Showing two linear functionals are the same up to constant “$g(A) = qtr(A)$”












1












$begingroup$


The original problem is this




If $g: M_2(mathbb{R}) to mathbb{R}$ satisfies $g(AB) = g(BA)$, then there is a constant $k$ such that $g(A) = ktr(A)$ where $tr$ is the trace.




So after reading a bit, this identification is given when messing a bit with their kernels



Is there a more obvious solution? It wasn't clear to me to look at this general case.






linear-algebra







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$endgroup$












  • $begingroup$
    I am just curious. Can you really show directly that the kernel of $g$ is the same as the kernel of $A$?
    $endgroup$
    – user9077
    Dec 13 '18 at 0:14










  • $begingroup$
    You mean the trace? G doesn’t have an explicit map, so I can’t imagine how
    $endgroup$
    – Hawk
    Dec 13 '18 at 4:59










  • $begingroup$
    I mean if we want to use the result from the link we need to show that g and tr have the same kernel right? So I thought you know how to show that the kernel of g is the set of all matrix of zero trace.
    $endgroup$
    – user9077
    Dec 13 '18 at 7:29
















1












$begingroup$


The original problem is this




If $g: M_2(mathbb{R}) to mathbb{R}$ satisfies $g(AB) = g(BA)$, then there is a constant $k$ such that $g(A) = ktr(A)$ where $tr$ is the trace.




So after reading a bit, this identification is given when messing a bit with their kernels



Is there a more obvious solution? It wasn't clear to me to look at this general case.






linear-algebra







share|cite|improve this question









$endgroup$












  • $begingroup$
    I am just curious. Can you really show directly that the kernel of $g$ is the same as the kernel of $A$?
    $endgroup$
    – user9077
    Dec 13 '18 at 0:14










  • $begingroup$
    You mean the trace? G doesn’t have an explicit map, so I can’t imagine how
    $endgroup$
    – Hawk
    Dec 13 '18 at 4:59










  • $begingroup$
    I mean if we want to use the result from the link we need to show that g and tr have the same kernel right? So I thought you know how to show that the kernel of g is the set of all matrix of zero trace.
    $endgroup$
    – user9077
    Dec 13 '18 at 7:29














1












1








1





$begingroup$


The original problem is this




If $g: M_2(mathbb{R}) to mathbb{R}$ satisfies $g(AB) = g(BA)$, then there is a constant $k$ such that $g(A) = ktr(A)$ where $tr$ is the trace.




So after reading a bit, this identification is given when messing a bit with their kernels



Is there a more obvious solution? It wasn't clear to me to look at this general case.






linear-algebra







share|cite|improve this question









$endgroup$




The original problem is this




If $g: M_2(mathbb{R}) to mathbb{R}$ satisfies $g(AB) = g(BA)$, then there is a constant $k$ such that $g(A) = ktr(A)$ where $tr$ is the trace.




So after reading a bit, this identification is given when messing a bit with their kernels



Is there a more obvious solution? It wasn't clear to me to look at this general case.






linear-algebra




linear-algebra






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 12 '18 at 0:28









HawkHawk

5,5151139104




5,5151139104












  • $begingroup$
    I am just curious. Can you really show directly that the kernel of $g$ is the same as the kernel of $A$?
    $endgroup$
    – user9077
    Dec 13 '18 at 0:14










  • $begingroup$
    You mean the trace? G doesn’t have an explicit map, so I can’t imagine how
    $endgroup$
    – Hawk
    Dec 13 '18 at 4:59










  • $begingroup$
    I mean if we want to use the result from the link we need to show that g and tr have the same kernel right? So I thought you know how to show that the kernel of g is the set of all matrix of zero trace.
    $endgroup$
    – user9077
    Dec 13 '18 at 7:29


















  • $begingroup$
    I am just curious. Can you really show directly that the kernel of $g$ is the same as the kernel of $A$?
    $endgroup$
    – user9077
    Dec 13 '18 at 0:14










  • $begingroup$
    You mean the trace? G doesn’t have an explicit map, so I can’t imagine how
    $endgroup$
    – Hawk
    Dec 13 '18 at 4:59










  • $begingroup$
    I mean if we want to use the result from the link we need to show that g and tr have the same kernel right? So I thought you know how to show that the kernel of g is the set of all matrix of zero trace.
    $endgroup$
    – user9077
    Dec 13 '18 at 7:29
















$begingroup$
I am just curious. Can you really show directly that the kernel of $g$ is the same as the kernel of $A$?
$endgroup$
– user9077
Dec 13 '18 at 0:14




$begingroup$
I am just curious. Can you really show directly that the kernel of $g$ is the same as the kernel of $A$?
$endgroup$
– user9077
Dec 13 '18 at 0:14












$begingroup$
You mean the trace? G doesn’t have an explicit map, so I can’t imagine how
$endgroup$
– Hawk
Dec 13 '18 at 4:59




$begingroup$
You mean the trace? G doesn’t have an explicit map, so I can’t imagine how
$endgroup$
– Hawk
Dec 13 '18 at 4:59












$begingroup$
I mean if we want to use the result from the link we need to show that g and tr have the same kernel right? So I thought you know how to show that the kernel of g is the set of all matrix of zero trace.
$endgroup$
– user9077
Dec 13 '18 at 7:29




$begingroup$
I mean if we want to use the result from the link we need to show that g and tr have the same kernel right? So I thought you know how to show that the kernel of g is the set of all matrix of zero trace.
$endgroup$
– user9077
Dec 13 '18 at 7:29










2 Answers
2






active

oldest

votes


















1












$begingroup$

Let $E_1=begin{pmatrix}1&0\0&0end{pmatrix},E_2=begin{pmatrix}0&1\0&0end{pmatrix},E_3=begin{pmatrix}0&0\1&0end{pmatrix}$ and $E_4=begin{pmatrix}0&0\0&1end{pmatrix}$.



