1 end{pmatrix} in X$$



But here the problem is, the appearance of $2$. I Know $$2=A_{11}+A_{22}+A_{33}$$



so I think in this case the set becomes bounded and closed



Any help?






general-topology







share|cite|improve this question











$endgroup$





Consider $$X=Big{A in M_3(Bbb R): rho_A(x)=x^3-3x^2+2x-1Big}$$ where $rho_A(x)$ is the characteristic polynomial of $A$ and $M_3(Bbb R)$ is the space of all $3 times 3$ matrices over $Bbb R$.



Is $X$ compact in $M_3(Bbb R)$ ?





My try: I confused with only the $2x$ term in $rho_A(x)$. Because , if the $2x$ term not appear , then $X$ becomes $$X={A in M_3(Bbb R): text{trace}(A)=3,det A=1} $$ which is unbounded, since $$(forall n in Bbb N):begin{pmatrix} 1&0 & n\0&1&0\0&0&1 end{pmatrix} in X$$



But here the problem is, the appearance of $2$. I Know $$2=A_{11}+A_{22}+A_{33}$$



so I think in this case the set becomes bounded and closed



Any help?






general-topology




general-topology






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 2:53







Chinnapparaj R

















asked Dec 10 '18 at 2:51









Chinnapparaj RChinnapparaj R

5,5072928




5,5072928












  • $begingroup$
    I hope my earlier incorrect comment did not mislead you. The characteristic polynomial has three distinct eigenvalues, so any $A in X$ must have the same eigenvalues.
    $endgroup$
    – copper.hat
    Dec 10 '18 at 3:00










  • $begingroup$
    If you were dealing with complex elements then the set $X$ would be unbounded (take an upper triangular matrix).
    $endgroup$
    – copper.hat
    Dec 10 '18 at 3:13


















  • $begingroup$
    I hope my earlier incorrect comment did not mislead you. The characteristic polynomial has three distinct eigenvalues, so any $A in X$ must have the same eigenvalues.
    $endgroup$
    – copper.hat
    Dec 10 '18 at 3:00










  • $begingroup$
    If you were dealing with complex elements then the set $X$ would be unbounded (take an upper triangular matrix).
    $endgroup$
    – copper.hat
    Dec 10 '18 at 3:13
















$begingroup$
I hope my earlier incorrect comment did not mislead you. The characteristic polynomial has three distinct eigenvalues, so any $A in X$ must have the same eigenvalues.
$endgroup$
– copper.hat
Dec 10 '18 at 3:00




$begingroup$
I hope my earlier incorrect comment did not mislead you. The characteristic polynomial has three distinct eigenvalues, so any $A in X$ must have the same eigenvalues.
$endgroup$
– copper.hat
Dec 10 '18 at 3:00












$begingroup$
If you were dealing with complex elements then the set $X$ would be unbounded (take an upper triangular matrix).
$endgroup$
– copper.hat
Dec 10 '18 at 3:13




$begingroup$
If you were dealing with complex elements then the set $X$ would be unbounded (take an upper triangular matrix).
$endgroup$
– copper.hat
Dec 10 '18 at 3:13










1 Answer
1






active

oldest

votes


















3












$begingroup$

For example,
$$ pmatrix{t & 0 & 1cr 1-2t & 0 & -2cr -t^2+3t & 1 & 3-tcr} $$
has that characteristic polynomial. This is $S^{-1} A S$ where
$$ A = pmatrix{0 & 0 & 1cr 1 & 0 & -2cr 0 & 1 & 3cr},
S = pmatrix{1 & 0 & 0cr 0 & 1 & 0cr t & 0 & 1cr}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Can you shed light on how this very nice answer might have been guessed -- and whether specifying the characteristic polynomial (or some annihilating polynomial) ever yields a compact set?
    $endgroup$
    – user25959
    Dec 10 '18 at 3:19










  • $begingroup$
    @Robert Israel: super! thanks! what is the intuition behind for finding this matrix ?
    $endgroup$
    – Chinnapparaj R
    Dec 10 '18 at 3:21








  • 1




    $begingroup$
    Start with a companion matrix for that characteristic polynomial. Choose an $S$ such that $S$ and $S^{-1}$ have entries that are polynomial in $t$...
    $endgroup$
    – Robert Israel
    Dec 10 '18 at 3:22










