Assess the Lie-bracket of random matrices.

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
3
down vote

favorite












Assume we have $A, B in mathbbS_n(mathbfR)$, where $mathbbS_n(mathbfR)$ is the set of all symmetric matrices with real entries and size $n$. The Lie-bracket (or, the commutator for a matrix group) is defined as follows
$$
[A, B] = AB - BA.
$$
Now assume $A, B$ are random matrices (feel free to assume any particular ensemble of matrices, say Gaussian ensembles). Are there such "special" random matrices that something could be said about $[A, B]$? Maybe some bounds on its operator norm?



Any other properties along with suggestions and comments will be highly appreciated.



EDIT (Aug 4, 2018): Notice that $(AB)_ij = (BA)_ji$ yielding the following simple fact: $[A, B]_ij = -[A, B]_ji$. The latter, in particular, implies that $[A, B]_ii = 0$ as noticed by daruma in comments.



So, the lie bracket $[A, B]$ can be rewritten as follows:
$$
[A, B] = beginpmatrix 0 & & S \ & ddots\ -S^T & & 0endpmatrix := S_1 - S_2,
$$
where $S$ a diagonal block and matrices $S_1$ and $S_2$ can be viewed as upper and lower triangular matrices with corresponding blocks of $S$ and $S^T$, respectively.




linear-algebra lie-groups symmetric-matrices random-matrices




share|cite|improve this question





















  • symmetric matrices are not necessarily commuting. consider $A = beginpmatrix 1 & 1 \ 1 & 0 endpmatrix$ and $B = beginpmatrix 1 & 2 \ 2 & 4 endpmatrix$. Then $$ [A, B] = AB - BA = beginpmatrix 3 & 6 \ 1 & 2 endpmatrix - beginpmatrix 3 & 1 \ 6 & 2 endpmatrix neq mathbf0_2 times 2. $$ EDIT: @daruma I've corrected my comment, sorry for typos.
    – pointguard0
    Aug 1 at 10:41











  • Yeah, sorry my mistake. Already deleted the comment before you said it. (But $B$ is not symmetric in that one.)
    – daruma
    Aug 1 at 10:42










  • You can however say that the diagonal elements should be $0$.
    – daruma
    Aug 1 at 10:42










  • $(AB)_iisum_ja_ij b_ji=sum_jb_jia_ji=(BA)_ii$
    – daruma
    Aug 1 at 10:44







  • 1




    Not about symmetric ones but general real valued ones: sciencedirect.com/science/article/pii/S0024379505000832.
    – Rudi_Birnbaum
    2 days ago














up vote
3
down vote

favorite












Assume we have $A, B in mathbbS_n(mathbfR)$, where $mathbbS_n(mathbfR)$ is the set of all symmetric matrices with real entries and size $n$. The Lie-bracket (or, the commutator for a matrix group) is defined as follows
$$
[A, B] = AB - BA.
$$
Now assume $A, B$ are random matrices (feel free to assume any particular ensemble of matrices, say Gaussian ensembles). Are there such "special" random matrices that something could be said about $[A, B]$? Maybe some bounds on its operator norm?



Any other properties along with suggestions and comments will be highly appreciated.



EDIT (Aug 4, 2018): Notice that $(AB)_ij = (BA)_ji$ yielding the following simple fact: $[A, B]_ij = -[A, B]_ji$. The latter, in particular, implies that $[A, B]_ii = 0$ as noticed by daruma in comments.



So, the lie bracket $[A, B]$ can be rewritten as follows:
$$
[A, B] = beginpmatrix 0 & & S \ & ddots\ -S^T & & 0endpmatrix := S_1 - S_2,
$$
where $S$ a diagonal block and matrices $S_1$ and $S_2$ can be viewed as upper and lower triangular matrices with corresponding blocks of $S$ and $S^T$, respectively.




linear-algebra lie-groups symmetric-matrices random-matrices




share|cite|improve this question





















  • symmetric matrices are not necessarily commuting. consider $A = beginpmatrix 1 & 1 \ 1 & 0 endpmatrix$ and $B = beginpmatrix 1 & 2 \ 2 & 4 endpmatrix$. Then $$ [A, B] = AB - BA = beginpmatrix 3 & 6 \ 1 & 2 endpmatrix - beginpmatrix 3 & 1 \ 6 & 2 endpmatrix neq mathbf0_2 times 2. $$ EDIT: @daruma I've corrected my comment, sorry for typos.
    – pointguard0
    Aug 1 at 10:41











  • Yeah, sorry my mistake. Already deleted the comment before you said it. (But $B$ is not symmetric in that one.)
    – daruma
    Aug 1 at 10:42










  • You can however say that the diagonal elements should be $0$.
    – daruma
    Aug 1 at 10:42










  • $(AB)_iisum_ja_ij b_ji=sum_jb_jia_ji=(BA)_ii$
    – daruma
    Aug 1 at 10:44







  • 1




    Not about symmetric ones but general real valued ones: sciencedirect.com/science/article/pii/S0024379505000832.
    – Rudi_Birnbaum
    2 days ago












up vote
3
down vote

favorite









up vote
3
down vote

favorite











Assume we have $A, B in mathbbS_n(mathbfR)$, where $mathbbS_n(mathbfR)$ is the set of all symmetric matrices with real entries and size $n$. The Lie-bracket (or, the commutator for a matrix group) is defined as follows
$$
[A, B] = AB - BA.
$$
Now assume $A, B$ are random matrices (feel free to assume any particular ensemble of matrices, say Gaussian ensembles). Are there such "special" random matrices that something could be said about $[A, B]$? Maybe some bounds on its operator norm?



