Factorising vector spaces
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Suppose I have some complex (unit) vector $V$ on a vector space $mathcal{H}$ of dimension $N$, where $N$ has prime factorisation $N=prod^d_i p_i$ (multiplicity is allowed- the $p_i$ need not be unique). $mathcal{H}$ is spanned by a certain choice of orthnormal basis ${mathbb{e}_i}$. How can one show whether there exists some factorisation $mathcal{H}=mathcal{h}_1otimesmathcal{h}_2otimes...otimesmathcal{h}_d$, where $h_i$ is a vector space of dimension $p_i$, such that $V$ and all ${mathbb{e}_i}$ are separable?
Alternative perspective: I have a set of $N$ complex numbers ${z_1, z_2,...,z_N}$. How does one determine whether this set is fully factorisable- that is, there exist sets of numbers ${a_1,...a_{p_1}}, {b_1,...b_{p_2}},...,{x_1,...x_{p_d}}$ such that each element $z_i$ can be expressed as $z_i=a_{k_1}b_{k_2}...x_{k_d}$ (for a unique set of indices ${k_1,k_2,...,k_d}$ for each $z_i$)?
vector-spaces hilbert-spaces several-complex-variables
asked Nov 27 '18 at 14:52
Alex DavisAlex Davis
11
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add a comment |
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Suppose I have some complex (unit) vector $V$ on a vector space $mathcal{H}$ of dimension $N$, where $N$ has prime factorisation $N=prod^d_i p_i$ (multiplicity is allowed- the $p_i$ need not be unique). $mathcal{H}$ is spanned by a certain choice of orthnormal basis ${mathbb{e}_i}$. How can one show whether there exists some factorisation $mathcal{H}=mathcal{h}_1otimesmathcal{h}_2otimes...otimesmathcal{h}_d$, where $h_i$ is a vector space of dimension $p_i$, such that $V$ and all ${mathbb{e}_i}$ are separable?
Alternative perspective: I have a set of $N$ complex numbers ${z_1, z_2,...,z_N}$. How does one determine whether this set is fully factorisable- that is, there exist sets of numbers ${a_1,...a_{p_1}}, {b_1,...b_{p_2}},...,{x_1,...x_{p_d}}$ such that each element $z_i$ can be expressed as $z_i=a_{k_1}b_{k_2}...x_{k_d}$ (for a unique set of indices ${k_1,k_2,...,k_d}$ for each $z_i$)?
vector-spaces hilbert-spaces several-complex-variables
asked Nov 27 '18 at 14:52
Alex DavisAlex Davis
11
$endgroup$
$begingroup$
How do you mean that a set of vectors is separable? And what does it have to do with the tensor composite of $mathcal H$?
$endgroup$
– Berci
Nov 27 '18 at 16:17
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A vector in $mathcal{H}$ is separable if it can be expressed as a single outer product of vectors that each lie in the factor spaces $h_i$. So for example, if $d=2$ then this question boils down to asking how one partitions $mathcal{H}$ into two factor spaces $h_1$ and $h_2$ such that the SVD of $V$ has only one nonzero singular value.
$endgroup$
– Alex Davis
Nov 27 '18 at 18:00
add a comment |
$begingroup$
Suppose I have some complex (unit) vector $V$ on a vector space $mathcal{H}$ of dimension $N$, where $N$ has prime factorisation $N=prod^d_i p_i$ (multiplicity is allowed- the $p_i$ need not be unique). $mathcal{H}$ is spanned by a certain choice of orthnormal basis ${mathbb{e}_i}$. How can one show whether there exists some factorisation $mathcal{H}=mathcal{h}_1otimesmathcal{h}_2otimes...otimesmathcal{h}_d$, where $h_i$ is a vector space of dimension $p_i$, such that $V$ and all ${mathbb{e}_i}$ are separable?
Alternative perspective: I have a set of $N$ complex numbers ${z_1, z_2,...,z_N}$. How does one determine whether this set is fully factorisable- that is, there exist sets of numbers ${a_1,...a_{p_1}}, {b_1,...b_{p_2}},...,{x_1,...x_{p_d}}$ such that each element $z_i$ can be expressed as $z_i=a_{k_1}b_{k_2}...x_{k_d}$ (for a unique set of indices ${k_1,k_2,...,k_d}$ for each $z_i$)?
vector-spaces hilbert-spaces several-complex-variables
asked Nov 27 '18 at 14:52
Alex DavisAlex Davis
11
$endgroup$
Suppose I have some complex (unit) vector $V$ on a vector space $mathcal{H}$ of dimension $N$, where $N$ has prime factorisation $N=prod^d_i p_i$ (multiplicity is allowed- the $p_i$ need not be unique). $mathcal{H}$ is spanned by a certain choice of orthnormal basis ${mathbb{e}_i}$. How can one show whether there exists some factorisation $mathcal{H}=mathcal{h}_1otimesmathcal{h}_2otimes...otimesmathcal{h}_d$, where $h_i$ is a vector space of dimension $p_i$, such that $V$ and all ${mathbb{e}_i}$ are separable?
