A module $B$ is flat if Tor $= 0$












1












$begingroup$


From Weibel's "An Introduction to Homological Algebra":



Exercise 3.2.1: An $R$-module $B$ is flat if Tor$_i^R(A,B) = 0$ for every $R$-module A.



It seems to me that the obvious way to do this would be to use the definition:



Tor$_i^R(A,B) = $H$_i(P.otimes B)$ where $P.$ is a projective resolution of A.



We need to show that for an exact sequence:



$...rightarrow A_{n+1} rightarrow A_{n} rightarrow A_{n-1} rightarrow ...$ ,



$...rightarrow A_{n+1} otimes B rightarrow A_{n} otimes B rightarrow A_{n-1} otimes B rightarrow ...$ is exact.



Since we can only access exactness via the definition of Tor, it seems to me that we might have to construct a projective resolution around this point in the sequence, however I am currently struggling.



Any help is appreciated.






homological-algebra exact-sequence







share|cite|improve this question









$endgroup$












  • $begingroup$
    I think it's enough to show that for any short exact sequence $$0to A_1to A_2to A_3to 0$$ we have $$0to A_1otimes Bto A_2otimes Bto A_3otimes Bto 0$$exact. At least, that is the definition of "flat" that I know (I seem to recall that the translation between the two isn't all that difficult, and short exact sequences are often nicer to work with)..
    $endgroup$
    – Arthur
    Jan 7 at 14:53












  • $begingroup$
    Yeah you're right, we can work with short exact sequences. Although I still have a problem! Thanks
    $endgroup$
    – Daven
    Jan 7 at 14:59






  • 1




    $begingroup$
    Check the horseshoe lemma, then the snake lemma.
    $endgroup$
    – Arthur
    Jan 7 at 15:01










  • $begingroup$
    That looks like it should do it! Every time I saw an instance of the question I asked it was always phrased in such a way that made it look like a nice simple question, however this proof is certainly not a simple rearranging of definitions.
    $endgroup$
    – Daven
    Jan 7 at 15:05










  • $begingroup$
    It is possible that there are more elementary solutions, but if those lemmas are available to use why not use them?
    $endgroup$
    – Arthur
    Jan 7 at 15:08


















1












$begingroup$


From Weibel's "An Introduction to Homological Algebra":



Exercise 3.2.1: An $R$-module $B$ is flat if Tor$_i^R(A,B) = 0$ for every $R$-module A.



It seems to me that the obvious way to do this would be to use the definition:



Tor$_i^R(A,B) = $H$_i(P.otimes B)$ where $P.$ is a projective resolution of A.



We need to show that for an exact sequence:



$...rightarrow A_{n+1} rightarrow A_{n} rightarrow A_{n-1} rightarrow ...$ ,



$...rightarrow A_{n+1} otimes B rightarrow A_{n} otimes B rightarrow A_{n-1} otimes B rightarrow ...$ is exact.



Since we can only access exactness via the definition of Tor, it seems to me that we might have to construct a projective resolution around this point in the sequence, however I am currently struggling.



Any help is appreciated.






homological-algebra exact-sequence







share|cite|improve this question









$endgroup$












  • $begingroup$
    I think it's enough to show that for any short exact sequence $$0to A_1to A_2to A_3to 0$$ we have $$0to A_1otimes Bto A_2otimes Bto A_3otimes Bto 0$$exact. At least, that is the definition of "flat" that I know (I seem to recall that the translation between the two isn't all that difficult, and short exact sequences are often nicer to work with)..
    $endgroup$
    – Arthur
    Jan 7 at 14:53












  • $begingroup$
    Yeah you're right, we can work with short exact sequences. Although I still have a problem! Thanks
    $endgroup$
    – Daven
    Jan 7 at 14:59






  • 1




    $begingroup$
    Check the horseshoe lemma, then the snake lemma.
    $endgroup$
    – Arthur
    Jan 7 at 15:01










