Mixing
Given two similar matrices $A$, $B$, is there a way to find an invertible matrix $P$ such that $A=P^{-1}BP$?
I was wondering if given two similar square matrices $A$ and $B$ would always be possible to find an matrix $Pin GL(n)$ such that $A=P^{-1}BP$.
thank you!
linear-algebra matrices
asked Dec 10 '13 at 20:50
giuliogiulio
411413
add a comment |
I was wondering if given two similar square matrices $A$ and $B$ would always be possible to find an matrix $Pin GL(n)$ such that $A=P^{-1}BP$.
thank you!
linear-algebra matrices
asked Dec 10 '13 at 20:50
giuliogiulio
411413
The definition of similarity between $A$ and $B$ is that there exists invertible $P$ such that $A = P^{-1}BP$.
– EuYu
Dec 10 '13 at 21:02
According to the definition of similarity one can say the following: When someone sells you two matrices $A$ and $B$, claiming that they are similar, he has to provide a $P$ with $A=P^{-1}B P$ to testify for this similarity.
– Christian Blatter
Dec 10 '13 at 21:02
There are a lot of "quantities" that are preserved by conjugation, for example if $A$ and $B$ have different ranks, determinants, sets of eigenvalues, and so on, then $A$ and $B$ can't be similar, meaning that no such $P$ exists in those cases.
– Jeppe Stig Nielsen
Dec 10 '13 at 21:02
I assume that $A$ and $B$ are similar and want to find $P$ that makes them similar
– giulio
Dec 10 '13 at 21:07
Through Smith normal form of the matrices, we are able to write an explicit algorithm producing such matrix $P$. This does not require the field to be algebraically closed.
– i707107
Nov 22 '18 at 2:51
add a comment |
I was wondering if given two similar square matrices $A$ and $B$ would always be possible to find an matrix $Pin GL(n)$ such that $A=P^{-1}BP$.
thank you!
linear-algebra matrices
asked Dec 10 '13 at 20:50
giuliogiulio
411413
I was wondering if given two similar square matrices $A$ and $B$ would always be possible to find an matrix $Pin GL(n)$ such that $A=P^{-1}BP$.
thank you!
linear-algebra matrices
linear-algebra matrices
asked Dec 10 '13 at 20:50
giuliogiulio
411413
asked Dec 10 '13 at 20:50
giuliogiulio
411413
edited Dec 10 '13 at 21:00
giulio
asked Dec 10 '13 at 20:50
giuliogiulio
411413
asked Dec 10 '13 at 20:50
giuliogiulio
411413
asked Dec 10 '13 at 20:50
giuliogiulio
411413
411413
The definition of similarity between $A$ and $B$ is that there exists invertible $P$ such that $A = P^{-1}BP$.
– EuYu
Dec 10 '13 at 21:02
According to the definition of similarity one can say the following: When someone sells you two matrices $A$ and $B$, claiming that they are similar, he has to provide a $P$ with $A=P^{-1}B P$ to testify for this similarity.
– Christian Blatter
Dec 10 '13 at 21:02
There are a lot of "quantities" that are preserved by conjugation, for example if $A$ and $B$ have different ranks, determinants, sets of eigenvalues, and so on, then $A$ and $B$ can't be similar, meaning that no such $P$ exists in those cases.
– Jeppe Stig Nielsen
Dec 10 '13 at 21:02
I assume that $A$ and $B$ are similar and want to find $P$ that makes them similar
– giulio
Dec 10 '13 at 21:07
Through Smith normal form of the matrices, we are able to write an explicit algorithm producing such matrix $P$. This does not require the field to be algebraically closed.
– i707107
Nov 22 '18 at 2:51
add a comment |
The definition of similarity between $A$ and $B$ is that there exists invertible $P$ such that $A = P^{-1}BP$.
– EuYu
Dec 10 '13 at 21:02
According to the definition of similarity one can say the following: When someone sells you two matrices $A$ and $B$, claiming that they are similar, he has to provide a $P$ with $A=P^{-1}B P$ to testify for this similarity.
– Christian Blatter
Dec 10 '13 at 21:02
There are a lot of "quantities" that are preserved by conjugation, for example if $A$ and $B$ have different ranks, determinants, sets of eigenvalues, and so on, then $A$ and $B$ can't be similar, meaning that no such $P$ exists in those cases.
