Tensor product of linear maps

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Let $phi:Urightarrow X$ and $psi:Vrightarrow Y$ be linear maps. Then we have $phiotimes psi:Uotimes Vrightarrow Xotimes Y$. The following (1 to 4) are all clear:

1. $kerleft(phiright)otimeskerleft(psiright)leqkerleft(phiotimespsiright)$


2. $textrmimleft(phiright)otimestextrmimleft(psiright)leqtextrmimleft(phiotimespsiright)$

and if $U=X$ and $V=Y$

3. $E_lambdaleft(phiright)otimes E_muleft(psiright)leq E_lambdamuleft(phiotimespsiright)$

and if $dimleft(Uright)=n$ and $dimleft(Vright)=m$

4. $chi_phiotimespsi=chi_phi^mchi_psi^n
   $

Where $E_lambdaleft(phiright)=kerleft(phi-lambda Iright)$ is the eigenspace, and $chi_phileft(xright)=detleft(phi-x Iright)$ is the characteristic polynomial.

The question then is which if any of these ought to hold in some reverse form. For instance, do we have $$kerleft(phiright)otimeskerleft(psiright)=kerleft(phiotimespsiright)$$ or $$textrmimleft(phiright)otimestextrmimleft(psiright)=textrmimleft(phiotimespsiright)$$

Futhermore, for the minimal polynomials $m_phiotimespsi$, $m_phi$ and $m_psi$, is there a relation between $m_phiotimespsi$ and $m_phi^mm_psi^n$ like we have for the characteristic polynomials? Essentially, do the properties of the constituent maps entirely determine the properties for the tensor-product map?

linear-algebra linear-transformations tensor-products

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asked Jul 16 at 17:26

Joshua Tilley

460212

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Let $phi:Urightarrow X$ and $psi:Vrightarrow Y$ be linear maps. Then we have $phiotimes psi:Uotimes Vrightarrow Xotimes Y$. The following (1 to 4) are all clear:

1. $kerleft(phiright)otimeskerleft(psiright)leqkerleft(phiotimespsiright)$


2. $textrmimleft(phiright)otimestextrmimleft(psiright)leqtextrmimleft(phiotimespsiright)$

and if $U=X$ and $V=Y$

3. $E_lambdaleft(phiright)otimes E_muleft(psiright)leq E_lambdamuleft(phiotimespsiright)$

and if $dimleft(Uright)=n$ and $dimleft(Vright)=m$

4. $chi_phiotimespsi=chi_phi^mchi_psi^n
   $

Where $E_lambdaleft(phiright)=kerleft(phi-lambda Iright)$ is the eigenspace, and $chi_phileft(xright)=detleft(phi-x Iright)$ is the characteristic polynomial.

The question then is which if any of these ought to hold in some reverse form. For instance, do we have $$kerleft(phiright)otimeskerleft(psiright)=kerleft(phiotimespsiright)$$ or $$textrmimleft(phiright)otimestextrmimleft(psiright)=textrmimleft(phiotimespsiright)$$

Futhermore, for the minimal polynomials $m_phiotimespsi$, $m_phi$ and $m_psi$, is there a relation between $m_phiotimespsi$ and $m_phi^mm_psi^n$ like we have for the characteristic polynomials? Essentially, do the properties of the constituent maps entirely determine the properties for the tensor-product map?

linear-algebra linear-transformations tensor-products

share|cite|improve this question

asked Jul 16 at 17:26

Joshua Tilley

460212

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Let $phi:Urightarrow X$ and $psi:Vrightarrow Y$ be linear maps. Then we have $phiotimes psi:Uotimes Vrightarrow Xotimes Y$. The following (1 to 4) are all clear:

1. $kerleft(phiright)otimeskerleft(psiright)leqkerleft(phiotimespsiright)$


2. $textrmimleft(phiright)otimestextrmimleft(psiright)leqtextrmimleft(phiotimespsiright)$

and if $U=X$ and $V=Y$

3. $E_lambdaleft(phiright)otimes E_muleft(psiright)leq E_lambdamuleft(phiotimespsiright)$

and if $dimleft(Uright)=n$ and $dimleft(Vright)=m$

4. $chi_phiotimespsi=chi_phi^mchi_psi^n
   $

Where $E_lambdaleft(phiright)=kerleft(phi-lambda Iright)$ is the eigenspace, and $chi_phileft(xright)=detleft(phi-x Iright)$ is the characteristic polynomial.