  • $begingroup$
    That looks like it should do it! Every time I saw an instance of the question I asked it was always phrased in such a way that made it look like a nice simple question, however this proof is certainly not a simple rearranging of definitions.
    $endgroup$
    – Daven
    Jan 7 at 15:05










  • $begingroup$
    It is possible that there are more elementary solutions, but if those lemmas are available to use why not use them?
    $endgroup$
    – Arthur
    Jan 7 at 15:08
















1












1








1


1



$begingroup$


From Weibel's "An Introduction to Homological Algebra":



Exercise 3.2.1: An $R$-module $B$ is flat if Tor$_i^R(A,B) = 0$ for every $R$-module A.



It seems to me that the obvious way to do this would be to use the definition:



Tor$_i^R(A,B) = $H$_i(P.otimes B)$ where $P.$ is a projective resolution of A.



We need to show that for an exact sequence:



$...rightarrow A_{n+1} rightarrow A_{n} rightarrow A_{n-1} rightarrow ...$ ,



$...rightarrow A_{n+1} otimes B rightarrow A_{n} otimes B rightarrow A_{n-1} otimes B rightarrow ...$ is exact.



Since we can only access exactness via the definition of Tor, it seems to me that we might have to construct a projective resolution around this point in the sequence, however I am currently struggling.



Any help is appreciated.






homological-algebra exact-sequence







share|cite|improve this question









$endgroup$




From Weibel's "An Introduction to Homological Algebra":



Exercise 3.2.1: An $R$-module $B$ is flat if Tor$_i^R(A,B) = 0$ for every $R$-module A.



It seems to me that the obvious way to do this would be to use the definition:



Tor$_i^R(A,B) = $H$_i(P.otimes B)$ where $P.$ is a projective resolution of A.



We need to show that for an exact sequence:



$...rightarrow A_{n+1} rightarrow A_{n} rightarrow A_{n-1} rightarrow ...$ ,



$...rightarrow A_{n+1} otimes B rightarrow A_{n} otimes B rightarrow A_{n-1} otimes B rightarrow ...$ is exact.



Since we can only access exactness via the definition of Tor, it seems to me that we might have to construct a projective resolution around this point in the sequence, however I am currently struggling.



Any help is appreciated.






homological-algebra exact-sequence




homological-algebra exact-sequence






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 7 at 14:40









DavenDaven

41229




41229












  • $begingroup$
    I think it's enough to show that for any short exact sequence $$0to A_1to A_2to A_3to 0$$ we have $$0to A_1otimes Bto A_2otimes Bto A_3otimes Bto 0$$exact. At least, that is the definition of "flat" that I know (I seem to recall that the translation between the two isn't all that difficult, and short exact sequences are often nicer to work with)..
    $endgroup$
    – Arthur
    Jan 7 at 14:53












  • $begingroup$
    Yeah you're right, we can work with short exact sequences. Although I still have a problem! Thanks
    $endgroup$
    – Daven
    Jan 7 at 14:59






  • 1




    $begingroup$
    Check the horseshoe lemma, then the snake lemma.
    $endgroup$
    – Arthur
    Jan 7 at 15:01










  • $begingroup$
    That looks like it should do it! Every time I saw an instance of the question I asked it was always phrased in such a way that made it look like a nice simple question, however this proof is certainly not a simple rearranging of definitions.
    $endgroup$
    – Daven
    Jan 7 at 15:05










  • $begingroup$
    It is possible that there are more elementary solutions, but if those lemmas are available to use why not use them?
    $endgroup$
    – Arthur
    Jan 7 at 15:08




















  • $begingroup$
    I think it's enough to show that for any short exact sequence $$0to A_1to A_2to A_3to 0$$ we have $$0to A_1otimes Bto A_2otimes Bto A_3otimes Bto 0$$exact. At least, that is the definition of "flat" that I know (I seem to recall that the translation between the two isn't all that difficult, and short exact sequences are often nicer to work with)..
    $endgroup$
    – Arthur
    Jan 7 at 14:53












