Is power distributable in product of two matrices? [closed]












-1












$begingroup$


given two matrices A and B, is the following equality correct?
$(AB)^n$=$A^n$$B^n$ for $nin$ $Bbb R$






matrices







share|cite|improve this question











$endgroup$



closed as off-topic by Servaes, Did, Adrian Keister, José Carlos Santos, Lord_Farin Jan 21 at 17:15


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Servaes, Did, Adrian Keister, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    No, it is not. Also, how do you take a non-integer power of a matrix?
    $endgroup$
    – Servaes
    Jan 21 at 12:58










  • $begingroup$
    "for $nin$ $Bbb R$" Sure about this?
    $endgroup$
    – Did
    Jan 21 at 12:58










  • $begingroup$
    yes for example n=0.5
    $endgroup$
    – Parsa Parvanehro
    Jan 21 at 13:05












  • $begingroup$
    Please don't just spam tags with "matrix" in them just because your question is about matrices.
    $endgroup$
    – Lord_Farin
    Jan 21 at 17:15
















-1












$begingroup$


given two matrices A and B, is the following equality correct?
$(AB)^n$=$A^n$$B^n$ for $nin$ $Bbb R$






matrices







share|cite|improve this question











$endgroup$



closed as off-topic by Servaes, Did, Adrian Keister, José Carlos Santos, Lord_Farin Jan 21 at 17:15


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Servaes, Did, Adrian Keister, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    No, it is not. Also, how do you take a non-integer power of a matrix?
    $endgroup$
    – Servaes
    Jan 21 at 12:58










  • $begingroup$
    "for $nin$ $Bbb R$" Sure about this?
    $endgroup$
    – Did
    Jan 21 at 12:58










  • $begingroup$
    yes for example n=0.5
    $endgroup$
    – Parsa Parvanehro
    Jan 21 at 13:05












  • $begingroup$
    Please don't just spam tags with "matrix" in them just because your question is about matrices.
    $endgroup$
    – Lord_Farin
    Jan 21 at 17:15














-1












-1








-1





$begingroup$


given two matrices A and B, is the following equality correct?
$(AB)^n$=$A^n$$B^n$ for $nin$ $Bbb R$






matrices







share|cite|improve this question











$endgroup$




given two matrices A and B, is the following equality correct?
$(AB)^n$=$A^n$$B^n$ for $nin$ $Bbb R$






matrices




matrices






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 21 at 17:15









Lord_Farin

15.6k636108




15.6k636108










asked Jan 21 at 12:56









Parsa ParvanehroParsa Parvanehro

103




103




closed as off-topic by Servaes, Did, Adrian Keister, José Carlos Santos, Lord_Farin Jan 21 at 17:15


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Servaes, Did, Adrian Keister, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by Servaes, Did, Adrian Keister, José Carlos Santos, Lord_Farin Jan 21 at 17:15


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Servaes, Did, Adrian Keister, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.












  • $begingroup$
    No, it is not. Also, how do you take a non-integer power of a matrix?
    $endgroup$
    – Servaes
    Jan 21 at 12:58










  • $begingroup$
    "for $nin$ $Bbb R$" Sure about this?
    $endgroup$
    – Did
    Jan 21 at 12:58










  • $begingroup$
    yes for example n=0.5
    $endgroup$
    – Parsa Parvanehro
    Jan 21 at 13:05












  • $begingroup$
    Please don't just spam tags with "matrix" in them just because your question is about matrices.
    $endgroup$
    – Lord_Farin
    Jan 21 at 17:15


















  • $begingroup$
    No, it is not. Also, how do you take a non-integer power of a matrix?
    $endgroup$
    – Servaes
    Jan 21 at 12:58










  • $begingroup$
    "for $nin$ $Bbb R$" Sure about this?
    $endgroup$
    – Did
    Jan 21 at 12:58










  • $begingroup$
    yes for example n=0.5
    $endgroup$
    – Parsa Parvanehro
    Jan 21 at 13:05












  • $begingroup$
    Please don't just spam tags with "matrix" in them just because your question is about matrices.
    $endgroup$
    – Lord_Farin
    Jan 21 at 17:15
















$begingroup$
No, it is not. Also, how do you take a non-integer power of a matrix?
$endgroup$
– Servaes
Jan 21 at 12:58




$begingroup$
No, it is not. Also, how do you take a non-integer power of a matrix?
$endgroup$
– Servaes
Jan 21 at 12:58












$begingroup$
"for $nin$ $Bbb R$" Sure about this?
$endgroup$
– Did
Jan 21 at 12:58




$begingroup$
"for $nin$ $Bbb R$" Sure about this?
$endgroup$
– Did
Jan 21 at 12:58












$begingroup$
yes for example n=0.5
$endgroup$
– Parsa Parvanehro
Jan 21 at 13:05






$begingroup$
yes for example n=0.5
$endgroup$
– Parsa Parvanehro
Jan 21 at 13:05














$begingroup$
Please don't just spam tags with "matrix" in them just because your question is about matrices.
$endgroup$
– Lord_Farin
Jan 21 at 17:15




$begingroup$
Please don't just spam tags with "matrix" in them just because your question is about matrices.
$endgroup$
– Lord_Farin
Jan 21 at 17:15










1 Answer
1






active

oldest

votes


















2












$begingroup$

No, it is true for $ninmathbb N$ if $AB=BA$. For $ninmathbb N$,
$$
(AB)^n=ABABdots AB.
$$



For $nin-mathbb N$ it would be the $|n|$ power of the inverse of $AB$, assuming it is invertible. For $ninmathbb Rbackslashmathbb Z$, this is no-longer well defined. For example matrices tend have many square roots.






share|cite|improve this answer









$endgroup$




















    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    No, it is true for $ninmathbb N$ if $AB=BA$. For $ninmathbb N$,
    $$
    (AB)^n=ABABdots AB.
    $$



    For $nin-mathbb N$ it would be the $|n|$ power of the inverse of $AB$, assuming it is invertible. For $ninmathbb Rbackslashmathbb Z$, this is no-longer well defined. For example matrices tend have many square roots.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      No, it is true for $ninmathbb N$ if $AB=BA$. For $ninmathbb N$,
      $$
      (AB)^n=ABABdots AB.
      $$



      For $nin-mathbb N$ it would be the $|n|$ power of the inverse of $AB$, assuming it is invertible. For $ninmathbb Rbackslashmathbb Z$, this is no-longer well defined. For example matrices tend have many square roots.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        No, it is true for $ninmathbb N$ if $AB=BA$. For $ninmathbb N$,
        $$
        (AB)^n=ABABdots AB.
        $$



        For $nin-mathbb N$ it would be the $|n|$ power of the inverse of $AB$, assuming it is invertible. For $ninmathbb Rbackslashmathbb Z$, this is no-longer well defined. For example matrices tend have many square roots.






        share|cite|improve this answer









        $endgroup$



        No, it is true for $ninmathbb N$ if $AB=BA$. For $ninmathbb N$,
        $$
        (AB)^n=ABABdots AB.
        $$



        For $nin-mathbb N$ it would be the $|n|$ power of the inverse of $AB$, assuming it is invertible. For $ninmathbb Rbackslashmathbb Z$, this is no-longer well defined. For example matrices tend have many square roots.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 21 at 12:58









        Alec B-GAlec B-G

        49519




        49519















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