Mixing
Characterising minors of diagonal matrices
$begingroup$
Let $k,d$ be positive integers, $1<k<d$. Let $lambda_I=lambda_{i_1,ldots,i_k}$ be real numbers, indexed by multi-indices $I=(i_1,ldots,i_k)$, where $1le i_1<ldots<i_k le d$.
Are there necessary and sufficient conditions on $lambda_{i_1,ldots,i_k}$ which are equivalent to the existence of $sigma_1,ldots,sigma_d in mathbb{R}$ such that $lambda_{i_1,ldots,i_k}=sigma_{i_1}cdot ldotscdotsigma_{i_k}$ holds for every multi-index $I$?
In other words, I am asking whether we can characterise which sequences of real numbers can arise as the $k$-minors of diagonal $d times d$ matrices?
I am interested mainly in the case where all the $lambda_{i_1,ldots,i_k}$ are non-zero.
I have heard that the general problem of recognizing $k$-minors of arbitrary square matrices is open, but I am hoping that for diagonal matrices, the situation maybe better understood.
I guess this should be easier by starting over $mathbb{C}$. What is known about that case?
Commnet: If I understand correctly, the Plucker relations only describe the minors of top-degree of a non-square matrix. Here I am talking about the minors of degree $k$, when $1<k<d$, i.e. non-top minors of a square matrix.
algebraic-geometry polynomials determinant exterior-algebra real-algebraic-geometry
asked Jan 27 at 12:36
Asaf ShacharAsaf Shachar
5,78431145
$endgroup$
add a comment |
$begingroup$
Let $k,d$ be positive integers, $1<k<d$. Let $lambda_I=lambda_{i_1,ldots,i_k}$ be real numbers, indexed by multi-indices $I=(i_1,ldots,i_k)$, where $1le i_1<ldots<i_k le d$.
Are there necessary and sufficient conditions on $lambda_{i_1,ldots,i_k}$ which are equivalent to the existence of $sigma_1,ldots,sigma_d in mathbb{R}$ such that $lambda_{i_1,ldots,i_k}=sigma_{i_1}cdot ldotscdotsigma_{i_k}$ holds for every multi-index $I$?
In other words, I am asking whether we can characterise which sequences of real numbers can arise as the $k$-minors of diagonal $d times d$ matrices?
I am interested mainly in the case where all the $lambda_{i_1,ldots,i_k}$ are non-zero.
I have heard that the general problem of recognizing $k$-minors of arbitrary square matrices is open, but I am hoping that for diagonal matrices, the situation maybe better understood.
I guess this should be easier by starting over $mathbb{C}$. What is known about that case?
Commnet: If I understand correctly, the Plucker relations only describe the minors of top-degree of a non-square matrix. Here I am talking about the minors of degree $k$, when $1<k<d$, i.e. non-top minors of a square matrix.
algebraic-geometry polynomials determinant exterior-algebra real-algebraic-geometry
asked Jan 27 at 12:36
Asaf ShacharAsaf Shachar
5,78431145
$endgroup$
$begingroup$
I guess you mean $lambda_{i_1, dots, i_k} = sigma_{i_1} cdot ... cdot sigma_{i_k}$?
$endgroup$
– Severin Schraven
Jan 27 at 12:51
$begingroup$
Yes, thank you! corrected now.
$endgroup$
– Asaf Shachar
Jan 27 at 13:03
add a comment |
$begingroup$
Let $k,d$ be positive integers, $1<k<d$. Let $lambda_I=lambda_{i_1,ldots,i_k}$ be real numbers, indexed by multi-indices $I=(i_1,ldots,i_k)$, where $1le i_1<ldots<i_k le d$.
Are there necessary and sufficient conditions on $lambda_{i_1,ldots,i_k}$ which are equivalent to the existence of $sigma_1,ldots,sigma_d in mathbb{R}$ such that $lambda_{i_1,ldots,i_k}=sigma_{i_1}cdot ldotscdotsigma_{i_k}$ holds for every multi-index $I$?
