A positive integer matrix with no integer eigenvalues
$begingroup$
Let $n in mathbb{N}$ and $Ain M_n(mathbb{N})$ with $Tr(A)=0$ and $A^3+A-2I_n=O_n$.Prove that $n$ is a multiple of $3$ and $det(A^2)=det(A^2+I_n)$.
I tried to find $A$'s eigenvalues, but the equation $x^3+x-2=0$ has only one integer root, $1$,and this contradicts $Tr(A) =0$.
EDIT:My approach doesn' t work as pointed out in the comments, how should this be solved?
linear-algebra matrices
asked Jan 11 at 15:15
JustAnAmateurJustAnAmateur
825
$endgroup$
|
show 3 more comments
$begingroup$
Let $n in mathbb{N}$ and $Ain M_n(mathbb{N})$ with $Tr(A)=0$ and $A^3+A-2I_n=O_n$.Prove that $n$ is a multiple of $3$ and $det(A^2)=det(A^2+I_n)$.
I tried to find $A$'s eigenvalues, but the equation $x^3+x-2=0$ has only one integer root, $1$,and this contradicts $Tr(A) =0$.
EDIT:My approach doesn' t work as pointed out in the comments, how should this be solved?
linear-algebra matrices
asked Jan 11 at 15:15
JustAnAmateurJustAnAmateur
825
$endgroup$
$begingroup$
Eigenvalues need not only be integers.
$endgroup$
– Theo Bendit
Jan 11 at 15:19
$begingroup$
Why not? The matrix is over $N$. I know that if it were over $C$ it would have eigenvalues.
$endgroup$
– JustAnAmateur
Jan 11 at 15:20
$begingroup$
The eigenvalues of a matrix with entries in, for instance $mathbb N$, are solutions to a polynomial equation whose coefficients are in $mathbb N$. These solutions need not (and in fact are often not) in $mathbb N$.
$endgroup$
– Dave
Jan 11 at 15:21
$begingroup$
I have edited my question, my approach clearly doesn't work.
$endgroup$
– JustAnAmateur
Jan 11 at 15:43
$begingroup$
I think, $M_n(Bbb N)$ is a typo.
$endgroup$
– Dietrich Burde
Jan 11 at 16:23
|
show 3 more comments
$begingroup$
Let $n in mathbb{N}$ and $Ain M_n(mathbb{N})$ with $Tr(A)=0$ and $A^3+A-2I_n=O_n$.Prove that $n$ is a multiple of $3$ and $det(A^2)=det(A^2+I_n)$.
I tried to find $A$'s eigenvalues, but the equation $x^3+x-2=0$ has only one integer root, $1$,and this contradicts $Tr(A) =0$.
EDIT:My approach doesn' t work as pointed out in the comments, how should this be solved?
linear-algebra matrices
asked Jan 11 at 15:15
JustAnAmateurJustAnAmateur
825
$endgroup$
Let $n in mathbb{N}$ and $Ain M_n(mathbb{N})$ with $Tr(A)=0$ and $A^3+A-2I_n=O_n$.Prove that $n$ is a multiple of $3$ and $det(A^2)=det(A^2+I_n)$.
I tried to find $A$'s eigenvalues, but the equation $x^3+x-2=0$ has only one integer root, $1$,and this contradicts $Tr(A) =0$.
EDIT:My approach doesn' t work as pointed out in the comments, how should this be solved?
linear-algebra matrices
linear-algebra matrices
asked Jan 11 at 15:15
JustAnAmateurJustAnAmateur
825
asked Jan 11 at 15:15
JustAnAmateurJustAnAmateur
825
edited Jan 11 at 15:42
JustAnAmateur
asked Jan 11 at 15:15
JustAnAmateurJustAnAmateur
825
asked Jan 11 at 15:15
JustAnAmateurJustAnAmateur
825
asked Jan 11 at 15:15
JustAnAmateurJustAnAmateur
825
825
$begingroup$
Eigenvalues need not only be integers.
$endgroup$
– Theo Bendit
Jan 11 at 15:19
$begingroup$
Why not? The matrix is over $N$. I know that if it were over $C$ it would have eigenvalues.
$endgroup$
– JustAnAmateur
Jan 11 at 15:20
$begingroup$
The eigenvalues of a matrix with entries in, for instance $mathbb N$, are solutions to a polynomial equation whose coefficients are in $mathbb N$. These solutions need not (and in fact are often not) in $mathbb N$.
$endgroup$
– Dave
Jan 11 at 15:21
$begingroup$
I have edited my question, my approach clearly doesn't work.
