If all eigenvalues of $A in operatornameM_2(mathbbZ)$ are rational, then they are integers.
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How to show that if $A$ is a $2times 2$ matrix with all integer entries and all eigenvalues are in $mathbbQ$, then all eigenvalue are integers?
I cannot find why eigenvalues in $mathbbQ$, in this case, will precisely belong to $mathbbZ$?
What is the proof?
linear-algebra eigenvalues-eigenvectors
edited Sep 7 at 13:11
Jendrik Stelzner
7,69121137
asked Sep 1 at 3:32
Priya Dey
393
add a comment |Â
up vote
3
down vote
favorite
How to show that if $A$ is a $2times 2$ matrix with all integer entries and all eigenvalues are in $mathbbQ$, then all eigenvalue are integers?
I cannot find why eigenvalues in $mathbbQ$, in this case, will precisely belong to $mathbbZ$?
What is the proof?
linear-algebra eigenvalues-eigenvectors
edited Sep 7 at 13:11
Jendrik Stelzner
7,69121137
asked Sep 1 at 3:32
Priya Dey
393
en.wikipedia.org/wiki/Rational_root_theorem
– Will Jagy
Sep 1 at 3:37
Are you familiar with the theorem that says a monic polynomial with integer coefficients cannot have rational roots that are not integers?
– spaceisdarkgreen
Sep 1 at 3:38
The characteristic polynomial of $A$ is monic of degree two, has integer entries, and rational roots...
– dan_fulea
Sep 1 at 3:38
how? please explain in answer not in comment spaceisdarkgreen
– Priya Dey
Sep 1 at 4:27
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
How to show that if $A$ is a $2times 2$ matrix with all integer entries and all eigenvalues are in $mathbbQ$, then all eigenvalue are integers?
I cannot find why eigenvalues in $mathbbQ$, in this case, will precisely belong to $mathbbZ$?
What is the proof?
linear-algebra eigenvalues-eigenvectors
edited Sep 7 at 13:11
Jendrik Stelzner
7,69121137
asked Sep 1 at 3:32
Priya Dey
393
How to show that if $A$ is a $2times 2$ matrix with all integer entries and all eigenvalues are in $mathbbQ$, then all eigenvalue are integers?
I cannot find why eigenvalues in $mathbbQ$, in this case, will precisely belong to $mathbbZ$?
What is the proof?
linear-algebra eigenvalues-eigenvectors
linear-algebra eigenvalues-eigenvectors
edited Sep 7 at 13:11
Jendrik Stelzner
7,69121137
asked Sep 1 at 3:32
Priya Dey
393
edited Sep 7 at 13:11
Jendrik Stelzner
7,69121137
asked Sep 1 at 3:32
Priya Dey
393
edited Sep 7 at 13:11
Jendrik Stelzner
7,69121137
edited Sep 7 at 13:11
Jendrik Stelzner
7,69121137
edited Sep 7 at 13:11
Jendrik Stelzner
7,69121137
7,69121137
asked Sep 1 at 3:32
Priya Dey
393
asked Sep 1 at 3:32
Priya Dey
393
asked Sep 1 at 3:32
Priya Dey
393
393
en.wikipedia.org/wiki/Rational_root_theorem
– Will Jagy
Sep 1 at 3:37
Are you familiar with the theorem that says a monic polynomial with integer coefficients cannot have rational roots that are not integers?
– spaceisdarkgreen
Sep 1 at 3:38
The characteristic polynomial of $A$ is monic of degree two, has integer entries, and rational roots...
– dan_fulea
Sep 1 at 3:38
how? please explain in answer not in comment spaceisdarkgreen
– Priya Dey
Sep 1 at 4:27
add a comment |Â
en.wikipedia.org/wiki/Rational_root_theorem
– Will Jagy
Sep 1 at 3:37
Are you familiar with the theorem that says a monic polynomial with integer coefficients cannot have rational roots that are not integers?