In other words, I am asking whether we can characterise which sequences of real numbers can arise as the $k$-minors of diagonal $d times d$ matrices?
I am interested mainly in the case where all the $lambda_{i_1,ldots,i_k}$ are non-zero.
I have heard that the general problem of recognizing $k$-minors of arbitrary square matrices is open, but I am hoping that for diagonal matrices, the situation maybe better understood.
I guess this should be easier by starting over $mathbb{C}$. What is known about that case?
Commnet: If I understand correctly, the Plucker relations only describe the minors of top-degree of a non-square matrix. Here I am talking about the minors of degree $k$, when $1<k<d$, i.e. non-top minors of a square matrix.
algebraic-geometry polynomials determinant exterior-algebra real-algebraic-geometry
asked Jan 27 at 12:36
Asaf ShacharAsaf Shachar
5,78431145
$endgroup$
Let $k,d$ be positive integers, $1<k<d$. Let $lambda_I=lambda_{i_1,ldots,i_k}$ be real numbers, indexed by multi-indices $I=(i_1,ldots,i_k)$, where $1le i_1<ldots<i_k le d$.
Are there necessary and sufficient conditions on $lambda_{i_1,ldots,i_k}$ which are equivalent to the existence of $sigma_1,ldots,sigma_d in mathbb{R}$ such that $lambda_{i_1,ldots,i_k}=sigma_{i_1}cdot ldotscdotsigma_{i_k}$ holds for every multi-index $I$?
In other words, I am asking whether we can characterise which sequences of real numbers can arise as the $k$-minors of diagonal $d times d$ matrices?
I am interested mainly in the case where all the $lambda_{i_1,ldots,i_k}$ are non-zero.
I have heard that the general problem of recognizing $k$-minors of arbitrary square matrices is open, but I am hoping that for diagonal matrices, the situation maybe better understood.
I guess this should be easier by starting over $mathbb{C}$. What is known about that case?
Commnet: If I understand correctly, the Plucker relations only describe the minors of top-degree of a non-square matrix. Here I am talking about the minors of degree $k$, when $1<k<d$, i.e. non-top minors of a square matrix.
algebraic-geometry polynomials determinant exterior-algebra real-algebraic-geometry
algebraic-geometry polynomials determinant exterior-algebra real-algebraic-geometry
asked Jan 27 at 12:36
Asaf ShacharAsaf Shachar
5,78431145
asked Jan 27 at 12:36
Asaf ShacharAsaf Shachar
5,78431145
edited Jan 27 at 13:03
Asaf Shachar
asked Jan 27 at 12:36
Asaf ShacharAsaf Shachar
5,78431145
asked Jan 27 at 12:36
Asaf ShacharAsaf Shachar
5,78431145
asked Jan 27 at 12:36
Asaf ShacharAsaf Shachar
5,78431145
5,78431145
$begingroup$
I guess you mean $lambda_{i_1, dots, i_k} = sigma_{i_1} cdot ... cdot sigma_{i_k}$?
$endgroup$
– Severin Schraven
Jan 27 at 12:51
$begingroup$
Yes, thank you! corrected now.
$endgroup$
– Asaf Shachar
Jan 27 at 13:03
add a comment |
$begingroup$
I guess you mean $lambda_{i_1, dots, i_k} = sigma_{i_1} cdot ... cdot sigma_{i_k}$?
$endgroup$
– Severin Schraven
Jan 27 at 12:51
$begingroup$
Yes, thank you! corrected now.
$endgroup$
– Asaf Shachar
Jan 27 at 13:03
$begingroup$
I guess you mean $lambda_{i_1, dots, i_k} = sigma_{i_1} cdot ... cdot sigma_{i_k}$?