– Jeppe Stig Nielsen
Dec 10 '13 at 21:02
I assume that $A$ and $B$ are similar and want to find $P$ that makes them similar
– giulio
Dec 10 '13 at 21:07
Through Smith normal form of the matrices, we are able to write an explicit algorithm producing such matrix $P$. This does not require the field to be algebraically closed.
– i707107
Nov 22 '18 at 2:51
The definition of similarity between $A$ and $B$ is that there exists invertible $P$ such that $A = P^{-1}BP$.
– EuYu
Dec 10 '13 at 21:02
The definition of similarity between $A$ and $B$ is that there exists invertible $P$ such that $A = P^{-1}BP$.
– EuYu
Dec 10 '13 at 21:02
According to the definition of similarity one can say the following: When someone sells you two matrices $A$ and $B$, claiming that they are similar, he has to provide a $P$ with $A=P^{-1}B P$ to testify for this similarity.
– Christian Blatter
Dec 10 '13 at 21:02
According to the definition of similarity one can say the following: When someone sells you two matrices $A$ and $B$, claiming that they are similar, he has to provide a $P$ with $A=P^{-1}B P$ to testify for this similarity.
– Christian Blatter
Dec 10 '13 at 21:02
There are a lot of "quantities" that are preserved by conjugation, for example if $A$ and $B$ have different ranks, determinants, sets of eigenvalues, and so on, then $A$ and $B$ can't be similar, meaning that no such $P$ exists in those cases.
– Jeppe Stig Nielsen
Dec 10 '13 at 21:02
There are a lot of "quantities" that are preserved by conjugation, for example if $A$ and $B$ have different ranks, determinants, sets of eigenvalues, and so on, then $A$ and $B$ can't be similar, meaning that no such $P$ exists in those cases.
– Jeppe Stig Nielsen
Dec 10 '13 at 21:02
I assume that $A$ and $B$ are similar and want to find $P$ that makes them similar
– giulio
Dec 10 '13 at 21:07
I assume that $A$ and $B$ are similar and want to find $P$ that makes them similar
– giulio
Dec 10 '13 at 21:07
Through Smith normal form of the matrices, we are able to write an explicit algorithm producing such matrix $P$. This does not require the field to be algebraically closed.
– i707107
Nov 22 '18 at 2:51
Through Smith normal form of the matrices, we are able to write an explicit algorithm producing such matrix $P$. This does not require the field to be algebraically closed.
– i707107
Nov 22 '18 at 2:51
add a comment |
3 Answers
3
active
oldest
votes
Let $J$ the Jordan canonical matrix similar to $A$ and $B$ so we can find the change basis matrices $Q$ and $S$ such that
$$A=QJQ^{-1}quad;quad B=SJS^{-1}$$
hence with $P=Q^{-1}S$ we have $A=P^{-1}BP$.
Thank you for your answers. But what if we consider matrices over non algebrally closed fields?
– giulio
Dec 10 '13 at 21:27
$ddotsmile{}{}$
– mrs
Dec 11 '13 at 5:21
For me, I found an appropriate Q for A and an appropriate S for B. However, for $$A = P^{-1} B P$$ to be true, I had to use $$P = S Q^{-1}$$ instead.
– RAH
Nov 17 '18 at 18:49
add a comment |
The short answer will be no. For example, let $Aneq I$ and $B=I$. Then clearly $P^{-1}BP=P^{-1}IP=Ineq A$ for all $Pin GL_n(F)$.
You are looking for matrix similarity, and you can read about the conditions in the link above.
--
Edit: As the question was edited, now the answer is yes: For every matrix $A$ one can find its Jordan canonical form, $J_A$. Find the base change matrix $P_A$. Since $A$, $B$ are similar if and only if $J_A=J_B$, we have $J=P_AAP_A^{-1}=P_BBP_B^{-1}$. So $A=(P_B^{-1}P_A)^{-1}B(P_B^{-1}P_A)$
answered Dec 10 '13 at 20:55
Dennis GulkoDennis Gulko
13.9k12655
sorry, I edited the text
– giulio
Dec 10 '13 at 21:00
add a comment |
Let $K$ be a subfield of $mathbb{C}$. We assume that $A,Bin M_n(K)$ are similar (that is, there is $Pin M_n(K)$ s.t. $det(P)not= 0$ and $PA=BP$). Note that we can always choose $P$ in $M_n(K)$.
Of course, it is important not to calculate Jordan's forms of $A,B$.