– spaceisdarkgreen
Sep 1 at 3:38
The characteristic polynomial of $A$ is monic of degree two, has integer entries, and rational roots...
– dan_fulea
Sep 1 at 3:38
how? please explain in answer not in comment spaceisdarkgreen
– Priya Dey
Sep 1 at 4:27
en.wikipedia.org/wiki/Rational_root_theorem
– Will Jagy
Sep 1 at 3:37
en.wikipedia.org/wiki/Rational_root_theorem
– Will Jagy
Sep 1 at 3:37
Are you familiar with the theorem that says a monic polynomial with integer coefficients cannot have rational roots that are not integers?
– spaceisdarkgreen
Sep 1 at 3:38
Are you familiar with the theorem that says a monic polynomial with integer coefficients cannot have rational roots that are not integers?
– spaceisdarkgreen
Sep 1 at 3:38
The characteristic polynomial of $A$ is monic of degree two, has integer entries, and rational roots...
– dan_fulea
Sep 1 at 3:38
The characteristic polynomial of $A$ is monic of degree two, has integer entries, and rational roots...
– dan_fulea
Sep 1 at 3:38
how? please explain in answer not in comment spaceisdarkgreen
– Priya Dey
Sep 1 at 4:27
how? please explain in answer not in comment spaceisdarkgreen
– Priya Dey
Sep 1 at 4:27
add a comment |Â
1 Answer
1
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oldest
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up vote
1
down vote
This follows from the rational root theorem, which implies that any rational root of a polynomial with integer coefficients and leading coefficient one is an integer. The characteristic polynomial of the matrix is a polynomial that meets this criterion.
This actually implies the result for an $ntimes n$ matrix, not just a $2times 2.$ So perhaps they want you to give a more direct proof of the special case. If the characteristic polynomial is $ lambda^2 + blambda + c,$ the roots are $$ lambda = frac-bpmsqrtb^2-4c2.$$ If these are to be rational, we need $b^2-4c$ to be a perfect square, since it is known that the square root of an integer is either irrational or an integer. (And thus $sqrtb^2-4c$ is an integer.)
Can you show that $-bpm sqrtb^2-4c$ is always even? I suggest splitting it up into the case where $b$ is even and odd.
answered Sep 1 at 3:53
spaceisdarkgreen
28.8k21548
what is the case when (b^2-4*a*c)^(1/2) is irrotational? spaceisdarkgreen
– Priya Dey
Sep 1 at 4:07
and how to show b^2>=4*c?
– Priya Dey
Sep 1 at 4:10
In those cases the solution won’t be rational... you don’t have to worry about them
– spaceisdarkgreen
Sep 1 at 4:15
what is the exact theorem says? then we cannot say any monic polynomial with integer cofficient has roots in integer.. then could be anhthing.. then what is the specility we have here?
– Priya Dey
Sep 1 at 4:22
yes I'm able to show what may be "b", numeretor always is even integer whenever the sq. root gives us a perfect squre
– Priya Dey
Sep 1 at 4:24
 |Â
show 4 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
This follows from the rational root theorem, which implies that any rational root of a polynomial with integer coefficients and leading coefficient one is an integer. The characteristic polynomial of the matrix is a polynomial that meets this criterion.
This actually implies the result for an $ntimes n$ matrix, not just a $2times 2.$ So perhaps they want you to give a more direct proof of the special case. If the characteristic polynomial is $ lambda^2 + blambda + c,$ the roots are $$ lambda = frac-bpmsqrtb^2-4c2.$$ If these are to be rational, we need $b^2-4c$ to be a perfect square, since it is known that the square root of an integer is either irrational or an integer. (And thus $sqrtb^2-4c$ is an integer.)
Can you show that $-bpm sqrtb^2-4c$ is always even? I suggest splitting it up into the case where $b$ is even and odd.
answered Sep 1 at 3:53
spaceisdarkgreen
28.8k21548
what is the case when (b^2-4*a*c)^(1/2) is irrotational? spaceisdarkgreen
– Priya Dey
Sep 1 at 4:07
and how to show b^2>=4*c?