  • $begingroup$
    Of course there's the trivial case where the minimal polynomial has degree $1$, so the matrix is a constant multiple of $I$.
    $endgroup$
    – Robert Israel
    Dec 10 '18 at 3:25










  • $begingroup$
    Thank you sir! :-)
    $endgroup$
    – Chinnapparaj R
    Dec 10 '18 at 3:28











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1 Answer
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1 Answer
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active

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active

oldest

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3












$begingroup$

For example,
$$ pmatrix{t & 0 & 1cr 1-2t & 0 & -2cr -t^2+3t & 1 & 3-tcr} $$
has that characteristic polynomial. This is $S^{-1} A S$ where
$$ A = pmatrix{0 & 0 & 1cr 1 & 0 & -2cr 0 & 1 & 3cr},
S = pmatrix{1 & 0 & 0cr 0 & 1 & 0cr t & 0 & 1cr}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Can you shed light on how this very nice answer might have been guessed -- and whether specifying the characteristic polynomial (or some annihilating polynomial) ever yields a compact set?
    $endgroup$
    – user25959
    Dec 10 '18 at 3:19










  • $begingroup$
    @Robert Israel: super! thanks! what is the intuition behind for finding this matrix ?
    $endgroup$
    – Chinnapparaj R
    Dec 10 '18 at 3:21








  • 1




    $begingroup$
    Start with a companion matrix for that characteristic polynomial. Choose an $S$ such that $S$ and $S^{-1}$ have entries that are polynomial in $t$...
    $endgroup$
    – Robert Israel
    Dec 10 '18 at 3:22










  • $begingroup$
    Of course there's the trivial case where the minimal polynomial has degree $1$, so the matrix is a constant multiple of $I$.
    $endgroup$
    – Robert Israel
    Dec 10 '18 at 3:25










  • $begingroup$
    Thank you sir! :-)
    $endgroup$
    – Chinnapparaj R
    Dec 10 '18 at 3:28
















3












$begingroup$

For example,
$$ pmatrix{t & 0 & 1cr 1-2t & 0 & -2cr -t^2+3t & 1 & 3-tcr} $$
has that characteristic polynomial. This is $S^{-1} A S$ where
$$ A = pmatrix{0 & 0 & 1cr 1 & 0 & -2cr 0 & 1 & 3cr},
S = pmatrix{1 & 0 & 0cr 0 & 1 & 0cr t & 0 & 1cr}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Can you shed light on how this very nice answer might have been guessed -- and whether specifying the characteristic polynomial (or some annihilating polynomial) ever yields a compact set?
    $endgroup$
    – user25959
    Dec 10 '18 at 3:19










  • $begingroup$
    @Robert Israel: super! thanks! what is the intuition behind for finding this matrix ?
    $endgroup$
    – Chinnapparaj R
    Dec 10 '18 at 3:21








  • 1




    $begingroup$
    Start with a companion matrix for that characteristic polynomial. Choose an $S$ such that $S$ and $S^{-1}$ have entries that are polynomial in $t$...
    $endgroup$
    – Robert Israel
    Dec 10 '18 at 3:22










  • $begingroup$
    Of course there's the trivial case where the minimal polynomial has degree $1$, so the matrix is a constant multiple of $I$.
    $endgroup$
    – Robert Israel
    Dec 10 '18 at 3:25










  • $begingroup$
    Thank you sir! :-)
    $endgroup$
    – Chinnapparaj R
    Dec 10 '18 at 3:28














3












3








3





$begingroup$

For example,
$$ pmatrix{t & 0 & 1cr 1-2t & 0 & -2cr -t^2+3t & 1 & 3-tcr} $$
has that characteristic polynomial. This is $S^{-1} A S$ where
$$ A = pmatrix{0 & 0 & 1cr 1 & 0 & -2cr 0 & 1 & 3cr},
S = pmatrix{1 & 0 & 0cr 0 & 1 & 0cr t & 0 & 1cr}$$






share|cite|improve this answer









$endgroup$



For example,
$$ pmatrix{t & 0 & 1cr 1-2t & 0 & -2cr -t^2+3t & 1 & 3-tcr} $$
has that characteristic polynomial. This is $S^{-1} A S$ where
$$ A = pmatrix{0 & 0 & 1cr 1 & 0 & -2cr 0 & 1 & 3cr},
S = pmatrix{1 & 0 & 0cr 0 & 1 & 0cr t & 0 & 1cr}$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 10 '18 at 3:15