$endgroup$
– Severin Schraven
Jan 27 at 12:51
$begingroup$
I guess you mean $lambda_{i_1, dots, i_k} = sigma_{i_1} cdot ... cdot sigma_{i_k}$?
$endgroup$
– Severin Schraven
Jan 27 at 12:51
$begingroup$
Yes, thank you! corrected now.
$endgroup$
– Asaf Shachar
Jan 27 at 13:03
$begingroup$
Yes, thank you! corrected now.
$endgroup$
– Asaf Shachar
Jan 27 at 13:03
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let me give an answer over complex numbers.
The conditions are the following: for each pair $i ne j$ and for each pair of subsets $I,J subset {1,dots,d} setminus {i,j}$ of cardinality $k - 1$ one has a relation
$$
lambda_{I sqcup {i}} cdot lambda_{J sqcup {j}} =
lambda_{I sqcup {j}} cdot lambda_{J sqcup {i}}.
$$
Clearly these relations are necessary.
Let us also show that they are sufficient. In fact, let me just explain how $sigma_i$ can be reconstructed. Take any set $J subset {1,dots,d-1}$ of cardinality $k$. Then set
$$
sigma_d = sqrt[k]{frac{prod_{j in J} lambda_{J setminus {j} sqcup {d}}}{lambda_J^{k-1}}}.
$$
After that for each $i in {1,dots,d-1}$ choose $J subset {1,dots,d-1} setminus {i}$ of cardinality $k-1$ and set
$$
sigma_i = frac{prod_{j in J} lambda_{J setminus {j} sqcup {i,d}}}{lambda_{J sqcup {i}}^{k-2}sigma_d^{k-1}}.
$$
One can check that this solves the required equations.
Over real numbers the only extra problem is the existence of the root.
answered Jan 27 at 13:05
SashaSasha
5,178139
$endgroup$
$begingroup$
I think we have $k$ fixed.
$endgroup$
– Severin Schraven
Jan 27 at 13:06
$begingroup$
@SeverinSchraven: You are right, I will change the answer.
$endgroup$
– Sasha
Jan 27 at 13:50
$begingroup$
@AsafShachar: Thanks for correction!
$endgroup$
– Sasha
Jan 27 at 14:53
$begingroup$
If $k$ is even, don't we have to be careful with the choice of the sign of $sigma_d$? How do you know that you can pick the positive? Respectively, that you can take the kth root of the expression?
$endgroup$
– Severin Schraven
Jan 27 at 14:57
1
$begingroup$
@SeverinSchraven I will just note that when you work over the complex numbers, you don't really have a problem: Instead of possible non-existence of roots (in the real case), you now have too many of them. However, it does not matter "which one you choose", since choosing a different $k$-th root only amounts to multiplying all the $sigma_i$ by some specific $k$-th root of unity, which is the only ambiguity we have here. (This was to be expected in advance, as I will elaborate in the question).
$endgroup$
– Asaf Shachar
Jan 27 at 15:08
|
show 2 more comments
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1 Answer
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1 Answer
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oldest
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oldest
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oldest
votes
$begingroup$
Let me give an answer over complex numbers.
The conditions are the following: for each pair $i ne j$ and for each pair of subsets $I,J subset {1,dots,d} setminus {i,j}$ of cardinality $k - 1$ one has a relation
$$
lambda_{I sqcup {i}} cdot lambda_{J sqcup {j}} =
lambda_{I sqcup {j}} cdot lambda_{J sqcup {i}}.
$$
Clearly these relations are necessary.
Let us also show that they are sufficient. In fact, let me just explain how $sigma_i$ can be reconstructed. Take any set $J subset {1,dots,d-1}$ of cardinality $k$. Then set
$$
sigma_d = sqrt[k]{frac{prod_{j in J} lambda_{J setminus {j} sqcup {d}}}{lambda_J^{k-1}}}.
$$
After that for each $i in {1,dots,d-1}$ choose $J subset {1,dots,d-1} setminus {i}$ of cardinality $k-1$ and set
$$
sigma_i = frac{prod_{j in J} lambda_{J setminus {j} sqcup {i,d}}}{lambda_{J sqcup {i}}^{k-2}sigma_d^{k-1}}.
$$
One can check that this solves the required equations.