  • $begingroup$
    Yeah you're right, we can work with short exact sequences. Although I still have a problem! Thanks
    $endgroup$
    – Daven
    Jan 7 at 14:59






  • 1




    $begingroup$
    Check the horseshoe lemma, then the snake lemma.
    $endgroup$
    – Arthur
    Jan 7 at 15:01










  • $begingroup$
    That looks like it should do it! Every time I saw an instance of the question I asked it was always phrased in such a way that made it look like a nice simple question, however this proof is certainly not a simple rearranging of definitions.
    $endgroup$
    – Daven
    Jan 7 at 15:05










  • $begingroup$
    It is possible that there are more elementary solutions, but if those lemmas are available to use why not use them?
    $endgroup$
    – Arthur
    Jan 7 at 15:08


















$begingroup$
I think it's enough to show that for any short exact sequence $$0to A_1to A_2to A_3to 0$$ we have $$0to A_1otimes Bto A_2otimes Bto A_3otimes Bto 0$$exact. At least, that is the definition of "flat" that I know (I seem to recall that the translation between the two isn't all that difficult, and short exact sequences are often nicer to work with)..
$endgroup$
– Arthur
Jan 7 at 14:53






$begingroup$
I think it's enough to show that for any short exact sequence $$0to A_1to A_2to A_3to 0$$ we have $$0to A_1otimes Bto A_2otimes Bto A_3otimes Bto 0$$exact. At least, that is the definition of "flat" that I know (I seem to recall that the translation between the two isn't all that difficult, and short exact sequences are often nicer to work with)..
$endgroup$
– Arthur
Jan 7 at 14:53














$begingroup$
Yeah you're right, we can work with short exact sequences. Although I still have a problem! Thanks
$endgroup$
– Daven
Jan 7 at 14:59




$begingroup$
Yeah you're right, we can work with short exact sequences. Although I still have a problem! Thanks
$endgroup$
– Daven
Jan 7 at 14:59




1




1




$begingroup$
Check the horseshoe lemma, then the snake lemma.
$endgroup$
– Arthur
Jan 7 at 15:01




$begingroup$
Check the horseshoe lemma, then the snake lemma.
$endgroup$
– Arthur
Jan 7 at 15:01












$begingroup$
That looks like it should do it! Every time I saw an instance of the question I asked it was always phrased in such a way that made it look like a nice simple question, however this proof is certainly not a simple rearranging of definitions.
$endgroup$
– Daven
Jan 7 at 15:05




$begingroup$
That looks like it should do it! Every time I saw an instance of the question I asked it was always phrased in such a way that made it look like a nice simple question, however this proof is certainly not a simple rearranging of definitions.
$endgroup$
– Daven
Jan 7 at 15:05












$begingroup$
It is possible that there are more elementary solutions, but if those lemmas are available to use why not use them?
$endgroup$
– Arthur
Jan 7 at 15:08






$begingroup$
It is possible that there are more elementary solutions, but if those lemmas are available to use why not use them?
$endgroup$
– Arthur
Jan 7 at 15:08












1 Answer
1






active

oldest

votes


















2












$begingroup$

We want to show that $_ otimes B$ is exact. Now Lets take for this a short exact sequence $$0 to X to Y to Z to 0$$
Now lets apply the functors $mathrm{Tor}^i(_,B)$ to that sequence. Now this gives by the definition of $mathrm{Tor}^i(_,B)$ as a homological functor a long exact sequence $$ ... to mathrm{Tor}^1(X,B) to mathrm{Tor}^1(Y,B) to mathrm{Tor}^1(Z,B) tomathrm{Tor}^0(X,B) to mathrm{Tor}^0(Y,B) to mathrm{Tor}^0(Z,B) to 0 $$
Now since we have a natural isomorphism $mathrm{Tor}^0(Z,B) cong Zotimes B$ we may rewrite the top sequence as:
$$ mathrm{Tor}^1(X,B) to mathrm{Tor}^1(Y,B) to mathrm{Tor}^1(Z,B) to Xotimes B to Yotimes B to Z otimes B to 0 $$
But since $ mathrm{Tor}^1(Z,B)=0$ this becomes:
$$0to Xotimes B to Yotimes B to Z otimes B to 0 $$
as desired.
(for the other direction of the implication just observe that if $B$ is flat, the projective resolution stays exact after tensoring and hence the higher $mathrm{Tor}$-terms vanish)