The question then is which if any of these ought to hold in some reverse form. For instance, do we have $$kerleft(phiright)otimeskerleft(psiright)=kerleft(phiotimespsiright)$$ or $$textrmimleft(phiright)otimestextrmimleft(psiright)=textrmimleft(phiotimespsiright)$$

Futhermore, for the minimal polynomials $m_phiotimespsi$, $m_phi$ and $m_psi$, is there a relation between $m_phiotimespsi$ and $m_phi^mm_psi^n$ like we have for the characteristic polynomials? Essentially, do the properties of the constituent maps entirely determine the properties for the tensor-product map?

linear-algebra linear-transformations tensor-products

share|cite|improve this question

asked Jul 16 at 17:26

Joshua Tilley

460212

Let $phi:Urightarrow X$ and $psi:Vrightarrow Y$ be linear maps. Then we have $phiotimes psi:Uotimes Vrightarrow Xotimes Y$. The following (1 to 4) are all clear:

1. $kerleft(phiright)otimeskerleft(psiright)leqkerleft(phiotimespsiright)$


2. $textrmimleft(phiright)otimestextrmimleft(psiright)leqtextrmimleft(phiotimespsiright)$

and if $U=X$ and $V=Y$

3. $E_lambdaleft(phiright)otimes E_muleft(psiright)leq E_lambdamuleft(phiotimespsiright)$

and if $dimleft(Uright)=n$ and $dimleft(Vright)=m$

4. $chi_phiotimespsi=chi_phi^mchi_psi^n
   $

Where $E_lambdaleft(phiright)=kerleft(phi-lambda Iright)$ is the eigenspace, and $chi_phileft(xright)=detleft(phi-x Iright)$ is the characteristic polynomial.

The question then is which if any of these ought to hold in some reverse form. For instance, do we have $$kerleft(phiright)otimeskerleft(psiright)=kerleft(phiotimespsiright)$$ or $$textrmimleft(phiright)otimestextrmimleft(psiright)=textrmimleft(phiotimespsiright)$$

Futhermore, for the minimal polynomials $m_phiotimespsi$, $m_phi$ and $m_psi$, is there a relation between $m_phiotimespsi$ and $m_phi^mm_psi^n$ like we have for the characteristic polynomials? Essentially, do the properties of the constituent maps entirely determine the properties for the tensor-product map?

linear-algebra linear-transformations tensor-products

share|cite|improve this question

asked Jul 16 at 17:26

Joshua Tilley

460212

share|cite|improve this question

share|cite|improve this question

asked Jul 16 at 17:26

Joshua Tilley

460212

asked Jul 16 at 17:26

Joshua Tilley

460212

asked Jul 16 at 17:26

Joshua Tilley

460212

460212

add a comment | 

add a comment | 

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In general, you won't have the first of these, since
$$kerphiotimes V+Uotimeskerpsisubseteqker(phiotimespsi).$$