Any other properties along with suggestions and comments will be highly appreciated.



EDIT (Aug 4, 2018): Notice that $(AB)_ij = (BA)_ji$ yielding the following simple fact: $[A, B]_ij = -[A, B]_ji$. The latter, in particular, implies that $[A, B]_ii = 0$ as noticed by daruma in comments.



So, the lie bracket $[A, B]$ can be rewritten as follows:
$$
[A, B] = beginpmatrix 0 & & S \ & ddots\ -S^T & & 0endpmatrix := S_1 - S_2,
$$
where $S$ a diagonal block and matrices $S_1$ and $S_2$ can be viewed as upper and lower triangular matrices with corresponding blocks of $S$ and $S^T$, respectively.




linear-algebra lie-groups symmetric-matrices random-matrices




share|cite|improve this question













Assume we have $A, B in mathbbS_n(mathbfR)$, where $mathbbS_n(mathbfR)$ is the set of all symmetric matrices with real entries and size $n$. The Lie-bracket (or, the commutator for a matrix group) is defined as follows
$$
[A, B] = AB - BA.
$$
Now assume $A, B$ are random matrices (feel free to assume any particular ensemble of matrices, say Gaussian ensembles). Are there such "special" random matrices that something could be said about $[A, B]$? Maybe some bounds on its operator norm?



Any other properties along with suggestions and comments will be highly appreciated.



EDIT (Aug 4, 2018): Notice that $(AB)_ij = (BA)_ji$ yielding the following simple fact: $[A, B]_ij = -[A, B]_ji$. The latter, in particular, implies that $[A, B]_ii = 0$ as noticed by daruma in comments.



So, the lie bracket $[A, B]$ can be rewritten as follows:
$$
[A, B] = beginpmatrix 0 & & S \ & ddots\ -S^T & & 0endpmatrix := S_1 - S_2,
$$
where $S$ a diagonal block and matrices $S_1$ and $S_2$ can be viewed as upper and lower triangular matrices with corresponding blocks of $S$ and $S^T$, respectively.





linear-algebra lie-groups symmetric-matrices random-matrices





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited 2 days ago
























asked Aug 1 at 10:17









pointguard0

705517




705517











  • symmetric matrices are not necessarily commuting. consider $A = beginpmatrix 1 & 1 \ 1 & 0 endpmatrix$ and $B = beginpmatrix 1 & 2 \ 2 & 4 endpmatrix$. Then $$ [A, B] = AB - BA = beginpmatrix 3 & 6 \ 1 & 2 endpmatrix - beginpmatrix 3 & 1 \ 6 & 2 endpmatrix neq mathbf0_2 times 2. $$ EDIT: @daruma I've corrected my comment, sorry for typos.
    – pointguard0
    Aug 1 at 10:41











  • Yeah, sorry my mistake. Already deleted the comment before you said it. (But $B$ is not symmetric in that one.)
    – daruma
    Aug 1 at 10:42










  • You can however say that the diagonal elements should be $0$.
    – daruma
    Aug 1 at 10:42










  • $(AB)_iisum_ja_ij b_ji=sum_jb_jia_ji=(BA)_ii$
    – daruma
    Aug 1 at 10:44







  • 1




    Not about symmetric ones but general real valued ones: sciencedirect.com/science/article/pii/S0024379505000832.
    – Rudi_Birnbaum
    2 days ago
















  • symmetric matrices are not necessarily commuting. consider $A = beginpmatrix 1 & 1 \ 1 & 0 endpmatrix$ and $B = beginpmatrix 1 & 2 \ 2 & 4 endpmatrix$. Then $$ [A, B] = AB - BA = beginpmatrix 3 & 6 \ 1 & 2 endpmatrix - beginpmatrix 3 & 1 \ 6 & 2 endpmatrix neq mathbf0_2 times 2. $$ EDIT: @daruma I've corrected my comment, sorry for typos.
    – pointguard0
    Aug 1 at 10:41











  • Yeah, sorry my mistake. Already deleted the comment before you said it. (But $B$ is not symmetric in that one.)
    – daruma
    Aug 1 at 10:42










  • You can however say that the diagonal elements should be $0$.
    – daruma
    Aug 1 at 10:42










