Eigenvalues of a Matrix Power over Arbitrary Fields Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Nilpotent matrix over a division algebraEigenvalues of a Matrix Using Diagonal EntriesInvertible matrix over a ring and its eigenvaluesEigenvalues of an orthogonal matrix over an arbitrary fieldBlock Matrix Determinant (Eigenvalues)Do endomorphisms of infinite-dimensional vector spaces over algebraically closed fields always have eigenvalues?My own version of definition of eigenvalues and eigenvectors for matrices.The characteristic polynomial of a square matrix whose eigenvalues are all 0Real eigenvalues of a matrix over a field of Complex Numbers$3$-by-$3$ Real Matrices with Repeated Eigenvalues
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Eigenvalues of a Matrix Power over Arbitrary Fields
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Nilpotent matrix over a division algebraEigenvalues of a Matrix Using Diagonal EntriesInvertible matrix over a ring and its eigenvaluesEigenvalues of an orthogonal matrix over an arbitrary fieldBlock Matrix Determinant (Eigenvalues)Do endomorphisms of infinite-dimensional vector spaces over algebraically closed fields always have eigenvalues?My own version of definition of eigenvalues and eigenvectors for matrices.The characteristic polynomial of a square matrix whose eigenvalues are all 0Real eigenvalues of a matrix over a field of Complex Numbers$3$-by-$3$ Real Matrices with Repeated Eigenvalues
$begingroup$
Let $A$ be a square matrix over some field $BbbF$. It can be shown that if $lambda$ is an eigenvalue of $A$ then $lambda^n$ is an eigenvalue of $A^n$. In my linear algebra class we were asked if all eigenvalues of $A^n$ can be written in this form.
In general, the answer is that they cannot. For example, if $BbbF = BbbR$ then the matrix
$$A =
beginbmatrix
0 & -1 \
1 & 0 \
endbmatrix$$
has no eigenvalues but $A^2$ has $-1$ as an eigenvalue. In fact, this argument can be generalized to show that if $BbbF$ is not closed under $n$th roots then the result does not hold over that field.
However, if $BbbF$ is algebraically closed and $A$ is as above, then it can actually be shown that all eigenvalues of $A^n$ are of the form $lambda^n$ for some eigenvalue $lambda$ of $A$.
Unfortunately, there are fields which are not algebraically closed but which are closed under $n$th roots. Because of this, I wanted to ask the following question:
For which fields $BbbF$ does it hold that if $A$ is a square matrix over $BbbF$ then all eigenvalues of $A^n$ are of the form $lambda^n$ for some eigenvalue $lambda$ of $A$?
linear-algebra matrices eigenvalues-eigenvectors
asked Apr 2 at 1:48
S. SenkoS. Senko
1335
$endgroup$
add a comment |
$begingroup$
Let $A$ be a square matrix over some field $BbbF$. It can be shown that if $lambda$ is an eigenvalue of $A$ then $lambda^n$ is an eigenvalue of $A^n$. In my linear algebra class we were asked if all eigenvalues of $A^n$ can be written in this form.
In general, the answer is that they cannot. For example, if $BbbF = BbbR$ then the matrix
$$A =
beginbmatrix
0 & -1 \
1 & 0 \
endbmatrix$$
has no eigenvalues but $A^2$ has $-1$ as an eigenvalue. In fact, this argument can be generalized to show that if $BbbF$ is not closed under $n$th roots then the result does not hold over that field.
However, if $BbbF$ is algebraically closed and $A$ is as above, then it can actually be shown that all eigenvalues of $A^n$ are of the form $lambda^n$ for some eigenvalue $lambda$ of $A$.
Unfortunately, there are fields which are not algebraically closed but which are closed under $n$th roots. Because of this, I wanted to ask the following question:
For which fields $BbbF$ does it hold that if $A$ is a square matrix over $BbbF$ then all eigenvalues of $A^n$ are of the form $lambda^n$ for some eigenvalue $lambda$ of $A$?
linear-algebra matrices eigenvalues-eigenvectors
asked Apr 2 at 1:48
S. SenkoS. Senko
1335
$endgroup$
1
$begingroup$
Does "closed under $n$th roots" mean "every element has at least one $n$th root" or "every element has a full set of $n$th roots"? Because the latter is enough, but the former is not.