Robert IsraelRobert Israel

324k23214468




324k23214468












  • $begingroup$
    Can you shed light on how this very nice answer might have been guessed -- and whether specifying the characteristic polynomial (or some annihilating polynomial) ever yields a compact set?
    $endgroup$
    – user25959
    Dec 10 '18 at 3:19










  • $begingroup$
    @Robert Israel: super! thanks! what is the intuition behind for finding this matrix ?
    $endgroup$
    – Chinnapparaj R
    Dec 10 '18 at 3:21








  • 1




    $begingroup$
    Start with a companion matrix for that characteristic polynomial. Choose an $S$ such that $S$ and $S^{-1}$ have entries that are polynomial in $t$...
    $endgroup$
    – Robert Israel
    Dec 10 '18 at 3:22










  • $begingroup$
    Of course there's the trivial case where the minimal polynomial has degree $1$, so the matrix is a constant multiple of $I$.
    $endgroup$
    – Robert Israel
    Dec 10 '18 at 3:25










  • $begingroup$
    Thank you sir! :-)
    $endgroup$
    – Chinnapparaj R
    Dec 10 '18 at 3:28


















  • $begingroup$
    Can you shed light on how this very nice answer might have been guessed -- and whether specifying the characteristic polynomial (or some annihilating polynomial) ever yields a compact set?
    $endgroup$
    – user25959
    Dec 10 '18 at 3:19










  • $begingroup$
    @Robert Israel: super! thanks! what is the intuition behind for finding this matrix ?
    $endgroup$
    – Chinnapparaj R
    Dec 10 '18 at 3:21








  • 1




    $begingroup$
    Start with a companion matrix for that characteristic polynomial. Choose an $S$ such that $S$ and $S^{-1}$ have entries that are polynomial in $t$...
    $endgroup$
    – Robert Israel
    Dec 10 '18 at 3:22










  • $begingroup$
    Of course there's the trivial case where the minimal polynomial has degree $1$, so the matrix is a constant multiple of $I$.
    $endgroup$
    – Robert Israel
    Dec 10 '18 at 3:25










  • $begingroup$
    Thank you sir! :-)
    $endgroup$
    – Chinnapparaj R
    Dec 10 '18 at 3:28
















$begingroup$
Can you shed light on how this very nice answer might have been guessed -- and whether specifying the characteristic polynomial (or some annihilating polynomial) ever yields a compact set?
$endgroup$
– user25959
Dec 10 '18 at 3:19




$begingroup$
Can you shed light on how this very nice answer might have been guessed -- and whether specifying the characteristic polynomial (or some annihilating polynomial) ever yields a compact set?
$endgroup$
– user25959
Dec 10 '18 at 3:19












$begingroup$
@Robert Israel: super! thanks! what is the intuition behind for finding this matrix ?
$endgroup$
– Chinnapparaj R
Dec 10 '18 at 3:21






$begingroup$
@Robert Israel: super! thanks! what is the intuition behind for finding this matrix ?
$endgroup$
– Chinnapparaj R
Dec 10 '18 at 3:21






1




1




$begingroup$
Start with a companion matrix for that characteristic polynomial. Choose an $S$ such that $S$ and $S^{-1}$ have entries that are polynomial in $t$...
$endgroup$
– Robert Israel
Dec 10 '18 at 3:22




$begingroup$
Start with a companion matrix for that characteristic polynomial. Choose an $S$ such that $S$ and $S^{-1}$ have entries that are polynomial in $t$...
$endgroup$
– Robert Israel
Dec 10 '18 at 3:22












$begingroup$
Of course there's the trivial case where the minimal polynomial has degree $1$, so the matrix is a constant multiple of $I$.
$endgroup$
– Robert Israel
Dec 10 '18 at 3:25




$begingroup$
Of course there's the trivial case where the minimal polynomial has degree $1$, so the matrix is a constant multiple of $I$.
$endgroup$
– Robert Israel
Dec 10 '18 at 3:25












$begingroup$
Thank you sir! :-)
$endgroup$
– Chinnapparaj R
Dec 10 '18 at 3:28




$begingroup$
Thank you sir! :-)
$endgroup$
– Chinnapparaj R
Dec 10 '18 at 3:28


















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