– Priya Dey
Sep 1 at 4:10
In those cases the solution won’t be rational... you don’t have to worry about them
– spaceisdarkgreen
Sep 1 at 4:15
what is the exact theorem says? then we cannot say any monic polynomial with integer cofficient has roots in integer.. then could be anhthing.. then what is the specility we have here?
– Priya Dey
Sep 1 at 4:22
yes I'm able to show what may be "b", numeretor always is even integer whenever the sq. root gives us a perfect squre
– Priya Dey
Sep 1 at 4:24
 |Â
show 4 more comments
up vote
1
down vote
This follows from the rational root theorem, which implies that any rational root of a polynomial with integer coefficients and leading coefficient one is an integer. The characteristic polynomial of the matrix is a polynomial that meets this criterion.
This actually implies the result for an $ntimes n$ matrix, not just a $2times 2.$ So perhaps they want you to give a more direct proof of the special case. If the characteristic polynomial is $ lambda^2 + blambda + c,$ the roots are $$ lambda = frac-bpmsqrtb^2-4c2.$$ If these are to be rational, we need $b^2-4c$ to be a perfect square, since it is known that the square root of an integer is either irrational or an integer. (And thus $sqrtb^2-4c$ is an integer.)
Can you show that $-bpm sqrtb^2-4c$ is always even? I suggest splitting it up into the case where $b$ is even and odd.
answered Sep 1 at 3:53
spaceisdarkgreen
28.8k21548
what is the case when (b^2-4*a*c)^(1/2) is irrotational? spaceisdarkgreen
– Priya Dey
Sep 1 at 4:07
and how to show b^2>=4*c?
– Priya Dey
Sep 1 at 4:10
In those cases the solution won’t be rational... you don’t have to worry about them
– spaceisdarkgreen
Sep 1 at 4:15
what is the exact theorem says? then we cannot say any monic polynomial with integer cofficient has roots in integer.. then could be anhthing.. then what is the specility we have here?
– Priya Dey
Sep 1 at 4:22
yes I'm able to show what may be "b", numeretor always is even integer whenever the sq. root gives us a perfect squre
– Priya Dey
Sep 1 at 4:24
 |Â
show 4 more comments
up vote
1
down vote
up vote
1
down vote
This follows from the rational root theorem, which implies that any rational root of a polynomial with integer coefficients and leading coefficient one is an integer. The characteristic polynomial of the matrix is a polynomial that meets this criterion.
This actually implies the result for an $ntimes n$ matrix, not just a $2times 2.$ So perhaps they want you to give a more direct proof of the special case. If the characteristic polynomial is $ lambda^2 + blambda + c,$ the roots are $$ lambda = frac-bpmsqrtb^2-4c2.$$ If these are to be rational, we need $b^2-4c$ to be a perfect square, since it is known that the square root of an integer is either irrational or an integer. (And thus $sqrtb^2-4c$ is an integer.)
Can you show that $-bpm sqrtb^2-4c$ is always even? I suggest splitting it up into the case where $b$ is even and odd.
answered Sep 1 at 3:53
spaceisdarkgreen
28.8k21548
This follows from the rational root theorem, which implies that any rational root of a polynomial with integer coefficients and leading coefficient one is an integer. The characteristic polynomial of the matrix is a polynomial that meets this criterion.
This actually implies the result for an $ntimes n$ matrix, not just a $2times 2.$ So perhaps they want you to give a more direct proof of the special case. If the characteristic polynomial is $ lambda^2 + blambda + c,$ the roots are $$ lambda = frac-bpmsqrtb^2-4c2.$$ If these are to be rational, we need $b^2-4c$ to be a perfect square, since it is known that the square root of an integer is either irrational or an integer. (And thus $sqrtb^2-4c$ is an integer.)