$endgroup$
– darij grinberg
Apr 2 at 1:58
$begingroup$
@darij grinberg To be more specific about my claim, if there is any polynomial of the form $x^n - a$ over $BbbF$ that has no roots then the result does not hold. I don't know about whether or not it holds if $x^n-a$ has at least $1$ but fewer than $n$ roots (and that would in retrospect perhaps be a better definition of closed under $n$th roots)
$endgroup$
– S. Senko
Apr 2 at 2:03
1
$begingroup$
E.g., if your field is the reals, and the eigenvalues of $A$ are the nonreal cube roots of 2, then $A^3$ has eigenvalue $2$ which is not $lambda^3$ for some (real) eigenvalue of of $A$.
$endgroup$
– Gerry Myerson
Apr 2 at 2:59
$begingroup$
@Gerry Myerson so that clearly implies that "each element has a single $n$th root" is not sufficient and I imagine that it shouldn't be hard to generalize this to show that it is necessary that each element has a full set of $n$th roots. However, it is still not immediately clear to me why this condition should be sufficient.
$endgroup$
– S. Senko
Apr 2 at 3:27
add a comment |
$begingroup$
Let $A$ be a square matrix over some field $BbbF$. It can be shown that if $lambda$ is an eigenvalue of $A$ then $lambda^n$ is an eigenvalue of $A^n$. In my linear algebra class we were asked if all eigenvalues of $A^n$ can be written in this form.
In general, the answer is that they cannot. For example, if $BbbF = BbbR$ then the matrix
$$A =
beginbmatrix
0 & -1 \
1 & 0 \
endbmatrix$$
has no eigenvalues but $A^2$ has $-1$ as an eigenvalue. In fact, this argument can be generalized to show that if $BbbF$ is not closed under $n$th roots then the result does not hold over that field.
However, if $BbbF$ is algebraically closed and $A$ is as above, then it can actually be shown that all eigenvalues of $A^n$ are of the form $lambda^n$ for some eigenvalue $lambda$ of $A$.
Unfortunately, there are fields which are not algebraically closed but which are closed under $n$th roots. Because of this, I wanted to ask the following question:
For which fields $BbbF$ does it hold that if $A$ is a square matrix over $BbbF$ then all eigenvalues of $A^n$ are of the form $lambda^n$ for some eigenvalue $lambda$ of $A$?
linear-algebra matrices eigenvalues-eigenvectors
asked Apr 2 at 1:48
S. SenkoS. Senko
1335
$endgroup$
Let $A$ be a square matrix over some field $BbbF$. It can be shown that if $lambda$ is an eigenvalue of $A$ then $lambda^n$ is an eigenvalue of $A^n$. In my linear algebra class we were asked if all eigenvalues of $A^n$ can be written in this form.
In general, the answer is that they cannot. For example, if $BbbF = BbbR$ then the matrix
$$A =
beginbmatrix
0 & -1 \
1 & 0 \
endbmatrix$$
has no eigenvalues but $A^2$ has $-1$ as an eigenvalue. In fact, this argument can be generalized to show that if $BbbF$ is not closed under $n$th roots then the result does not hold over that field.
However, if $BbbF$ is algebraically closed and $A$ is as above, then it can actually be shown that all eigenvalues of $A^n$ are of the form $lambda^n$ for some eigenvalue $lambda$ of $A$.
Unfortunately, there are fields which are not algebraically closed but which are closed under $n$th roots. Because of this, I wanted to ask the following question:
For which fields $BbbF$ does it hold that if $A$ is a square matrix over $BbbF$ then all eigenvalues of $A^n$ are of the form $lambda^n$ for some eigenvalue $lambda$ of $A$?
linear-algebra matrices eigenvalues-eigenvectors
linear-algebra matrices eigenvalues-eigenvectors
asked Apr 2 at 1:48
S. SenkoS. Senko
1335
asked Apr 2 at 1:48
S. SenkoS. Senko
1335
asked Apr 2 at 1:48
S. SenkoS. Senko
1335
asked Apr 2 at 1:48
S. SenkoS. Senko
1335
asked Apr 2 at 1:48
S. SenkoS. Senko
1335
1335
1
$begingroup$
Does "closed under $n$th roots" mean "every element has at least one $n$th root" or "every element has a full set of $n$th roots"? Because the latter is enough, but the former is not.