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This is the kind of nice answer I was looking for, thanks so much!
    $endgroup$
    – Daven
    Jan 7 at 17:04










  • $begingroup$
    you are very welcome! one just has to love long exact sequences!
    $endgroup$
    – Enkidu
    Jan 8 at 8:06






  • 1




    $begingroup$
    @Daven In some sense, the $operatorname{Tor}$ functor's purpose is to measure (an to some extent fix) the fact that tensoring isn't left-exact. So, when you have a module where tensoring in fact is left exact, $operatorname{Tor}$ gives you $0$. Similarily, the $operatorname{Ext}$ functor measures the non-exactness of $operatorname{hom}(B, -)$, in the other direction (and even for $operatorname{hom}(-, B)$, if you just remember to turn all the arrows the right way).
    $endgroup$
    – Arthur
    Jan 8 at 11:44














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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









2












$begingroup$

We want to show that $_ otimes B$ is exact. Now Lets take for this a short exact sequence $$0 to X to Y to Z to 0$$
Now lets apply the functors $mathrm{Tor}^i(_,B)$ to that sequence. Now this gives by the definition of $mathrm{Tor}^i(_,B)$ as a homological functor a long exact sequence $$ ... to mathrm{Tor}^1(X,B) to mathrm{Tor}^1(Y,B) to mathrm{Tor}^1(Z,B) tomathrm{Tor}^0(X,B) to mathrm{Tor}^0(Y,B) to mathrm{Tor}^0(Z,B) to 0 $$
Now since we have a natural isomorphism $mathrm{Tor}^0(Z,B) cong Zotimes B$ we may rewrite the top sequence as:
$$ mathrm{Tor}^1(X,B) to mathrm{Tor}^1(Y,B) to mathrm{Tor}^1(Z,B) to Xotimes B to Yotimes B to Z otimes B to 0 $$
But since $ mathrm{Tor}^1(Z,B)=0$ this becomes:
$$0to Xotimes B to Yotimes B to Z otimes B to 0 $$
as desired.
(for the other direction of the implication just observe that if $B$ is flat, the projective resolution stays exact after tensoring and hence the higher $mathrm{Tor}$-terms vanish)






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This is the kind of nice answer I was looking for, thanks so much!
    $endgroup$
    – Daven
    Jan 7 at 17:04










  • $begingroup$
    you are very welcome! one just has to love long exact sequences!
    $endgroup$
    – Enkidu
    Jan 8 at 8:06






  • 1




    $begingroup$
    @Daven In some sense, the $operatorname{Tor}$ functor's purpose is to measure (an to some extent fix) the fact that tensoring isn't left-exact. So, when you have a module where tensoring in fact is left exact, $operatorname{Tor}$ gives you $0$. Similarily, the $operatorname{Ext}$ functor measures the non-exactness of $operatorname{hom}(B, -)$, in the other direction (and even for $operatorname{hom}(-, B)$, if you just remember to turn all the arrows the right way).
    $endgroup$
    – Arthur
    Jan 8 at 11:44


