Alternative perspective: I have a set of $N$ complex numbers ${z_1, z_2,...,z_N}$. How does one determine whether this set is fully factorisable- that is, there exist sets of numbers ${a_1,...a_{p_1}}, {b_1,...b_{p_2}},...,{x_1,...x_{p_d}}$ such that each element $z_i$ can be expressed as $z_i=a_{k_1}b_{k_2}...x_{k_d}$ (for a unique set of indices ${k_1,k_2,...,k_d}$ for each $z_i$)?
vector-spaces hilbert-spaces several-complex-variables
vector-spaces hilbert-spaces several-complex-variables
asked Nov 27 '18 at 14:52
Alex DavisAlex Davis
11
asked Nov 27 '18 at 14:52
Alex DavisAlex Davis
11
edited Nov 27 '18 at 15:20
Alex Davis
asked Nov 27 '18 at 14:52
Alex DavisAlex Davis
11
asked Nov 27 '18 at 14:52
Alex DavisAlex Davis
11
asked Nov 27 '18 at 14:52
Alex DavisAlex Davis
11
11
$begingroup$
How do you mean that a set of vectors is separable? And what does it have to do with the tensor composite of $mathcal H$?
$endgroup$
– Berci
Nov 27 '18 at 16:17
$begingroup$
A vector in $mathcal{H}$ is separable if it can be expressed as a single outer product of vectors that each lie in the factor spaces $h_i$. So for example, if $d=2$ then this question boils down to asking how one partitions $mathcal{H}$ into two factor spaces $h_1$ and $h_2$ such that the SVD of $V$ has only one nonzero singular value.
$endgroup$
– Alex Davis
Nov 27 '18 at 18:00
add a comment |
$begingroup$
How do you mean that a set of vectors is separable? And what does it have to do with the tensor composite of $mathcal H$?
$endgroup$
– Berci
Nov 27 '18 at 16:17
$begingroup$
A vector in $mathcal{H}$ is separable if it can be expressed as a single outer product of vectors that each lie in the factor spaces $h_i$. So for example, if $d=2$ then this question boils down to asking how one partitions $mathcal{H}$ into two factor spaces $h_1$ and $h_2$ such that the SVD of $V$ has only one nonzero singular value.
$endgroup$
– Alex Davis
Nov 27 '18 at 18:00
$begingroup$
How do you mean that a set of vectors is separable? And what does it have to do with the tensor composite of $mathcal H$?
$endgroup$
– Berci
Nov 27 '18 at 16:17
$begingroup$
How do you mean that a set of vectors is separable? And what does it have to do with the tensor composite of $mathcal H$?
$endgroup$
– Berci
Nov 27 '18 at 16:17
$begingroup$
A vector in $mathcal{H}$ is separable if it can be expressed as a single outer product of vectors that each lie in the factor spaces $h_i$. So for example, if $d=2$ then this question boils down to asking how one partitions $mathcal{H}$ into two factor spaces $h_1$ and $h_2$ such that the SVD of $V$ has only one nonzero singular value.
$endgroup$
– Alex Davis
Nov 27 '18 at 18:00
$begingroup$
A vector in $mathcal{H}$ is separable if it can be expressed as a single outer product of vectors that each lie in the factor spaces $h_i$. So for example, if $d=2$ then this question boils down to asking how one partitions $mathcal{H}$ into two factor spaces $h_1$ and $h_2$ such that the SVD of $V$ has only one nonzero singular value.
$endgroup$
– Alex Davis
Nov 27 '18 at 18:00
add a comment |
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$begingroup$
How do you mean that a set of vectors is separable? And what does it have to do with the tensor composite of $mathcal H$?
$endgroup$
– Berci
Nov 27 '18 at 16:17
$begingroup$
A vector in $mathcal{H}$ is separable if it can be expressed as a single outer product of vectors that each lie in the factor spaces $h_i$. So for example, if $d=2$ then this question boils down to asking how one partitions $mathcal{H}$ into two factor spaces $h_1$ and $h_2$ such that the SVD of $V$ has only one nonzero singular value.
$endgroup$
– Alex Davis
Nov 27 '18 at 18:00