$endgroup$
– darij grinberg
Apr 2 at 1:58
$begingroup$
@darij grinberg To be more specific about my claim, if there is any polynomial of the form $x^n - a$ over $BbbF$ that has no roots then the result does not hold. I don't know about whether or not it holds if $x^n-a$ has at least $1$ but fewer than $n$ roots (and that would in retrospect perhaps be a better definition of closed under $n$th roots)
$endgroup$
– S. Senko
Apr 2 at 2:03
1
$begingroup$
E.g., if your field is the reals, and the eigenvalues of $A$ are the nonreal cube roots of 2, then $A^3$ has eigenvalue $2$ which is not $lambda^3$ for some (real) eigenvalue of of $A$.
$endgroup$
– Gerry Myerson
Apr 2 at 2:59
$begingroup$
@Gerry Myerson so that clearly implies that "each element has a single $n$th root" is not sufficient and I imagine that it shouldn't be hard to generalize this to show that it is necessary that each element has a full set of $n$th roots. However, it is still not immediately clear to me why this condition should be sufficient.
$endgroup$
– S. Senko
Apr 2 at 3:27
add a comment |
1
$begingroup$
Does "closed under $n$th roots" mean "every element has at least one $n$th root" or "every element has a full set of $n$th roots"? Because the latter is enough, but the former is not.
$endgroup$
– darij grinberg
Apr 2 at 1:58
$begingroup$
@darij grinberg To be more specific about my claim, if there is any polynomial of the form $x^n - a$ over $BbbF$ that has no roots then the result does not hold. I don't know about whether or not it holds if $x^n-a$ has at least $1$ but fewer than $n$ roots (and that would in retrospect perhaps be a better definition of closed under $n$th roots)
$endgroup$
– S. Senko
Apr 2 at 2:03
1
$begingroup$
E.g., if your field is the reals, and the eigenvalues of $A$ are the nonreal cube roots of 2, then $A^3$ has eigenvalue $2$ which is not $lambda^3$ for some (real) eigenvalue of of $A$.
$endgroup$
– Gerry Myerson
Apr 2 at 2:59
$begingroup$
@Gerry Myerson so that clearly implies that "each element has a single $n$th root" is not sufficient and I imagine that it shouldn't be hard to generalize this to show that it is necessary that each element has a full set of $n$th roots. However, it is still not immediately clear to me why this condition should be sufficient.
$endgroup$
– S. Senko
Apr 2 at 3:27
1
1
$begingroup$
Does "closed under $n$th roots" mean "every element has at least one $n$th root" or "every element has a full set of $n$th roots"? Because the latter is enough, but the former is not.
$endgroup$
– darij grinberg
Apr 2 at 1:58
$begingroup$
Does "closed under $n$th roots" mean "every element has at least one $n$th root" or "every element has a full set of $n$th roots"? Because the latter is enough, but the former is not.
$endgroup$
– darij grinberg
Apr 2 at 1:58
$begingroup$
@darij grinberg To be more specific about my claim, if there is any polynomial of the form $x^n - a$ over $BbbF$ that has no roots then the result does not hold. I don't know about whether or not it holds if $x^n-a$ has at least $1$ but fewer than $n$ roots (and that would in retrospect perhaps be a better definition of closed under $n$th roots)
$endgroup$
– S. Senko
Apr 2 at 2:03
$begingroup$
@darij grinberg To be more specific about my claim, if there is any polynomial of the form $x^n - a$ over $BbbF$ that has no roots then the result does not hold. I don't know about whether or not it holds if $x^n-a$ has at least $1$ but fewer than $n$ roots (and that would in retrospect perhaps be a better definition of closed under $n$th roots)
$endgroup$
– S. Senko
Apr 2 at 2:03
1
1
$begingroup$
E.g., if your field is the reals, and the eigenvalues of $A$ are the nonreal cube roots of 2, then $A^3$ has eigenvalue $2$ which is not $lambda^3$ for some (real) eigenvalue of of $A$.