2












$begingroup$

We want to show that $_ otimes B$ is exact. Now Lets take for this a short exact sequence $$0 to X to Y to Z to 0$$
Now lets apply the functors $mathrm{Tor}^i(_,B)$ to that sequence. Now this gives by the definition of $mathrm{Tor}^i(_,B)$ as a homological functor a long exact sequence $$ ... to mathrm{Tor}^1(X,B) to mathrm{Tor}^1(Y,B) to mathrm{Tor}^1(Z,B) tomathrm{Tor}^0(X,B) to mathrm{Tor}^0(Y,B) to mathrm{Tor}^0(Z,B) to 0 $$
Now since we have a natural isomorphism $mathrm{Tor}^0(Z,B) cong Zotimes B$ we may rewrite the top sequence as:
$$ mathrm{Tor}^1(X,B) to mathrm{Tor}^1(Y,B) to mathrm{Tor}^1(Z,B) to Xotimes B to Yotimes B to Z otimes B to 0 $$
But since $ mathrm{Tor}^1(Z,B)=0$ this becomes:
$$0to Xotimes B to Yotimes B to Z otimes B to 0 $$
as desired.
(for the other direction of the implication just observe that if $B$ is flat, the projective resolution stays exact after tensoring and hence the higher $mathrm{Tor}$-terms vanish)






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This is the kind of nice answer I was looking for, thanks so much!
    $endgroup$
    – Daven
    Jan 7 at 17:04










  • $begingroup$
    you are very welcome! one just has to love long exact sequences!
    $endgroup$
    – Enkidu
    Jan 8 at 8:06






  • 1




    $begingroup$
    @Daven In some sense, the $operatorname{Tor}$ functor's purpose is to measure (an to some extent fix) the fact that tensoring isn't left-exact. So, when you have a module where tensoring in fact is left exact, $operatorname{Tor}$ gives you $0$. Similarily, the $operatorname{Ext}$ functor measures the non-exactness of $operatorname{hom}(B, -)$, in the other direction (and even for $operatorname{hom}(-, B)$, if you just remember to turn all the arrows the right way).
    $endgroup$
    – Arthur
    Jan 8 at 11:44
















2












2








2





$begingroup$

We want to show that $_ otimes B$ is exact. Now Lets take for this a short exact sequence $$0 to X to Y to Z to 0$$
Now lets apply the functors $mathrm{Tor}^i(_,B)$ to that sequence. Now this gives by the definition of $mathrm{Tor}^i(_,B)$ as a homological functor a long exact sequence $$ ... to mathrm{Tor}^1(X,B) to mathrm{Tor}^1(Y,B) to mathrm{Tor}^1(Z,B) tomathrm{Tor}^0(X,B) to mathrm{Tor}^0(Y,B) to mathrm{Tor}^0(Z,B) to 0 $$
Now since we have a natural isomorphism $mathrm{Tor}^0(Z,B) cong Zotimes B$ we may rewrite the top sequence as:
$$ mathrm{Tor}^1(X,B) to mathrm{Tor}^1(Y,B) to mathrm{Tor}^1(Z,B) to Xotimes B to Yotimes B to Z otimes B to 0 $$
But since $ mathrm{Tor}^1(Z,B)=0$ this becomes:
$$0to Xotimes B to Yotimes B to Z otimes B to 0 $$
as desired.
(for the other direction of the implication just observe that if $B$ is flat, the projective resolution stays exact after tensoring and hence the higher $mathrm{Tor}$-terms vanish)






share|cite|improve this answer











$endgroup$



We want to show that $_ otimes B$ is exact. Now Lets take for this a short exact sequence $$0 to X to Y to Z to 0$$
Now lets apply the functors $mathrm{Tor}^i(_,B)$ to that sequence. Now this gives by the definition of $mathrm{Tor}^i(_,B)$ as a homological functor a long exact sequence $$ ... to mathrm{Tor}^1(X,B) to mathrm{Tor}^1(Y,B) to mathrm{Tor}^1(Z,B) tomathrm{Tor}^0(X,B) to mathrm{Tor}^0(Y,B) to mathrm{Tor}^0(Z,B) to 0 $$
Now since we have a natural isomorphism $mathrm{Tor}^0(Z,B) cong Zotimes B$ we may rewrite the top sequence as:
$$ mathrm{Tor}^1(X,B) to mathrm{Tor}^1(Y,B) to mathrm{Tor}^1(Z,B) to Xotimes B to Yotimes B to Z otimes B to 0 $$
But since $ mathrm{Tor}^1(Z,B)=0$ this becomes:
$$0to Xotimes B to Yotimes B to Z otimes B to 0 $$
as desired.
(for the other direction of the implication just observe that if $B$ is flat, the projective resolution stays exact after tensoring and hence the higher $mathrm{Tor}$-terms vanish)