Over real numbers the only extra problem is the existence of the root.
answered Jan 27 at 13:05
SashaSasha
5,178139
$endgroup$
$begingroup$
I think we have $k$ fixed.
$endgroup$
– Severin Schraven
Jan 27 at 13:06
$begingroup$
@SeverinSchraven: You are right, I will change the answer.
$endgroup$
– Sasha
Jan 27 at 13:50
$begingroup$
@AsafShachar: Thanks for correction!
$endgroup$
– Sasha
Jan 27 at 14:53
$begingroup$
If $k$ is even, don't we have to be careful with the choice of the sign of $sigma_d$? How do you know that you can pick the positive? Respectively, that you can take the kth root of the expression?
$endgroup$
– Severin Schraven
Jan 27 at 14:57
1
$begingroup$
@SeverinSchraven I will just note that when you work over the complex numbers, you don't really have a problem: Instead of possible non-existence of roots (in the real case), you now have too many of them. However, it does not matter "which one you choose", since choosing a different $k$-th root only amounts to multiplying all the $sigma_i$ by some specific $k$-th root of unity, which is the only ambiguity we have here. (This was to be expected in advance, as I will elaborate in the question).
$endgroup$
– Asaf Shachar
Jan 27 at 15:08
|
show 2 more comments
$begingroup$
Let me give an answer over complex numbers.
The conditions are the following: for each pair $i ne j$ and for each pair of subsets $I,J subset {1,dots,d} setminus {i,j}$ of cardinality $k - 1$ one has a relation
$$
lambda_{I sqcup {i}} cdot lambda_{J sqcup {j}} =
lambda_{I sqcup {j}} cdot lambda_{J sqcup {i}}.
$$
Clearly these relations are necessary.
Let us also show that they are sufficient. In fact, let me just explain how $sigma_i$ can be reconstructed. Take any set $J subset {1,dots,d-1}$ of cardinality $k$. Then set
$$
sigma_d = sqrt[k]{frac{prod_{j in J} lambda_{J setminus {j} sqcup {d}}}{lambda_J^{k-1}}}.
$$
After that for each $i in {1,dots,d-1}$ choose $J subset {1,dots,d-1} setminus {i}$ of cardinality $k-1$ and set
$$
sigma_i = frac{prod_{j in J} lambda_{J setminus {j} sqcup {i,d}}}{lambda_{J sqcup {i}}^{k-2}sigma_d^{k-1}}.
$$
One can check that this solves the required equations.
Over real numbers the only extra problem is the existence of the root.
answered Jan 27 at 13:05
SashaSasha
5,178139
$endgroup$
$begingroup$
I think we have $k$ fixed.
$endgroup$
– Severin Schraven
Jan 27 at 13:06
$begingroup$
@SeverinSchraven: You are right, I will change the answer.
$endgroup$
– Sasha
Jan 27 at 13:50
$begingroup$
@AsafShachar: Thanks for correction!
$endgroup$
– Sasha
Jan 27 at 14:53
$begingroup$
If $k$ is even, don't we have to be careful with the choice of the sign of $sigma_d$? How do you know that you can pick the positive? Respectively, that you can take the kth root of the expression?
$endgroup$
– Severin Schraven
Jan 27 at 14:57
1
$begingroup$
@SeverinSchraven I will just note that when you work over the complex numbers, you don't really have a problem: Instead of possible non-existence of roots (in the real case), you now have too many of them. However, it does not matter "which one you choose", since choosing a different $k$-th root only amounts to multiplying all the $sigma_i$ by some specific $k$-th root of unity, which is the only ambiguity we have here. (This was to be expected in advance, as I will elaborate in the question).