METHOD 1.
i) Solve the linear system $PA=BP$; the set of solutions $E$ is a sub-vector space of $M_n(K)$ of dimension $k=dim(C(A))geq n$. Note that, using the tensor product, there are methods that permit to solve such a system with complexity in $O(n^3)$ (and not in $O(n^6)$).
Since there is $Pin E$ s.t. $det(P)not= 0$, ${Pin GL_n(K);PA=BP}$ is Zariski open dense in $E$.
ii) Thus, it suffices to randomly choose the $k$ parameters that define an element of $E$ to obtain a required solution with probability $1$.
METHOD 2.
We can also use the Frobenius form over $K$. Indeed, $A,B$ have same Frobenius form $F$:
$A=QFQ^{-1},B=RFR^{-1}$ and the sequel is easy.
EDIT. Here is a quick history of the progress made in calculating the Frobenius form.
i) in "Nearly optimal algorithms for canonical matrix forms. SIAM Journal on Computing, 24:948–969, 1995"
M. Giesbrech gives a randomized method: a Las Vegas algorithm can be used to compute both
the Frobenius form and a Frobenius transition matrix for a given $ntimes n$ matrix over a field $F$ using an expected number of operations over $F$ that is in $O(n^3)$, with standard matrix and polynomial arithmetic, whenever $F$ has at least $n^2$ elements (to ensure a probability of success at least
$1/2$), and using an expected number of
operations in $O(n^3log_q(n))$ if $F$ is a finite field with size $q$.
ii) Storjohann [1997] exposed a deterministic method whose complexity is $O(n^3)$. cf.
http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.28.2505
However, the previous method does not give, at the same time, the matrix of change of basis.
iii) In "Algorithms for similarity transforms. In
Seventh Rhine Work-shop on Computer Algebra
, Bregenz, Austria, March 2000"
A. Storjohann and G. Villard obtain (with the same complexity), at the same time, the Frobenius form $F$ and the transform matrix $V$ .
There is a gap. None of the previous methods allow to use the fast multiplication of Strassen (in $O(n^{2.81})$).
iv) W. Eberly, in "Asymptotically efficient algorithms for the
Frobenius form. Technical report, Department of
Computer Science, University of Calgary, 2000"
gives $F,V$ thanks to a Las Vegas algorithm (using Strassen) that has complexity $O(n^{2.81}log(n))$.
Note that, unlike an urban legend, the methods of Coppersmith-Winograd and of Le Gall are practically useless.
v) In 2007, C. Pernet and A. Storjohann, in "Faster Algorithms for the Characteristic Polynomial. ISSAC"
give $F$ (but not $V$) with a Las Vegas algorithm in $O(n^{2.81})$.
This is the state of play in 2011-2013. I don't know if an algorithm giving $F, V$ in $O(n^{2.81})$ was found in recent years.
answered Nov 22 '18 at 13:45
loup blancloup blanc
22.5k21850
I wonder if there are improvements in $O(n^3)$ since 1997, and the open questions they put in their paper?
– i707107
Nov 23 '18 at 23:07
@i707107 , read my edit.
– loup blanc
Nov 27 '18 at 17:05
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
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active
oldest
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oldest
votes
Let $J$ the Jordan canonical matrix similar to $A$ and $B$ so we can find the change basis matrices $Q$ and $S$ such that
$$A=QJQ^{-1}quad;quad B=SJS^{-1}$$
hence with $P=Q^{-1}S$ we have $A=P^{-1}BP$.
Thank you for your answers. But what if we consider matrices over non algebrally closed fields?
– giulio
Dec 10 '13 at 21:27
$ddotsmile{}{}$
– mrs
Dec 11 '13 at 5:21
For me, I found an appropriate Q for A and an appropriate S for B. However, for $$A = P^{-1} B P$$ to be true, I had to use $$P = S Q^{-1}$$ instead.
– RAH
Nov 17 '18 at 18:49
add a comment |
Let $J$ the Jordan canonical matrix similar to $A$ and $B$ so we can find the change basis matrices $Q$ and $S$ such that
$$A=QJQ^{-1}quad;quad B=SJS^{-1}$$
hence with $P=Q^{-1}S$ we have $A=P^{-1}BP$.
Thank you for your answers. But what if we consider matrices over non algebrally closed fields?
– giulio
Dec 10 '13 at 21:27
$ddotsmile{}{}$
– mrs
Dec 11 '13 at 5:21
For me, I found an appropriate Q for A and an appropriate S for B. However, for $$A = P^{-1} B P$$ to be true, I had to use $$P = S Q^{-1}$$ instead.