$endgroup$
– JustAnAmateur
Jan 11 at 15:43
$begingroup$
I think, $M_n(Bbb N)$ is a typo.
$endgroup$
– Dietrich Burde
Jan 11 at 16:23
|
show 3 more comments
$begingroup$
Eigenvalues need not only be integers.
$endgroup$
– Theo Bendit
Jan 11 at 15:19
$begingroup$
Why not? The matrix is over $N$. I know that if it were over $C$ it would have eigenvalues.
$endgroup$
– JustAnAmateur
Jan 11 at 15:20
$begingroup$
The eigenvalues of a matrix with entries in, for instance $mathbb N$, are solutions to a polynomial equation whose coefficients are in $mathbb N$. These solutions need not (and in fact are often not) in $mathbb N$.
$endgroup$
– Dave
Jan 11 at 15:21
$begingroup$
I have edited my question, my approach clearly doesn't work.
$endgroup$
– JustAnAmateur
Jan 11 at 15:43
$begingroup$
I think, $M_n(Bbb N)$ is a typo.
$endgroup$
– Dietrich Burde
Jan 11 at 16:23
$begingroup$
Eigenvalues need not only be integers.
$endgroup$
– Theo Bendit
Jan 11 at 15:19
$begingroup$
Eigenvalues need not only be integers.
$endgroup$
– Theo Bendit
Jan 11 at 15:19
$begingroup$
Why not? The matrix is over $N$. I know that if it were over $C$ it would have eigenvalues.
$endgroup$
– JustAnAmateur
Jan 11 at 15:20
$begingroup$
Why not? The matrix is over $N$. I know that if it were over $C$ it would have eigenvalues.
$endgroup$
– JustAnAmateur
Jan 11 at 15:20
$begingroup$
The eigenvalues of a matrix with entries in, for instance $mathbb N$, are solutions to a polynomial equation whose coefficients are in $mathbb N$. These solutions need not (and in fact are often not) in $mathbb N$.
$endgroup$
– Dave
Jan 11 at 15:21
$begingroup$
The eigenvalues of a matrix with entries in, for instance $mathbb N$, are solutions to a polynomial equation whose coefficients are in $mathbb N$. These solutions need not (and in fact are often not) in $mathbb N$.
$endgroup$
– Dave
Jan 11 at 15:21
$begingroup$
I have edited my question, my approach clearly doesn't work.
$endgroup$
– JustAnAmateur
Jan 11 at 15:43
$begingroup$
I have edited my question, my approach clearly doesn't work.
$endgroup$
– JustAnAmateur
Jan 11 at 15:43
$begingroup$
I think, $M_n(Bbb N)$ is a typo.
$endgroup$
– Dietrich Burde
Jan 11 at 16:23
$begingroup$
I think, $M_n(Bbb N)$ is a typo.
$endgroup$
– Dietrich Burde
Jan 11 at 16:23
|
show 3 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Since $A^3 + A - 2 I = 0$, all eigenvalues of $A$ are roots of the polynomial $x^2 + x - 2$, thus $1$ or $-1/2 pm sqrt{7} i/2$. Since it's a real matrix, the two non-real eigenvalues have equal algebraic multiplicities. Since the trace (which is the sum of the eigenvalues) must be $0$,
all three eigenvalues have equal multiplicities. Thus if this multiplicity is $k$,
there are $3k$ eigenvalues counted by algebraic multiplicity, i.e. $n=3k$.
answered Jan 11 at 16:34
Robert IsraelRobert Israel
321k23210462
$endgroup$
$begingroup$
Thank you for your solution ! I only have one question : you said this is a real matrix, but the problem states it is a positive integer matrix. Are we still allowed to use the fact that the trace is the sum of the eigenvalues? I know that this works only over algebraically closed fields.
$endgroup$
– JustAnAmateur
Jan 12 at 12:01
$begingroup$
The matrix might have entries in $mathbb N$, but the eigenvalues are complex numbers. Yes, this is allowed.
$endgroup$
– Robert Israel
Jan 13 at 6:46
add a comment |
$begingroup$
Take the matrix in $GL_2(Bbb Z)$
$$
begin{pmatrix}
1 & 1 \
1 & 0
end{pmatrix},
$$
it has not integers as eigenvalues, but the golden ratio $frac{1pm sqrt{5}}{2} $. If you want to construct matrices with integer eigenvalues, then see for example here:
Constructing regular integer matrices with distinct integer eigenvalues
answered Jan 11 at 15:27
Dietrich BurdeDietrich Burde
78.5k64386
$endgroup$
add a comment |
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2 Answers
2
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oldest
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2 Answers
2
active
oldest
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oldest
votes
$begingroup$
Since $A^3 + A - 2 I = 0$, all eigenvalues of $A$ are roots of the polynomial $x^2 + x - 2$, thus $1$ or $-1/2 pm sqrt{7} i/2$. Since it's a real matrix, the two non-real eigenvalues have equal algebraic multiplicities. Since the trace (which is the sum of the eigenvalues) must be $0$,
all three eigenvalues have equal multiplicities. Thus if this multiplicity is $k$,
there are $3k$ eigenvalues counted by algebraic multiplicity, i.e. $n=3k$.
