Whether the set $A$ is a compact subset of $M_3(Bbb R)$.
$begingroup$
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
asked Dec 10 '18 at 2:51
Chinnapparaj RChinnapparaj R
5,3641828
$endgroup$
add a comment |
$begingroup$
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
asked Dec 10 '18 at 2:51
Chinnapparaj RChinnapparaj R
5,3641828
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
asked Dec 10 '18 at 2:51
Chinnapparaj RChinnapparaj R
5,3641828
$endgroup$
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
asked Dec 10 '18 at 2:51
Chinnapparaj RChinnapparaj R
5,3641828
asked Dec 10 '18 at 2:51
Chinnapparaj RChinnapparaj R
5,3641828
edited Dec 10 '18 at 2:53
Chinnapparaj R
asked Dec 10 '18 at 2:51
Chinnapparaj RChinnapparaj R
5,3641828
asked Dec 10 '18 at 2:51
Chinnapparaj RChinnapparaj R
5,3641828
asked Dec 10 '18 at 2:51
Chinnapparaj RChinnapparaj R
5,3641828
5,3641828
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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
$$ A = pmatrix{0 & 0 & 1cr 1 & 0 & -2cr 0 & 1 & 3cr},
S = pmatrix{1 & 0 & 0cr 0 & 1 & 0cr t & 0 & 1cr}$$
answered Dec 10 '18 at 3:15
Robert IsraelRobert Israel
321k23210462
$endgroup$
$begingroup$
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
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
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
$$ A = pmatrix{0 & 0 & 1cr 1 & 0 & -2cr 0 & 1 & 3cr},
S = pmatrix{1 & 0 & 0cr 0 & 1 & 0cr t & 0 & 1cr}$$
answered Dec 10 '18 at 3:15
Robert IsraelRobert Israel
321k23210462
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
$$ A = pmatrix{0 & 0 & 1cr 1 & 0 & -2cr 0 & 1 & 3cr},
S = pmatrix{1 & 0 & 0cr 0 & 1 & 0cr t & 0 & 1cr}$$
answered Dec 10 '18 at 3:15
Robert IsraelRobert Israel
321k23210462
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
$$ A = pmatrix{0 & 0 & 1cr 1 & 0 & -2cr 0 & 1 & 3cr},
S = pmatrix{1 & 0 & 0cr 0 & 1 & 0cr t & 0 & 1cr}$$
answered Dec 10 '18 at 3:15
Robert IsraelRobert Israel
321k23210462
$endgroup$
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
$$ A = pmatrix{0 & 0 & 1cr 1 & 0 & -2cr 0 & 1 & 3cr},
S = pmatrix{1 & 0 & 0cr 0 & 1 & 0cr t & 0 & 1cr}$$
answered Dec 10 '18 at 3:15
Robert IsraelRobert Israel
321k23210462
answered Dec 10 '18 at 3:15
Robert IsraelRobert Israel
321k23210462
answered Dec 10 '18 at 3:15
Robert IsraelRobert Israel
321k23210462
answered Dec 10 '18 at 3:15
Robert IsraelRobert Israel
321k23210462
321k23210462
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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