– RAH
Nov 17 '18 at 18:49
add a comment |
Let $J$ the Jordan canonical matrix similar to $A$ and $B$ so we can find the change basis matrices $Q$ and $S$ such that
$$A=QJQ^{-1}quad;quad B=SJS^{-1}$$
hence with $P=Q^{-1}S$ we have $A=P^{-1}BP$.
Let $J$ the Jordan canonical matrix similar to $A$ and $B$ so we can find the change basis matrices $Q$ and $S$ such that
$$A=QJQ^{-1}quad;quad B=SJS^{-1}$$
hence with $P=Q^{-1}S$ we have $A=P^{-1}BP$.
answered Dec 10 '13 at 21:00
user63181
Thank you for your answers. But what if we consider matrices over non algebrally closed fields?
– giulio
Dec 10 '13 at 21:27
$ddotsmile{}{}$
– mrs
Dec 11 '13 at 5:21
For me, I found an appropriate Q for A and an appropriate S for B. However, for $$A = P^{-1} B P$$ to be true, I had to use $$P = S Q^{-1}$$ instead.
– RAH
Nov 17 '18 at 18:49
add a comment |
Thank you for your answers. But what if we consider matrices over non algebrally closed fields?
– giulio
Dec 10 '13 at 21:27
$ddotsmile{}{}$
– mrs
Dec 11 '13 at 5:21
For me, I found an appropriate Q for A and an appropriate S for B. However, for $$A = P^{-1} B P$$ to be true, I had to use $$P = S Q^{-1}$$ instead.
– RAH
Nov 17 '18 at 18:49
Thank you for your answers. But what if we consider matrices over non algebrally closed fields?
– giulio
Dec 10 '13 at 21:27
Thank you for your answers. But what if we consider matrices over non algebrally closed fields?
– giulio
Dec 10 '13 at 21:27
$ddotsmile{}{}$
– mrs
Dec 11 '13 at 5:21
$ddotsmile{}{}$
– mrs
Dec 11 '13 at 5:21
For me, I found an appropriate Q for A and an appropriate S for B. However, for $$A = P^{-1} B P$$ to be true, I had to use $$P = S Q^{-1}$$ instead.
– RAH
Nov 17 '18 at 18:49
For me, I found an appropriate Q for A and an appropriate S for B. However, for $$A = P^{-1} B P$$ to be true, I had to use $$P = S Q^{-1}$$ instead.
– RAH
Nov 17 '18 at 18:49
add a comment |
The short answer will be no. For example, let $Aneq I$ and $B=I$. Then clearly $P^{-1}BP=P^{-1}IP=Ineq A$ for all $Pin GL_n(F)$.
You are looking for matrix similarity, and you can read about the conditions in the link above.
--
Edit: As the question was edited, now the answer is yes: For every matrix $A$ one can find its Jordan canonical form, $J_A$. Find the base change matrix $P_A$. Since $A$, $B$ are similar if and only if $J_A=J_B$, we have $J=P_AAP_A^{-1}=P_BBP_B^{-1}$. So $A=(P_B^{-1}P_A)^{-1}B(P_B^{-1}P_A)$
answered Dec 10 '13 at 20:55
Dennis GulkoDennis Gulko
13.9k12655
sorry, I edited the text
– giulio
Dec 10 '13 at 21:00
add a comment |
The short answer will be no. For example, let $Aneq I$ and $B=I$. Then clearly $P^{-1}BP=P^{-1}IP=Ineq A$ for all $Pin GL_n(F)$.
You are looking for matrix similarity, and you can read about the conditions in the link above.
--
Edit: As the question was edited, now the answer is yes: For every matrix $A$ one can find its Jordan canonical form, $J_A$. Find the base change matrix $P_A$. Since $A$, $B$ are similar if and only if $J_A=J_B$, we have $J=P_AAP_A^{-1}=P_BBP_B^{-1}$. So $A=(P_B^{-1}P_A)^{-1}B(P_B^{-1}P_A)$
answered Dec 10 '13 at 20:55
Dennis GulkoDennis Gulko
13.9k12655
sorry, I edited the text
– giulio
Dec 10 '13 at 21:00
add a comment |
The short answer will be no. For example, let $Aneq I$ and $B=I$. Then clearly $P^{-1}BP=P^{-1}IP=Ineq A$ for all $Pin GL_n(F)$.
You are looking for matrix similarity, and you can read about the conditions in the link above.
