$mathbb{S} = {MX - XM mid X in mathbb{R}^{ntimes n}}$ gives $dim mathbb{S} leq n^{2} - n$ [closed]












0












$begingroup$


Fix $Minmathbb{R}^{ntimes n}$ and define the vector space
begin{align*}
mathbb{S} = left{MX - XM mid X in mathbb{R}^{ntimes n}right}.
end{align*}
Show that $dim mathbb{S} leq n^{2} - n$.






linear-algebra matrices vector-spaces







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closed as off-topic by anomaly, KReiser, Leucippus, user91500, José Carlos Santos Jan 6 at 11:57


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." – anomaly, KReiser, Leucippus, user91500, José Carlos Santos

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  • $begingroup$
    Well, what do you think about the problem?
    $endgroup$
    – anomaly
    Jan 5 at 20:39










  • $begingroup$
    Still thinking.
    $endgroup$
    – BasicUser
    Jan 5 at 20:45
















0












$begingroup$


Fix $Minmathbb{R}^{ntimes n}$ and define the vector space
begin{align*}
mathbb{S} = left{MX - XM mid X in mathbb{R}^{ntimes n}right}.
end{align*}
Show that $dim mathbb{S} leq n^{2} - n$.






linear-algebra matrices vector-spaces







share|cite|improve this question











$endgroup$



closed as off-topic by anomaly, KReiser, Leucippus, user91500, José Carlos Santos Jan 6 at 11:57


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." – anomaly, KReiser, Leucippus, user91500, José Carlos Santos

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
















  • $begingroup$
    Well, what do you think about the problem?
    $endgroup$
    – anomaly
    Jan 5 at 20:39










  • $begingroup$
    Still thinking.
    $endgroup$
    – BasicUser
    Jan 5 at 20:45














0












0








0





$begingroup$


Fix $Minmathbb{R}^{ntimes n}$ and define the vector space
begin{align*}
mathbb{S} = left{MX - XM mid X in mathbb{R}^{ntimes n}right}.
end{align*}
Show that $dim mathbb{S} leq n^{2} - n$.






linear-algebra matrices vector-spaces







share|cite|improve this question











$endgroup$




Fix $Minmathbb{R}^{ntimes n}$ and define the vector space
begin{align*}
mathbb{S} = left{MX - XM mid X in mathbb{R}^{ntimes n}right}.
end{align*}
Show that $dim mathbb{S} leq n^{2} - n$.






linear-algebra matrices vector-spaces




linear-algebra matrices vector-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 5 at 20:56









Henning Makholm

241k17308549




241k17308549










asked Jan 5 at 20:33









BasicUserBasicUser

16114




16114




closed as off-topic by anomaly, KReiser, Leucippus, user91500, José Carlos Santos Jan 6 at 11:57


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." – anomaly, KReiser, Leucippus, user91500, 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 anomaly, KReiser, Leucippus, user91500, José Carlos Santos Jan 6 at 11:57


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." – anomaly, KReiser, Leucippus, user91500, José Carlos Santos

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












  • $begingroup$
    Well, what do you think about the problem?
    $endgroup$
    – anomaly
    Jan 5 at 20:39










  • $begingroup$
    Still thinking.
    $endgroup$
    – BasicUser
    Jan 5 at 20:45


















  • $begingroup$
    Well, what do you think about the problem?
    $endgroup$
    – anomaly
    Jan 5 at 20:39










  • $begingroup$
    Still thinking.
    $endgroup$
    – BasicUser
    Jan 5 at 20:45
















$begingroup$
Well, what do you think about the problem?
$endgroup$
– anomaly
Jan 5 at 20:39




$begingroup$
Well, what do you think about the problem?
$endgroup$
– anomaly
Jan 5 at 20:39












$begingroup$
Still thinking.
$endgroup$
– BasicUser
Jan 5 at 20:45




$begingroup$
Still thinking.
$endgroup$
– BasicUser
Jan 5 at 20:45










1 Answer
1






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$begingroup$

Sketch of proof: Define the linear map $T:Bbb C^{n times n} to Bbb C^{n times n}$ by $T(X) = MX - XM$. We see that $ker(T) = {X mid MX = XM}$. By this post or this post, we see that $dim ker (T) geq n$. By the rank-nullity theorem, we have $$
dim(Bbb S) = dim(operatorname{im(T)}) = n^2 - dim ker(T) leq n^2-n
$$
as desired.






share|cite|improve this answer









$endgroup$




















    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    Sketch of proof: Define the linear map $T:Bbb C^{n times n} to Bbb C^{n times n}$ by $T(X) = MX - XM$. We see that $ker(T) = {X mid MX = XM}$. By this post or this post, we see that $dim ker (T) geq n$. By the rank-nullity theorem, we have $$
    dim(Bbb S) = dim(operatorname{im(T)}) = n^2 - dim ker(T) leq n^2-n
    $$
    as desired.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Sketch of proof: Define the linear map $T:Bbb C^{n times n} to Bbb C^{n times n}$ by $T(X) = MX - XM$. We see that $ker(T) = {X mid MX = XM}$. By this post or this post, we see that $dim ker (T) geq n$. By the rank-nullity theorem, we have $$
      dim(Bbb S) = dim(operatorname{im(T)}) = n^2 - dim ker(T) leq n^2-n
      $$
      as desired.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Sketch of proof: Define the linear map $T:Bbb C^{n times n} to Bbb C^{n times n}$ by $T(X) = MX - XM$. We see that $ker(T) = {X mid MX = XM}$. By this post or this post, we see that $dim ker (T) geq n$. By the rank-nullity theorem, we have $$
        dim(Bbb S) = dim(operatorname{im(T)}) = n^2 - dim ker(T) leq n^2-n
        $$
        as desired.






        share|cite|improve this answer









        $endgroup$



        Sketch of proof: Define the linear map $T:Bbb C^{n times n} to Bbb C^{n times n}$ by $T(X) = MX - XM$. We see that $ker(T) = {X mid MX = XM}$. By this post or this post, we see that $dim ker (T) geq n$. By the rank-nullity theorem, we have $$
        dim(Bbb S) = dim(operatorname{im(T)}) = n^2 - dim ker(T) leq n^2-n
        $$
        as desired.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 5 at 20:54









        OmnomnomnomOmnomnomnom

        128k791186




        128k791186















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