share|cite|improve this answer

answered Jul 16 at 17:28

Lord Shark the Unknown

85.6k951112

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good point. Do you think that that might be an equality? You can show that $ker left(phi otimes 1right)=ker left(phi right)otimes V$, and since $phi otimes psi = phi otimes 1 circ 1 otimes psi$ does it follow?
  – Joshua Tilley
  Jul 17 at 14:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JoshuaTilley I'm sure this is an equality, certainly for finite-dimensional spaces.
  – Lord Shark the Unknown
  Jul 17 at 15:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You can prove both that $ker left(phi otimes psi right)=ker left(phi right)otimes V+Uotimes ker left(psi right)$ and $textrmimleft(phiright)otimestextrmimleft(psiright)=textrmimleft(phiotimespsiright)$ by rank nullity. Given the inclusion you noted, and the inclusion for the image that I noted, you get lower bounds for the rank and nullity of the tensor product map, but they add up to the dimension of $Uotimes V$, and hence are equalities.
  – Joshua Tilley
  Jul 17 at 16:20
  
  

  
  
  
  

add a comment | 

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1 Answer
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active

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votes

1 Answer
1

active

oldest

votes

active

oldest

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active

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votes

up vote
0
down vote

In general, you won't have the first of these, since
$$kerphiotimes V+Uotimeskerpsisubseteqker(phiotimespsi).$$

share|cite|improve this answer

answered Jul 16 at 17:28

Lord Shark the Unknown

85.6k951112

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good point. Do you think that that might be an equality? You can show that $ker left(phi otimes 1right)=ker left(phi right)otimes V$, and since $phi otimes psi = phi otimes 1 circ 1 otimes psi$ does it follow?
  – Joshua Tilley
  Jul 17 at 14:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JoshuaTilley I'm sure this is an equality, certainly for finite-dimensional spaces.
  – Lord Shark the Unknown
  Jul 17 at 15:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You can prove both that $ker left(phi otimes psi right)=ker left(phi right)otimes V+Uotimes ker left(psi right)$ and $textrmimleft(phiright)otimestextrmimleft(psiright)=textrmimleft(phiotimespsiright)$ by rank nullity. Given the inclusion you noted, and the inclusion for the image that I noted, you get lower bounds for the rank and nullity of the tensor product map, but they add up to the dimension of $Uotimes V$, and hence are equalities.
  – Joshua Tilley
  Jul 17 at 16:20
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

In general, you won't have the first of these, since
$$kerphiotimes V+Uotimeskerpsisubseteqker(phiotimespsi).$$

share|cite|improve this answer

answered Jul 16 at 17:28

Lord Shark the Unknown

85.6k951112

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good point. Do you think that that might be an equality? You can show that $ker left(phi otimes 1right)=ker left(phi right)otimes V$, and since $phi otimes psi = phi otimes 1 circ 1 otimes psi$ does it follow?
  – Joshua Tilley
  Jul 17 at 14:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JoshuaTilley I'm sure this is an equality, certainly for finite-dimensional spaces.
  – Lord Shark the Unknown
  Jul 17 at 15:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You can prove both that $ker left(phi otimes psi right)=ker left(phi right)otimes V+Uotimes ker left(psi right)$ and $textrmimleft(phiright)otimestextrmimleft(psiright)=textrmimleft(phiotimespsiright)$ by rank nullity. Given the inclusion you noted, and the inclusion for the image that I noted, you get lower bounds for the rank and nullity of the tensor product map, but they add up to the dimension of $Uotimes V$, and hence are equalities.
  – Joshua Tilley
  Jul 17 at 16:20
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

In general, you won't have the first of these, since
$$kerphiotimes V+Uotimeskerpsisubseteqker(phiotimespsi).$$

share|cite|improve this answer

answered Jul 16 at 17:28

Lord Shark the Unknown

85.6k951112

In general, you won't have the first of these, since
$$kerphiotimes V+Uotimeskerpsisubseteqker(phiotimespsi).$$