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 8 at 8:06

























answered Jan 7 at 15:48









EnkiduEnkidu

1,44429




1,44429












  • $begingroup$
    This is the kind of nice answer I was looking for, thanks so much!
    $endgroup$
    – Daven
    Jan 7 at 17:04










  • $begingroup$
    you are very welcome! one just has to love long exact sequences!
    $endgroup$
    – Enkidu
    Jan 8 at 8:06






  • 1




    $begingroup$
    @Daven In some sense, the $operatorname{Tor}$ functor's purpose is to measure (an to some extent fix) the fact that tensoring isn't left-exact. So, when you have a module where tensoring in fact is left exact, $operatorname{Tor}$ gives you $0$. Similarily, the $operatorname{Ext}$ functor measures the non-exactness of $operatorname{hom}(B, -)$, in the other direction (and even for $operatorname{hom}(-, B)$, if you just remember to turn all the arrows the right way).
    $endgroup$
    – Arthur
    Jan 8 at 11:44




















  • $begingroup$
    This is the kind of nice answer I was looking for, thanks so much!
    $endgroup$
    – Daven
    Jan 7 at 17:04










  • $begingroup$
    you are very welcome! one just has to love long exact sequences!
    $endgroup$
    – Enkidu
    Jan 8 at 8:06






  • 1




    $begingroup$
    @Daven In some sense, the $operatorname{Tor}$ functor's purpose is to measure (an to some extent fix) the fact that tensoring isn't left-exact. So, when you have a module where tensoring in fact is left exact, $operatorname{Tor}$ gives you $0$. Similarily, the $operatorname{Ext}$ functor measures the non-exactness of $operatorname{hom}(B, -)$, in the other direction (and even for $operatorname{hom}(-, B)$, if you just remember to turn all the arrows the right way).
    $endgroup$
    – Arthur
    Jan 8 at 11:44


















$begingroup$
This is the kind of nice answer I was looking for, thanks so much!
$endgroup$
– Daven
Jan 7 at 17:04




$begingroup$
This is the kind of nice answer I was looking for, thanks so much!
$endgroup$
– Daven
Jan 7 at 17:04












$begingroup$
you are very welcome! one just has to love long exact sequences!
$endgroup$
– Enkidu
Jan 8 at 8:06




$begingroup$
you are very welcome! one just has to love long exact sequences!
$endgroup$
– Enkidu
Jan 8 at 8:06




1




1




$begingroup$
@Daven In some sense, the $operatorname{Tor}$ functor's purpose is to measure (an to some extent fix) the fact that tensoring isn't left-exact. So, when you have a module where tensoring in fact is left exact, $operatorname{Tor}$ gives you $0$. Similarily, the $operatorname{Ext}$ functor measures the non-exactness of $operatorname{hom}(B, -)$, in the other direction (and even for $operatorname{hom}(-, B)$, if you just remember to turn all the arrows the right way).
$endgroup$
– Arthur
Jan 8 at 11:44






$begingroup$
@Daven In some sense, the $operatorname{Tor}$ functor's purpose is to measure (an to some extent fix) the fact that tensoring isn't left-exact. So, when you have a module where tensoring in fact is left exact, $operatorname{Tor}$ gives you $0$. Similarily, the $operatorname{Ext}$ functor measures the non-exactness of $operatorname{hom}(B, -)$, in the other direction (and even for $operatorname{hom}(-, B)$, if you just remember to turn all the arrows the right way).
$endgroup$
– Arthur
Jan 8 at 11:44




















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Probability when a professor distributes a quiz and homework assignment to a class of n students.

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0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
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Aardman Animations

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Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
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Are they similar matrix

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0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
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