--
Edit: As the question was edited, now the answer is yes: For every matrix $A$ one can find its Jordan canonical form, $J_A$. Find the base change matrix $P_A$. Since $A$, $B$ are similar if and only if $J_A=J_B$, we have $J=P_AAP_A^{-1}=P_BBP_B^{-1}$. So $A=(P_B^{-1}P_A)^{-1}B(P_B^{-1}P_A)$
answered Dec 10 '13 at 20:55
Dennis GulkoDennis Gulko
13.9k12655
The short answer will be no. For example, let $Aneq I$ and $B=I$. Then clearly $P^{-1}BP=P^{-1}IP=Ineq A$ for all $Pin GL_n(F)$.
You are looking for matrix similarity, and you can read about the conditions in the link above.
--
Edit: As the question was edited, now the answer is yes: For every matrix $A$ one can find its Jordan canonical form, $J_A$. Find the base change matrix $P_A$. Since $A$, $B$ are similar if and only if $J_A=J_B$, we have $J=P_AAP_A^{-1}=P_BBP_B^{-1}$. So $A=(P_B^{-1}P_A)^{-1}B(P_B^{-1}P_A)$
answered Dec 10 '13 at 20:55
Dennis GulkoDennis Gulko
13.9k12655
edited Dec 10 '13 at 21:05
answered Dec 10 '13 at 20:55
Dennis GulkoDennis Gulko
13.9k12655
answered Dec 10 '13 at 20:55
Dennis GulkoDennis Gulko
13.9k12655
answered Dec 10 '13 at 20:55
Dennis GulkoDennis Gulko
13.9k12655
13.9k12655
sorry, I edited the text
– giulio
Dec 10 '13 at 21:00
add a comment |
sorry, I edited the text
– giulio
Dec 10 '13 at 21:00
sorry, I edited the text
– giulio
Dec 10 '13 at 21:00
sorry, I edited the text
– giulio
Dec 10 '13 at 21:00
add a comment |
Let $K$ be a subfield of $mathbb{C}$. We assume that $A,Bin M_n(K)$ are similar (that is, there is $Pin M_n(K)$ s.t. $det(P)not= 0$ and $PA=BP$). Note that we can always choose $P$ in $M_n(K)$.
Of course, it is important not to calculate Jordan's forms of $A,B$.
METHOD 1.
i) Solve the linear system $PA=BP$; the set of solutions $E$ is a sub-vector space of $M_n(K)$ of dimension $k=dim(C(A))geq n$. Note that, using the tensor product, there are methods that permit to solve such a system with complexity in $O(n^3)$ (and not in $O(n^6)$).
Since there is $Pin E$ s.t. $det(P)not= 0$, ${Pin GL_n(K);PA=BP}$ is Zariski open dense in $E$.
ii) Thus, it suffices to randomly choose the $k$ parameters that define an element of $E$ to obtain a required solution with probability $1$.
METHOD 2.
We can also use the Frobenius form over $K$. Indeed, $A,B$ have same Frobenius form $F$:
$A=QFQ^{-1},B=RFR^{-1}$ and the sequel is easy.
EDIT. Here is a quick history of the progress made in calculating the Frobenius form.
i) in "Nearly optimal algorithms for canonical matrix forms. SIAM Journal on Computing, 24:948–969, 1995"
M. Giesbrech gives a randomized method: a Las Vegas algorithm can be used to compute both
the Frobenius form and a Frobenius transition matrix for a given $ntimes n$ matrix over a field $F$ using an expected number of operations over $F$ that is in $O(n^3)$, with standard matrix and polynomial arithmetic, whenever $F$ has at least $n^2$ elements (to ensure a probability of success at least
$1/2$), and using an expected number of
operations in $O(n^3log_q(n))$ if $F$ is a finite field with size $q$.
ii) Storjohann [1997] exposed a deterministic method whose complexity is $O(n^3)$. cf.
http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.28.2505
However, the previous method does not give, at the same time, the matrix of change of basis.
iii) In "Algorithms for similarity transforms. In
Seventh Rhine Work-shop on Computer Algebra
, Bregenz, Austria, March 2000"
A. Storjohann and G. Villard obtain (with the same complexity), at the same time, the Frobenius form $F$ and the transform matrix $V$ .