answered Jan 11 at 16:34
Robert IsraelRobert Israel
321k23210462
$endgroup$
$begingroup$
Thank you for your solution ! I only have one question : you said this is a real matrix, but the problem states it is a positive integer matrix. Are we still allowed to use the fact that the trace is the sum of the eigenvalues? I know that this works only over algebraically closed fields.
$endgroup$
– JustAnAmateur
Jan 12 at 12:01
$begingroup$
The matrix might have entries in $mathbb N$, but the eigenvalues are complex numbers. Yes, this is allowed.
$endgroup$
– Robert Israel
Jan 13 at 6:46
add a comment |
$begingroup$
Since $A^3 + A - 2 I = 0$, all eigenvalues of $A$ are roots of the polynomial $x^2 + x - 2$, thus $1$ or $-1/2 pm sqrt{7} i/2$. Since it's a real matrix, the two non-real eigenvalues have equal algebraic multiplicities. Since the trace (which is the sum of the eigenvalues) must be $0$,
all three eigenvalues have equal multiplicities. Thus if this multiplicity is $k$,
there are $3k$ eigenvalues counted by algebraic multiplicity, i.e. $n=3k$.
answered Jan 11 at 16:34
Robert IsraelRobert Israel
321k23210462
$endgroup$
$begingroup$
Thank you for your solution ! I only have one question : you said this is a real matrix, but the problem states it is a positive integer matrix. Are we still allowed to use the fact that the trace is the sum of the eigenvalues? I know that this works only over algebraically closed fields.
$endgroup$
– JustAnAmateur
Jan 12 at 12:01
$begingroup$
The matrix might have entries in $mathbb N$, but the eigenvalues are complex numbers. Yes, this is allowed.
$endgroup$
– Robert Israel
Jan 13 at 6:46
add a comment |
$begingroup$
Since $A^3 + A - 2 I = 0$, all eigenvalues of $A$ are roots of the polynomial $x^2 + x - 2$, thus $1$ or $-1/2 pm sqrt{7} i/2$. Since it's a real matrix, the two non-real eigenvalues have equal algebraic multiplicities. Since the trace (which is the sum of the eigenvalues) must be $0$,
all three eigenvalues have equal multiplicities. Thus if this multiplicity is $k$,
there are $3k$ eigenvalues counted by algebraic multiplicity, i.e. $n=3k$.
answered Jan 11 at 16:34
Robert IsraelRobert Israel
321k23210462
$endgroup$
Since $A^3 + A - 2 I = 0$, all eigenvalues of $A$ are roots of the polynomial $x^2 + x - 2$, thus $1$ or $-1/2 pm sqrt{7} i/2$. Since it's a real matrix, the two non-real eigenvalues have equal algebraic multiplicities. Since the trace (which is the sum of the eigenvalues) must be $0$,
all three eigenvalues have equal multiplicities. Thus if this multiplicity is $k$,
there are $3k$ eigenvalues counted by algebraic multiplicity, i.e. $n=3k$.
answered Jan 11 at 16:34
Robert IsraelRobert Israel
321k23210462
answered Jan 11 at 16:34
Robert IsraelRobert Israel
321k23210462
answered Jan 11 at 16:34
Robert IsraelRobert Israel
321k23210462
answered Jan 11 at 16:34
Robert IsraelRobert Israel
321k23210462
321k23210462
$begingroup$
Thank you for your solution ! I only have one question : you said this is a real matrix, but the problem states it is a positive integer matrix. Are we still allowed to use the fact that the trace is the sum of the eigenvalues? I know that this works only over algebraically closed fields.
$endgroup$
– JustAnAmateur
Jan 12 at 12:01
$begingroup$
The matrix might have entries in $mathbb N$, but the eigenvalues are complex numbers. Yes, this is allowed.
$endgroup$
– Robert Israel
Jan 13 at 6:46
add a comment |
$begingroup$
Thank you for your solution ! I only have one question : you said this is a real matrix, but the problem states it is a positive integer matrix. Are we still allowed to use the fact that the trace is the sum of the eigenvalues? I know that this works only over algebraically closed fields.