$endgroup$
– Gerry Myerson
Apr 2 at 2:59
$begingroup$
E.g., if your field is the reals, and the eigenvalues of $A$ are the nonreal cube roots of 2, then $A^3$ has eigenvalue $2$ which is not $lambda^3$ for some (real) eigenvalue of of $A$.
$endgroup$
– Gerry Myerson
Apr 2 at 2:59
$begingroup$
@Gerry Myerson so that clearly implies that "each element has a single $n$th root" is not sufficient and I imagine that it shouldn't be hard to generalize this to show that it is necessary that each element has a full set of $n$th roots. However, it is still not immediately clear to me why this condition should be sufficient.
$endgroup$
– S. Senko
Apr 2 at 3:27
$begingroup$
@Gerry Myerson so that clearly implies that "each element has a single $n$th root" is not sufficient and I imagine that it shouldn't be hard to generalize this to show that it is necessary that each element has a full set of $n$th roots. However, it is still not immediately clear to me why this condition should be sufficient.
$endgroup$
– S. Senko
Apr 2 at 3:27
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $F$ be a field that contains all the $n$th roots of each of its elements. Let $A$ be a square matrix with entries in $F$. Let $mu$ be an eigenvalue of $A^n$ with corresponding eigenvector $vne0$, so $(A^n-mu I)v=0$. Then we have $$(A-lambda_1I)(A-lambda_2I)cdots(A-lambda_nI)v=0$$ where, for each $i$, $lambda_i^n=mu$. By hypothesis, each $lambda_i$ is in $F$. Then there exists $j$ such that $$(A-lambda_jI)(A-lambda_j+1I)cdots(A-lambda_n)v=0,quad(A-lambda_j+1I)cdots(A-lambda_nI)vne0$$ Then $(A-lambda_jI)w=0$, where $$w=(A-lambda_j+1I)cdots(A-lambda_nI)vne0$$ so $lambda_j$ is an eigenvalue of $A$. That is, $A$ has an eigenvalue $lambda=lambda_j$ such that $lambda^n=mu$.
answered Apr 2 at 5:23
Gerry MyersonGerry Myerson
148k8152306
$endgroup$
add a comment |
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$begingroup$
Let $F$ be a field that contains all the $n$th roots of each of its elements. Let $A$ be a square matrix with entries in $F$. Let $mu$ be an eigenvalue of $A^n$ with corresponding eigenvector $vne0$, so $(A^n-mu I)v=0$. Then we have $$(A-lambda_1I)(A-lambda_2I)cdots(A-lambda_nI)v=0$$ where, for each $i$, $lambda_i^n=mu$. By hypothesis, each $lambda_i$ is in $F$. Then there exists $j$ such that $$(A-lambda_jI)(A-lambda_j+1I)cdots(A-lambda_n)v=0,quad(A-lambda_j+1I)cdots(A-lambda_nI)vne0$$ Then $(A-lambda_jI)w=0$, where $$w=(A-lambda_j+1I)cdots(A-lambda_nI)vne0$$ so $lambda_j$ is an eigenvalue of $A$. That is, $A$ has an eigenvalue $lambda=lambda_j$ such that $lambda^n=mu$.
answered Apr 2 at 5:23
Gerry MyersonGerry Myerson
148k8152306
$endgroup$
add a comment |
$begingroup$
Let $F$ be a field that contains all the $n$th roots of each of its elements. Let $A$ be a square matrix with entries in $F$. Let $mu$ be an eigenvalue of $A^n$ with corresponding eigenvector $vne0$, so $(A^n-mu I)v=0$. Then we have $$(A-lambda_1I)(A-lambda_2I)cdots(A-lambda_nI)v=0$$ where, for each $i$, $lambda_i^n=mu$. By hypothesis, each $lambda_i$ is in $F$. Then there exists $j$ such that $$(A-lambda_jI)(A-lambda_j+1I)cdots(A-lambda_n)v=0,quad(A-lambda_j+1I)cdots(A-lambda_nI)vne0$$ Then $(A-lambda_jI)w=0$, where $$w=(A-lambda_j+1I)cdots(A-lambda_nI)vne0$$ so $lambda_j$ is an eigenvalue of $A$. That is, $A$ has an eigenvalue $lambda=lambda_j$ such that $lambda^n=mu$.