$endgroup$
– Asaf Shachar
Jan 27 at 15:08
|
show 2 more comments
$begingroup$
Let me give an answer over complex numbers.
The conditions are the following: for each pair $i ne j$ and for each pair of subsets $I,J subset {1,dots,d} setminus {i,j}$ of cardinality $k - 1$ one has a relation
$$
lambda_{I sqcup {i}} cdot lambda_{J sqcup {j}} =
lambda_{I sqcup {j}} cdot lambda_{J sqcup {i}}.
$$
Clearly these relations are necessary.
Let us also show that they are sufficient. In fact, let me just explain how $sigma_i$ can be reconstructed. Take any set $J subset {1,dots,d-1}$ of cardinality $k$. Then set
$$
sigma_d = sqrt[k]{frac{prod_{j in J} lambda_{J setminus {j} sqcup {d}}}{lambda_J^{k-1}}}.
$$
After that for each $i in {1,dots,d-1}$ choose $J subset {1,dots,d-1} setminus {i}$ of cardinality $k-1$ and set
$$
sigma_i = frac{prod_{j in J} lambda_{J setminus {j} sqcup {i,d}}}{lambda_{J sqcup {i}}^{k-2}sigma_d^{k-1}}.
$$
One can check that this solves the required equations.
Over real numbers the only extra problem is the existence of the root.
answered Jan 27 at 13:05
SashaSasha
5,178139
$endgroup$
Let me give an answer over complex numbers.
The conditions are the following: for each pair $i ne j$ and for each pair of subsets $I,J subset {1,dots,d} setminus {i,j}$ of cardinality $k - 1$ one has a relation
$$
lambda_{I sqcup {i}} cdot lambda_{J sqcup {j}} =
lambda_{I sqcup {j}} cdot lambda_{J sqcup {i}}.
$$
Clearly these relations are necessary.
Let us also show that they are sufficient. In fact, let me just explain how $sigma_i$ can be reconstructed. Take any set $J subset {1,dots,d-1}$ of cardinality $k$. Then set
$$
sigma_d = sqrt[k]{frac{prod_{j in J} lambda_{J setminus {j} sqcup {d}}}{lambda_J^{k-1}}}.
$$
After that for each $i in {1,dots,d-1}$ choose $J subset {1,dots,d-1} setminus {i}$ of cardinality $k-1$ and set
$$
sigma_i = frac{prod_{j in J} lambda_{J setminus {j} sqcup {i,d}}}{lambda_{J sqcup {i}}^{k-2}sigma_d^{k-1}}.
$$
One can check that this solves the required equations.
Over real numbers the only extra problem is the existence of the root.
answered Jan 27 at 13:05
SashaSasha
5,178139
edited Jan 27 at 14:53
answered Jan 27 at 13:05
SashaSasha
5,178139
answered Jan 27 at 13:05
SashaSasha
5,178139
answered Jan 27 at 13:05
SashaSasha
5,178139
5,178139
$begingroup$
I think we have $k$ fixed.
$endgroup$
– Severin Schraven
Jan 27 at 13:06
$begingroup$
@SeverinSchraven: You are right, I will change the answer.
$endgroup$
– Sasha
Jan 27 at 13:50
$begingroup$
@AsafShachar: Thanks for correction!
$endgroup$
– Sasha
Jan 27 at 14:53
$begingroup$
If $k$ is even, don't we have to be careful with the choice of the sign of $sigma_d$? How do you know that you can pick the positive? Respectively, that you can take the kth root of the expression?
$endgroup$
– Severin Schraven
Jan 27 at 14:57
1
$begingroup$
@SeverinSchraven I will just note that when you work over the complex numbers, you don't really have a problem: Instead of possible non-existence of roots (in the real case), you now have too many of them. However, it does not matter "which one you choose", since choosing a different $k$-th root only amounts to multiplying all the $sigma_i$ by some specific $k$-th root of unity, which is the only ambiguity we have here. (This was to be expected in advance, as I will elaborate in the question).