There is a gap. None of the previous methods allow to use the fast multiplication of Strassen (in $O(n^{2.81})$).
iv) W. Eberly, in "Asymptotically efficient algorithms for the
Frobenius form. Technical report, Department of
Computer Science, University of Calgary, 2000"
gives $F,V$ thanks to a Las Vegas algorithm (using Strassen) that has complexity $O(n^{2.81}log(n))$.
Note that, unlike an urban legend, the methods of Coppersmith-Winograd and of Le Gall are practically useless.
v) In 2007, C. Pernet and A. Storjohann, in "Faster Algorithms for the Characteristic Polynomial. ISSAC"
give $F$ (but not $V$) with a Las Vegas algorithm in $O(n^{2.81})$.
This is the state of play in 2011-2013. I don't know if an algorithm giving $F, V$ in $O(n^{2.81})$ was found in recent years.
answered Nov 22 '18 at 13:45
loup blancloup blanc
22.5k21850
I wonder if there are improvements in $O(n^3)$ since 1997, and the open questions they put in their paper?
– i707107
Nov 23 '18 at 23:07
@i707107 , read my edit.
– loup blanc
Nov 27 '18 at 17:05
add a comment |
Let $K$ be a subfield of $mathbb{C}$. We assume that $A,Bin M_n(K)$ are similar (that is, there is $Pin M_n(K)$ s.t. $det(P)not= 0$ and $PA=BP$). Note that we can always choose $P$ in $M_n(K)$.
Of course, it is important not to calculate Jordan's forms of $A,B$.
METHOD 1.
i) Solve the linear system $PA=BP$; the set of solutions $E$ is a sub-vector space of $M_n(K)$ of dimension $k=dim(C(A))geq n$. Note that, using the tensor product, there are methods that permit to solve such a system with complexity in $O(n^3)$ (and not in $O(n^6)$).
Since there is $Pin E$ s.t. $det(P)not= 0$, ${Pin GL_n(K);PA=BP}$ is Zariski open dense in $E$.
ii) Thus, it suffices to randomly choose the $k$ parameters that define an element of $E$ to obtain a required solution with probability $1$.
METHOD 2.
We can also use the Frobenius form over $K$. Indeed, $A,B$ have same Frobenius form $F$:
$A=QFQ^{-1},B=RFR^{-1}$ and the sequel is easy.
EDIT. Here is a quick history of the progress made in calculating the Frobenius form.
i) in "Nearly optimal algorithms for canonical matrix forms. SIAM Journal on Computing, 24:948–969, 1995"
M. Giesbrech gives a randomized method: a Las Vegas algorithm can be used to compute both
the Frobenius form and a Frobenius transition matrix for a given $ntimes n$ matrix over a field $F$ using an expected number of operations over $F$ that is in $O(n^3)$, with standard matrix and polynomial arithmetic, whenever $F$ has at least $n^2$ elements (to ensure a probability of success at least
$1/2$), and using an expected number of
operations in $O(n^3log_q(n))$ if $F$ is a finite field with size $q$.
ii) Storjohann [1997] exposed a deterministic method whose complexity is $O(n^3)$. cf.
http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.28.2505
However, the previous method does not give, at the same time, the matrix of change of basis.
iii) In "Algorithms for similarity transforms. In
Seventh Rhine Work-shop on Computer Algebra
, Bregenz, Austria, March 2000"
A. Storjohann and G. Villard obtain (with the same complexity), at the same time, the Frobenius form $F$ and the transform matrix $V$ .
There is a gap. None of the previous methods allow to use the fast multiplication of Strassen (in $O(n^{2.81})$).
iv) W. Eberly, in "Asymptotically efficient algorithms for the
Frobenius form. Technical report, Department of
Computer Science, University of Calgary, 2000"
gives $F,V$ thanks to a Las Vegas algorithm (using Strassen) that has complexity $O(n^{2.81}log(n))$.
Note that, unlike an urban legend, the methods of Coppersmith-Winograd and of Le Gall are practically useless.
v) In 2007, C. Pernet and A. Storjohann, in "Faster Algorithms for the Characteristic Polynomial. ISSAC"
give $F$ (but not $V$) with a Las Vegas algorithm in $O(n^{2.81})$.
This is the state of play in 2011-2013. I don't know if an algorithm giving $F, V$ in $O(n^{2.81})$ was found in recent years.
answered Nov 22 '18 at 13:45
loup blancloup blanc
22.5k21850
I wonder if there are improvements in $O(n^3)$ since 1997, and the open questions they put in their paper?
– i707107
Nov 23 '18 at 23:07
@i707107 , read my edit.