share|cite|improve this answer

answered Jul 16 at 17:28

Lord Shark the Unknown

85.6k951112

share|cite|improve this answer

share|cite|improve this answer

answered Jul 16 at 17:28

Lord Shark the Unknown

85.6k951112

answered Jul 16 at 17:28

Lord Shark the Unknown

85.6k951112

answered Jul 16 at 17:28

Lord Shark the Unknown

85.6k951112

85.6k951112

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good point. Do you think that that might be an equality? You can show that $ker left(phi otimes 1right)=ker left(phi right)otimes V$, and since $phi otimes psi = phi otimes 1 circ 1 otimes psi$ does it follow?
  – Joshua Tilley
  Jul 17 at 14:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JoshuaTilley I'm sure this is an equality, certainly for finite-dimensional spaces.
  – Lord Shark the Unknown
  Jul 17 at 15:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You can prove both that $ker left(phi otimes psi right)=ker left(phi right)otimes V+Uotimes ker left(psi right)$ and $textrmimleft(phiright)otimestextrmimleft(psiright)=textrmimleft(phiotimespsiright)$ by rank nullity. Given the inclusion you noted, and the inclusion for the image that I noted, you get lower bounds for the rank and nullity of the tensor product map, but they add up to the dimension of $Uotimes V$, and hence are equalities.
  – Joshua Tilley
  Jul 17 at 16:20
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good point. Do you think that that might be an equality? You can show that $ker left(phi otimes 1right)=ker left(phi right)otimes V$, and since $phi otimes psi = phi otimes 1 circ 1 otimes psi$ does it follow?
  – Joshua Tilley
  Jul 17 at 14:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JoshuaTilley I'm sure this is an equality, certainly for finite-dimensional spaces.
  – Lord Shark the Unknown
  Jul 17 at 15:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You can prove both that $ker left(phi otimes psi right)=ker left(phi right)otimes V+Uotimes ker left(psi right)$ and $textrmimleft(phiright)otimestextrmimleft(psiright)=textrmimleft(phiotimespsiright)$ by rank nullity. Given the inclusion you noted, and the inclusion for the image that I noted, you get lower bounds for the rank and nullity of the tensor product map, but they add up to the dimension of $Uotimes V$, and hence are equalities.
  – Joshua Tilley
  Jul 17 at 16:20
  
  

  
  
  
  

Good point. Do you think that that might be an equality? You can show that $ker left(phi otimes 1right)=ker left(phi right)otimes V$, and since $phi otimes psi = phi otimes 1 circ 1 otimes psi$ does it follow?
– Joshua Tilley
Jul 17 at 14:32

Good point. Do you think that that might be an equality? You can show that $ker left(phi otimes 1right)=ker left(phi right)otimes V$, and since $phi otimes psi = phi otimes 1 circ 1 otimes psi$ does it follow?
– Joshua Tilley
Jul 17 at 14:32

@JoshuaTilley I'm sure this is an equality, certainly for finite-dimensional spaces.
– Lord Shark the Unknown
Jul 17 at 15:14

@JoshuaTilley I'm sure this is an equality, certainly for finite-dimensional spaces.
– Lord Shark the Unknown
Jul 17 at 15:14

You can prove both that $ker left(phi otimes psi right)=ker left(phi right)otimes V+Uotimes ker left(psi right)$ and $textrmimleft(phiright)otimestextrmimleft(psiright)=textrmimleft(phiotimespsiright)$ by rank nullity. Given the inclusion you noted, and the inclusion for the image that I noted, you get lower bounds for the rank and nullity of the tensor product map, but they add up to the dimension of $Uotimes V$, and hence are equalities.
– Joshua Tilley
Jul 17 at 16:20

You can prove both that $ker left(phi otimes psi right)=ker left(phi right)otimes V+Uotimes ker left(psi right)$ and $textrmimleft(phiright)otimestextrmimleft(psiright)=textrmimleft(phiotimespsiright)$ by rank nullity. Given the inclusion you noted, and the inclusion for the image that I noted, you get lower bounds for the rank and nullity of the tensor product map, but they add up to the dimension of $Uotimes V$, and hence are equalities.
– Joshua Tilley
Jul 17 at 16:20

add a comment | 

 

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