  • $(AB)_iisum_ja_ij b_ji=sum_jb_jia_ji=(BA)_ii$
    – daruma
    Aug 1 at 10:44







  • 1




    Not about symmetric ones but general real valued ones: sciencedirect.com/science/article/pii/S0024379505000832.
    – Rudi_Birnbaum
    2 days ago















symmetric matrices are not necessarily commuting. consider $A = beginpmatrix 1 & 1 \ 1 & 0 endpmatrix$ and $B = beginpmatrix 1 & 2 \ 2 & 4 endpmatrix$. Then $$ [A, B] = AB - BA = beginpmatrix 3 & 6 \ 1 & 2 endpmatrix - beginpmatrix 3 & 1 \ 6 & 2 endpmatrix neq mathbf0_2 times 2. $$ EDIT: @daruma I've corrected my comment, sorry for typos.
– pointguard0
Aug 1 at 10:41





symmetric matrices are not necessarily commuting. consider $A = beginpmatrix 1 & 1 \ 1 & 0 endpmatrix$ and $B = beginpmatrix 1 & 2 \ 2 & 4 endpmatrix$. Then $$ [A, B] = AB - BA = beginpmatrix 3 & 6 \ 1 & 2 endpmatrix - beginpmatrix 3 & 1 \ 6 & 2 endpmatrix neq mathbf0_2 times 2. $$ EDIT: @daruma I've corrected my comment, sorry for typos.
– pointguard0
Aug 1 at 10:41













Yeah, sorry my mistake. Already deleted the comment before you said it. (But $B$ is not symmetric in that one.)
– daruma
Aug 1 at 10:42




Yeah, sorry my mistake. Already deleted the comment before you said it. (But $B$ is not symmetric in that one.)
– daruma
Aug 1 at 10:42












You can however say that the diagonal elements should be $0$.
– daruma
Aug 1 at 10:42




You can however say that the diagonal elements should be $0$.
– daruma
Aug 1 at 10:42












$(AB)_iisum_ja_ij b_ji=sum_jb_jia_ji=(BA)_ii$
– daruma
Aug 1 at 10:44





$(AB)_iisum_ja_ij b_ji=sum_jb_jia_ji=(BA)_ii$
– daruma
Aug 1 at 10:44





1




1




Not about symmetric ones but general real valued ones: sciencedirect.com/science/article/pii/S0024379505000832.
– Rudi_Birnbaum
2 days ago




Not about symmetric ones but general real valued ones: sciencedirect.com/science/article/pii/S0024379505000832.
– Rudi_Birnbaum
2 days ago















active

oldest

votes











Your Answer




StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);








 

draft saved


draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2868922%2fassess-the-lie-bracket-of-random-matrices%23new-answer', 'question_page');

);

Post as a guest






















active

oldest

votes













active

oldest

votes









active

oldest

votes






active

oldest

votes










 

draft saved


draft discarded


























 


draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2868922%2fassess-the-lie-bracket-of-random-matrices%23new-answer', 'question_page');

);

Post as a guest
































































Comments

Post a Comment

Popular posts from this blog

What is the equation of a 3D cone with generalised tilt?

Image
Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite What is the equation of a 3D cone with generalized tilt? I've noticed that in most equations given to represent a cone, there is no parameter which defines the tilt of the cone in 3D space and that most of them have their point at the origin $(0,0,0)$ - I was wondering if anyone could give me a more generalized cone equation for a cone in any position in 3D space 3d share | cite | improve this question edited Jul 25 '16 at 0:05 ervx 9,425 3 13 36 asked Jul 24 '16 at 15:38 Charlene 11 2 2 Welcome to Math.SE! Could you please explain a few things to help people give an answer useful to you: 1. What do you mean by "generalized tilt"? 2. Are you asking about a right circular cone? Does the angle at the vertex matter? 3. What form of answer (implicit, parametric...?) are you looking for? 4. If this is homework, what tools do you have available, an...
Read more

Color the edges and diagonals of a regular polygon

Image
Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 1 Here is the problem: For what $n$ is it possible to color the edges and diagonals of an $n$-side regular polygon with $dfracbinomn23$ colors, such that you use every color exactly three times and for every color the three segments (edges or diagonals) with that color form a triangle? Trivially $nequiv 0 text or 1 pmod3$. I can also prove that if the statement is true for $k$ than it is true for $3k$ as well. How to finish? Please help! Thanks combinatorial-geometry share | cite | improve this question edited Aug 6 at 21:57 alcana 118 4 asked Aug 6 at 21:15 Leo Gardner 356 11 Even assuming "polyhpn" in the title is a typo for polygon , it is unclear what you mean by "the edges and sides". Aren't the side of a regular polygon the same as the edges? A regular $n$-sided polygon has $n$ edges. – hardmath Aug 6 at 21:20 3 $n$ ne...
Read more

Relationship between determinant of matrix and determinant of adjoint?

Image
Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it. linear-algebra determinant adjoint-operators share | cite | improve this question asked Nov 17 '17 at 17:32 CluelessCoder 32 5 For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants). – Prasun Biswas Nov 17 '17 at 17:35 1 @CluelessCoder: see here: math.stackexchange.com/questions/516127/…...
Read more