answered Apr 2 at 5:23
Gerry MyersonGerry Myerson
148k8152306
$endgroup$
add a comment |
$begingroup$
Let $F$ be a field that contains all the $n$th roots of each of its elements. Let $A$ be a square matrix with entries in $F$. Let $mu$ be an eigenvalue of $A^n$ with corresponding eigenvector $vne0$, so $(A^n-mu I)v=0$. Then we have $$(A-lambda_1I)(A-lambda_2I)cdots(A-lambda_nI)v=0$$ where, for each $i$, $lambda_i^n=mu$. By hypothesis, each $lambda_i$ is in $F$. Then there exists $j$ such that $$(A-lambda_jI)(A-lambda_j+1I)cdots(A-lambda_n)v=0,quad(A-lambda_j+1I)cdots(A-lambda_nI)vne0$$ Then $(A-lambda_jI)w=0$, where $$w=(A-lambda_j+1I)cdots(A-lambda_nI)vne0$$ so $lambda_j$ is an eigenvalue of $A$. That is, $A$ has an eigenvalue $lambda=lambda_j$ such that $lambda^n=mu$.
answered Apr 2 at 5:23
Gerry MyersonGerry Myerson
148k8152306
$endgroup$
Let $F$ be a field that contains all the $n$th roots of each of its elements. Let $A$ be a square matrix with entries in $F$. Let $mu$ be an eigenvalue of $A^n$ with corresponding eigenvector $vne0$, so $(A^n-mu I)v=0$. Then we have $$(A-lambda_1I)(A-lambda_2I)cdots(A-lambda_nI)v=0$$ where, for each $i$, $lambda_i^n=mu$. By hypothesis, each $lambda_i$ is in $F$. Then there exists $j$ such that $$(A-lambda_jI)(A-lambda_j+1I)cdots(A-lambda_n)v=0,quad(A-lambda_j+1I)cdots(A-lambda_nI)vne0$$ Then $(A-lambda_jI)w=0$, where $$w=(A-lambda_j+1I)cdots(A-lambda_nI)vne0$$ so $lambda_j$ is an eigenvalue of $A$. That is, $A$ has an eigenvalue $lambda=lambda_j$ such that $lambda^n=mu$.
answered Apr 2 at 5:23
Gerry MyersonGerry Myerson
148k8152306
answered Apr 2 at 5:23
Gerry MyersonGerry Myerson
148k8152306
answered Apr 2 at 5:23
Gerry MyersonGerry Myerson
148k8152306
answered Apr 2 at 5:23
Gerry MyersonGerry Myerson
148k8152306
148k8152306
add a comment |
add a comment |
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$begingroup$
Does "closed under $n$th roots" mean "every element has at least one $n$th root" or "every element has a full set of $n$th roots"? Because the latter is enough, but the former is not.
$endgroup$
– darij grinberg
Apr 2 at 1:58
$begingroup$
@darij grinberg To be more specific about my claim, if there is any polynomial of the form $x^n - a$ over $BbbF$ that has no roots then the result does not hold. I don't know about whether or not it holds if $x^n-a$ has at least $1$ but fewer than $n$ roots (and that would in retrospect perhaps be a better definition of closed under $n$th roots)
$endgroup$
– S. Senko
Apr 2 at 2:03
1
$begingroup$
E.g., if your field is the reals, and the eigenvalues of $A$ are the nonreal cube roots of 2, then $A^3$ has eigenvalue $2$ which is not $lambda^3$ for some (real) eigenvalue of of $A$.
$endgroup$
– Gerry Myerson
Apr 2 at 2:59
$begingroup$
@Gerry Myerson so that clearly implies that "each element has a single $n$th root" is not sufficient and I imagine that it shouldn't be hard to generalize this to show that it is necessary that each element has a full set of $n$th roots. However, it is still not immediately clear to me why this condition should be sufficient.
$endgroup$
– S. Senko
Apr 2 at 3:27