– loup blanc
Nov 27 '18 at 17:05
add a comment |
Let $K$ be a subfield of $mathbb{C}$. We assume that $A,Bin M_n(K)$ are similar (that is, there is $Pin M_n(K)$ s.t. $det(P)not= 0$ and $PA=BP$). Note that we can always choose $P$ in $M_n(K)$.
Of course, it is important not to calculate Jordan's forms of $A,B$.
METHOD 1.
i) Solve the linear system $PA=BP$; the set of solutions $E$ is a sub-vector space of $M_n(K)$ of dimension $k=dim(C(A))geq n$. Note that, using the tensor product, there are methods that permit to solve such a system with complexity in $O(n^3)$ (and not in $O(n^6)$).
Since there is $Pin E$ s.t. $det(P)not= 0$, ${Pin GL_n(K);PA=BP}$ is Zariski open dense in $E$.
ii) Thus, it suffices to randomly choose the $k$ parameters that define an element of $E$ to obtain a required solution with probability $1$.
METHOD 2.
We can also use the Frobenius form over $K$. Indeed, $A,B$ have same Frobenius form $F$:
$A=QFQ^{-1},B=RFR^{-1}$ and the sequel is easy.
EDIT. Here is a quick history of the progress made in calculating the Frobenius form.
i) in "Nearly optimal algorithms for canonical matrix forms. SIAM Journal on Computing, 24:948–969, 1995"
M. Giesbrech gives a randomized method: a Las Vegas algorithm can be used to compute both
the Frobenius form and a Frobenius transition matrix for a given $ntimes n$ matrix over a field $F$ using an expected number of operations over $F$ that is in $O(n^3)$, with standard matrix and polynomial arithmetic, whenever $F$ has at least $n^2$ elements (to ensure a probability of success at least
$1/2$), and using an expected number of
operations in $O(n^3log_q(n))$ if $F$ is a finite field with size $q$.
ii) Storjohann [1997] exposed a deterministic method whose complexity is $O(n^3)$. cf.
http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.28.2505
However, the previous method does not give, at the same time, the matrix of change of basis.
iii) In "Algorithms for similarity transforms. In
Seventh Rhine Work-shop on Computer Algebra
, Bregenz, Austria, March 2000"
A. Storjohann and G. Villard obtain (with the same complexity), at the same time, the Frobenius form $F$ and the transform matrix $V$ .
There is a gap. None of the previous methods allow to use the fast multiplication of Strassen (in $O(n^{2.81})$).
iv) W. Eberly, in "Asymptotically efficient algorithms for the
Frobenius form. Technical report, Department of
Computer Science, University of Calgary, 2000"
gives $F,V$ thanks to a Las Vegas algorithm (using Strassen) that has complexity $O(n^{2.81}log(n))$.
Note that, unlike an urban legend, the methods of Coppersmith-Winograd and of Le Gall are practically useless.
v) In 2007, C. Pernet and A. Storjohann, in "Faster Algorithms for the Characteristic Polynomial. ISSAC"
give $F$ (but not $V$) with a Las Vegas algorithm in $O(n^{2.81})$.
This is the state of play in 2011-2013. I don't know if an algorithm giving $F, V$ in $O(n^{2.81})$ was found in recent years.
answered Nov 22 '18 at 13:45
loup blancloup blanc
22.5k21850
Let $K$ be a subfield of $mathbb{C}$. We assume that $A,Bin M_n(K)$ are similar (that is, there is $Pin M_n(K)$ s.t. $det(P)not= 0$ and $PA=BP$). Note that we can always choose $P$ in $M_n(K)$.
Of course, it is important not to calculate Jordan's forms of $A,B$.
METHOD 1.
i) Solve the linear system $PA=BP$; the set of solutions $E$ is a sub-vector space of $M_n(K)$ of dimension $k=dim(C(A))geq n$. Note that, using the tensor product, there are methods that permit to solve such a system with complexity in $O(n^3)$ (and not in $O(n^6)$).
Since there is $Pin E$ s.t. $det(P)not= 0$, ${Pin GL_n(K);PA=BP}$ is Zariski open dense in $E$.
ii) Thus, it suffices to randomly choose the $k$ parameters that define an element of $E$ to obtain a required solution with probability $1$.
METHOD 2.
We can also use the Frobenius form over $K$. Indeed, $A,B$ have same Frobenius form $F$:
$A=QFQ^{-1},B=RFR^{-1}$ and the sequel is easy.