Can you show that $-bpm sqrtb^2-4c$ is always even? I suggest splitting it up into the case where $b$ is even and odd.
answered Sep 1 at 3:53
spaceisdarkgreen
28.8k21548
edited Sep 1 at 3:58
answered Sep 1 at 3:53
spaceisdarkgreen
28.8k21548
answered Sep 1 at 3:53
spaceisdarkgreen
28.8k21548
answered Sep 1 at 3:53
spaceisdarkgreen
28.8k21548
28.8k21548
what is the case when (b^2-4*a*c)^(1/2) is irrotational? spaceisdarkgreen
– Priya Dey
Sep 1 at 4:07
and how to show b^2>=4*c?
– Priya Dey
Sep 1 at 4:10
In those cases the solution won’t be rational... you don’t have to worry about them
– spaceisdarkgreen
Sep 1 at 4:15
what is the exact theorem says? then we cannot say any monic polynomial with integer cofficient has roots in integer.. then could be anhthing.. then what is the specility we have here?
– Priya Dey
Sep 1 at 4:22
yes I'm able to show what may be "b", numeretor always is even integer whenever the sq. root gives us a perfect squre
– Priya Dey
Sep 1 at 4:24
 |Â
show 4 more comments
what is the case when (b^2-4*a*c)^(1/2) is irrotational? spaceisdarkgreen
– Priya Dey
Sep 1 at 4:07
and how to show b^2>=4*c?
– Priya Dey
Sep 1 at 4:10
In those cases the solution won’t be rational... you don’t have to worry about them
– spaceisdarkgreen
Sep 1 at 4:15
what is the exact theorem says? then we cannot say any monic polynomial with integer cofficient has roots in integer.. then could be anhthing.. then what is the specility we have here?
– Priya Dey
Sep 1 at 4:22
yes I'm able to show what may be "b", numeretor always is even integer whenever the sq. root gives us a perfect squre
– Priya Dey
Sep 1 at 4:24
what is the case when (b^2-4*a*c)^(1/2) is irrotational? spaceisdarkgreen
– Priya Dey
Sep 1 at 4:07
what is the case when (b^2-4*a*c)^(1/2) is irrotational? spaceisdarkgreen
– Priya Dey
Sep 1 at 4:07
and how to show b^2>=4*c?
– Priya Dey
Sep 1 at 4:10
and how to show b^2>=4*c?
– Priya Dey
Sep 1 at 4:10
In those cases the solution won’t be rational... you don’t have to worry about them
– spaceisdarkgreen
Sep 1 at 4:15
In those cases the solution won’t be rational... you don’t have to worry about them
– spaceisdarkgreen
Sep 1 at 4:15
what is the exact theorem says? then we cannot say any monic polynomial with integer cofficient has roots in integer.. then could be anhthing.. then what is the specility we have here?
– Priya Dey
Sep 1 at 4:22
what is the exact theorem says? then we cannot say any monic polynomial with integer cofficient has roots in integer.. then could be anhthing.. then what is the specility we have here?
– Priya Dey
Sep 1 at 4:22
yes I'm able to show what may be "b", numeretor always is even integer whenever the sq. root gives us a perfect squre
– Priya Dey
Sep 1 at 4:24
yes I'm able to show what may be "b", numeretor always is even integer whenever the sq. root gives us a perfect squre
– Priya Dey
Sep 1 at 4:24
 |Â
show 4 more comments
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en.wikipedia.org/wiki/Rational_root_theorem
– Will Jagy
Sep 1 at 3:37
Are you familiar with the theorem that says a monic polynomial with integer coefficients cannot have rational roots that are not integers?
– spaceisdarkgreen
Sep 1 at 3:38
The characteristic polynomial of $A$ is monic of degree two, has integer entries, and rational roots...
– dan_fulea
Sep 1 at 3:38
how? please explain in answer not in comment spaceisdarkgreen
– Priya Dey
Sep 1 at 4:27