$endgroup$
– JustAnAmateur
Jan 12 at 12:01
$begingroup$
The matrix might have entries in $mathbb N$, but the eigenvalues are complex numbers. Yes, this is allowed.
$endgroup$
– Robert Israel
Jan 13 at 6:46
$begingroup$
Thank you for your solution ! I only have one question : you said this is a real matrix, but the problem states it is a positive integer matrix. Are we still allowed to use the fact that the trace is the sum of the eigenvalues? I know that this works only over algebraically closed fields.
$endgroup$
– JustAnAmateur
Jan 12 at 12:01
$begingroup$
Thank you for your solution ! I only have one question : you said this is a real matrix, but the problem states it is a positive integer matrix. Are we still allowed to use the fact that the trace is the sum of the eigenvalues? I know that this works only over algebraically closed fields.
$endgroup$
– JustAnAmateur
Jan 12 at 12:01
$begingroup$
The matrix might have entries in $mathbb N$, but the eigenvalues are complex numbers. Yes, this is allowed.
$endgroup$
– Robert Israel
Jan 13 at 6:46
$begingroup$
The matrix might have entries in $mathbb N$, but the eigenvalues are complex numbers. Yes, this is allowed.
$endgroup$
– Robert Israel
Jan 13 at 6:46
add a comment |
$begingroup$
Take the matrix in $GL_2(Bbb Z)$
$$
begin{pmatrix}
1 & 1 \
1 & 0
end{pmatrix},
$$
it has not integers as eigenvalues, but the golden ratio $frac{1pm sqrt{5}}{2} $. If you want to construct matrices with integer eigenvalues, then see for example here:
Constructing regular integer matrices with distinct integer eigenvalues
answered Jan 11 at 15:27
Dietrich BurdeDietrich Burde
78.5k64386
$endgroup$
add a comment |
$begingroup$
Take the matrix in $GL_2(Bbb Z)$
$$
begin{pmatrix}
1 & 1 \
1 & 0
end{pmatrix},
$$
it has not integers as eigenvalues, but the golden ratio $frac{1pm sqrt{5}}{2} $. If you want to construct matrices with integer eigenvalues, then see for example here:
Constructing regular integer matrices with distinct integer eigenvalues
answered Jan 11 at 15:27
Dietrich BurdeDietrich Burde
78.5k64386
$endgroup$
add a comment |
$begingroup$
Take the matrix in $GL_2(Bbb Z)$
$$
begin{pmatrix}
1 & 1 \
1 & 0
end{pmatrix},
$$
it has not integers as eigenvalues, but the golden ratio $frac{1pm sqrt{5}}{2} $. If you want to construct matrices with integer eigenvalues, then see for example here:
Constructing regular integer matrices with distinct integer eigenvalues
answered Jan 11 at 15:27
Dietrich BurdeDietrich Burde
78.5k64386
$endgroup$
Take the matrix in $GL_2(Bbb Z)$
$$
begin{pmatrix}
1 & 1 \
1 & 0
end{pmatrix},
$$
it has not integers as eigenvalues, but the golden ratio $frac{1pm sqrt{5}}{2} $. If you want to construct matrices with integer eigenvalues, then see for example here:
Constructing regular integer matrices with distinct integer eigenvalues
answered Jan 11 at 15:27
Dietrich BurdeDietrich Burde
78.5k64386
edited Jan 11 at 15:33
answered Jan 11 at 15:27
Dietrich BurdeDietrich Burde
78.5k64386
answered Jan 11 at 15:27
Dietrich BurdeDietrich Burde
78.5k64386
answered Jan 11 at 15:27
Dietrich BurdeDietrich Burde
78.5k64386
78.5k64386
add a comment |
add a comment |
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$begingroup$
Eigenvalues need not only be integers.
$endgroup$
– Theo Bendit
Jan 11 at 15:19
$begingroup$
Why not? The matrix is over $N$. I know that if it were over $C$ it would have eigenvalues.
$endgroup$
– JustAnAmateur
Jan 11 at 15:20
$begingroup$
The eigenvalues of a matrix with entries in, for instance $mathbb N$, are solutions to a polynomial equation whose coefficients are in $mathbb N$. These solutions need not (and in fact are often not) in $mathbb N$.
$endgroup$
– Dave
Jan 11 at 15:21
$begingroup$
I have edited my question, my approach clearly doesn't work.
$endgroup$
– JustAnAmateur
Jan 11 at 15:43
$begingroup$
I think, $M_n(Bbb N)$ is a typo.
$endgroup$
– Dietrich Burde
Jan 11 at 16:23