EDIT. Here is a quick history of the progress made in calculating the Frobenius form.
i) in "Nearly optimal algorithms for canonical matrix forms. SIAM Journal on Computing, 24:948–969, 1995"
M. Giesbrech gives a randomized method: a Las Vegas algorithm can be used to compute both
the Frobenius form and a Frobenius transition matrix for a given $ntimes n$ matrix over a field $F$ using an expected number of operations over $F$ that is in $O(n^3)$, with standard matrix and polynomial arithmetic, whenever $F$ has at least $n^2$ elements (to ensure a probability of success at least
$1/2$), and using an expected number of
operations in $O(n^3log_q(n))$ if $F$ is a finite field with size $q$.
ii) Storjohann [1997] exposed a deterministic method whose complexity is $O(n^3)$. cf.
http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.28.2505
However, the previous method does not give, at the same time, the matrix of change of basis.
iii) In "Algorithms for similarity transforms. In
Seventh Rhine Work-shop on Computer Algebra
, Bregenz, Austria, March 2000"
A. Storjohann and G. Villard obtain (with the same complexity), at the same time, the Frobenius form $F$ and the transform matrix $V$ .
There is a gap. None of the previous methods allow to use the fast multiplication of Strassen (in $O(n^{2.81})$).
iv) W. Eberly, in "Asymptotically efficient algorithms for the
Frobenius form. Technical report, Department of
Computer Science, University of Calgary, 2000"
gives $F,V$ thanks to a Las Vegas algorithm (using Strassen) that has complexity $O(n^{2.81}log(n))$.
Note that, unlike an urban legend, the methods of Coppersmith-Winograd and of Le Gall are practically useless.
v) In 2007, C. Pernet and A. Storjohann, in "Faster Algorithms for the Characteristic Polynomial. ISSAC"
give $F$ (but not $V$) with a Las Vegas algorithm in $O(n^{2.81})$.
This is the state of play in 2011-2013. I don't know if an algorithm giving $F, V$ in $O(n^{2.81})$ was found in recent years.
answered Nov 22 '18 at 13:45
loup blancloup blanc
22.5k21850
edited Nov 27 '18 at 17:05
answered Nov 22 '18 at 13:45
loup blancloup blanc
22.5k21850
answered Nov 22 '18 at 13:45
loup blancloup blanc
22.5k21850
answered Nov 22 '18 at 13:45
loup blancloup blanc
22.5k21850
22.5k21850
I wonder if there are improvements in $O(n^3)$ since 1997, and the open questions they put in their paper?
– i707107
Nov 23 '18 at 23:07
@i707107 , read my edit.
– loup blanc
Nov 27 '18 at 17:05
add a comment |
I wonder if there are improvements in $O(n^3)$ since 1997, and the open questions they put in their paper?
– i707107
Nov 23 '18 at 23:07
@i707107 , read my edit.
– loup blanc
Nov 27 '18 at 17:05
I wonder if there are improvements in $O(n^3)$ since 1997, and the open questions they put in their paper?
– i707107
Nov 23 '18 at 23:07
I wonder if there are improvements in $O(n^3)$ since 1997, and the open questions they put in their paper?
– i707107
Nov 23 '18 at 23:07
@i707107 , read my edit.
– loup blanc
Nov 27 '18 at 17:05
@i707107 , read my edit.
– loup blanc
Nov 27 '18 at 17:05
add a comment |
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The definition of similarity between $A$ and $B$ is that there exists invertible $P$ such that $A = P^{-1}BP$.
– EuYu
Dec 10 '13 at 21:02
According to the definition of similarity one can say the following: When someone sells you two matrices $A$ and $B$, claiming that they are similar, he has to provide a $P$ with $A=P^{-1}B P$ to testify for this similarity.
– Christian Blatter
Dec 10 '13 at 21:02
There are a lot of "quantities" that are preserved by conjugation, for example if $A$ and $B$ have different ranks, determinants, sets of eigenvalues, and so on, then $A$ and $B$ can't be similar, meaning that no such $P$ exists in those cases.
– Jeppe Stig Nielsen
Dec 10 '13 at 21:02
I assume that $A$ and $B$ are similar and want to find $P$ that makes them similar
– giulio
Dec 10 '13 at 21:07
Through Smith normal form of the matrices, we are able to write an explicit algorithm producing such matrix $P$. This does not require the field to be algebraically closed.
– i707107
Nov 22 '18 at 2:51