$endgroup$
– Asaf Shachar
Jan 27 at 15:08
|
show 2 more comments
$begingroup$
I think we have $k$ fixed.
$endgroup$
– Severin Schraven
Jan 27 at 13:06
$begingroup$
@SeverinSchraven: You are right, I will change the answer.
$endgroup$
– Sasha
Jan 27 at 13:50
$begingroup$
@AsafShachar: Thanks for correction!
$endgroup$
– Sasha
Jan 27 at 14:53
$begingroup$
If $k$ is even, don't we have to be careful with the choice of the sign of $sigma_d$? How do you know that you can pick the positive? Respectively, that you can take the kth root of the expression?
$endgroup$
– Severin Schraven
Jan 27 at 14:57
1
$begingroup$
@SeverinSchraven I will just note that when you work over the complex numbers, you don't really have a problem: Instead of possible non-existence of roots (in the real case), you now have too many of them. However, it does not matter "which one you choose", since choosing a different $k$-th root only amounts to multiplying all the $sigma_i$ by some specific $k$-th root of unity, which is the only ambiguity we have here. (This was to be expected in advance, as I will elaborate in the question).
$endgroup$
– Asaf Shachar
Jan 27 at 15:08
$begingroup$
I think we have $k$ fixed.
$endgroup$
– Severin Schraven
Jan 27 at 13:06
$begingroup$
I think we have $k$ fixed.
$endgroup$
– Severin Schraven
Jan 27 at 13:06
$begingroup$
@SeverinSchraven: You are right, I will change the answer.
$endgroup$
– Sasha
Jan 27 at 13:50
$begingroup$
@SeverinSchraven: You are right, I will change the answer.
$endgroup$
– Sasha
Jan 27 at 13:50
$begingroup$
@AsafShachar: Thanks for correction!
$endgroup$
– Sasha
Jan 27 at 14:53
$begingroup$
@AsafShachar: Thanks for correction!
$endgroup$
– Sasha
Jan 27 at 14:53
$begingroup$
If $k$ is even, don't we have to be careful with the choice of the sign of $sigma_d$? How do you know that you can pick the positive? Respectively, that you can take the kth root of the expression?
$endgroup$
– Severin Schraven
Jan 27 at 14:57
$begingroup$
If $k$ is even, don't we have to be careful with the choice of the sign of $sigma_d$? How do you know that you can pick the positive? Respectively, that you can take the kth root of the expression?
$endgroup$
– Severin Schraven
Jan 27 at 14:57
1
1
$begingroup$
@SeverinSchraven I will just note that when you work over the complex numbers, you don't really have a problem: Instead of possible non-existence of roots (in the real case), you now have too many of them. However, it does not matter "which one you choose", since choosing a different $k$-th root only amounts to multiplying all the $sigma_i$ by some specific $k$-th root of unity, which is the only ambiguity we have here. (This was to be expected in advance, as I will elaborate in the question).
$endgroup$
– Asaf Shachar
Jan 27 at 15:08
$begingroup$
@SeverinSchraven I will just note that when you work over the complex numbers, you don't really have a problem: Instead of possible non-existence of roots (in the real case), you now have too many of them. However, it does not matter "which one you choose", since choosing a different $k$-th root only amounts to multiplying all the $sigma_i$ by some specific $k$-th root of unity, which is the only ambiguity we have here. (This was to be expected in advance, as I will elaborate in the question).
$endgroup$
– Asaf Shachar
Jan 27 at 15:08
|
show 2 more comments
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I guess you mean $lambda_{i_1, dots, i_k} = sigma_{i_1} cdot ... cdot sigma_{i_k}$?
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– Severin Schraven
Jan 27 at 12:51
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Yes, thank you! corrected now.
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– Asaf Shachar
Jan 27 at 13:03