If we can show that $$g(E_1)=g(E_4)=k text{ and } g(E_2)=g(E_3)=0,quad (*)$$ then
$$gbegin{pmatrix}a&b\c&dend{pmatrix}=ag(E_1)+bg(E_2)+cg(E_3)+dg(E_4)=k(a+d)=kcdottext{tr}(A)$$
as we want. Hence it is enough to show that $g$ satisfies $(*)$.



Note that by the condition $g(AB)=g(BA)$ we have
$$g(E_2)=g(E_1E_2)=g(E_2E_1)=g(0)=0$$
and
$$g(E_3)=g(E_4E_3)=g(E_3E_4)=g(0)=0 $$
Now
$(E_1+E_3)(E_1+E_2)=E_1+E_2+E_3+E_4$ and $(E_1+E_2)(E_1+E_3)=2E_1$. Then
$$g(2E_1)=g(E_1+E_2+E_3+E_4)=g(E_1)+g(E_4)$$ which implies that
$$g(E_1)=g(E_4)=k text{ for some } kin mathbb{R}.$$
Hence we are done.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So I just checked again and apparently the original problem was stated for $M_n$, I just wrote it $2$ because that's what I wrote on my paper when solving the question and translate it incorrectly.
    $endgroup$
    – Hawk
    Dec 12 '18 at 11:50










  • $begingroup$
    Oh no. Maybe we can use the same idea?
    $endgroup$
    – user9077
    Dec 12 '18 at 12:04










  • $begingroup$
    Well i ruled that out since the columns and rows can get large. For $n =4$ you have to do 16 cases...for case $n$ it is general to talk about, but I found this problem in an exam paper...so something tells me our method here won't be applicable.
    $endgroup$
    – Hawk
    Dec 12 '18 at 12:25










  • $begingroup$
    Well, it can be done, I just tried it. We just need to choose $A,B$ wisely when using $g(AB)=g(BA)$.
    $endgroup$
    – user9077
    Dec 12 '18 at 12:53



















0












$begingroup$

We can use the same idea if $g:M_n(mathbb{R})tomathbb{R}$. The idea is that if $B$ is a basis of $M_n$, also $g$ and $text{tr}$ agree on $B$ up to some constant multiple $alpha$, then $$g(A)=alphacdot text{tr}(A) $$ for all $Ain M_n$.



Let $E_{i,j}$ be the matrix where all the entries are zero except it is 1 at the position $(i,j)$. Let $T_k=text{diag}(underbrace{1,1,ldots,1}_k,0,ldots,0)$. Notice that for $ineq j$ we have $E_{i,j}T_j=E_{i,j}$ while $T_jE_{i,j}=0$. Therefore for all $ineq j$ we have $g(E_{i,j})=0$.



Now let $J_k$ be a $ktimes n$ matrix where all the entries are 1. Define an $ntimes n$ matrix $S_k$ as a block matrix $$S_k:=begin{pmatrix}J_k\0end{pmatrix}$$.



Note that since $g(E_{i,j})=0$ for $ineq j$, we have $g(S_k)=g(T_k)$ for all $k$.



Now
begin{align*}
S_kS_1&=S_k \ S_1S_k&=kS_1
end{align*}



Hence $g(T_k)=g(S_k)=g(kS_1)=kg(S_1)=kg(T_1)$. Notice that the trace map also satisfies $text{tr}(T_k)=k$. By taking $alpha=g(T_1)$ then $g$ and $alpha cdot text{tr}$ are two linear maps that agree on $B={E_{i,j}mid ineq j}cup {T_kmid k=1,..,n}$ which is a basis of $M_n$. Hence $g=alphacdot text{tr}.$






share|cite|improve this answer









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    2 Answers
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    2 Answers
    2






    active

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    active

    oldest

    votes






    active

    oldest

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    1












    $begingroup$

    Let $E_1=begin{pmatrix}1&0\0&0end{pmatrix},E_2=begin{pmatrix}0&1\0&0end{pmatrix},E_3=begin{pmatrix}0&0\1&0end{pmatrix}$ and $E_4=begin{pmatrix}0&0\0&1end{pmatrix}$.



    If we can show that $$g(E_1)=g(E_4)=k text{ and } g(E_2)=g(E_3)=0,quad (*)$$ then
    $$gbegin{pmatrix}a&b\c&dend{pmatrix}=ag(E_1)+bg(E_2)+cg(E_3)+dg(E_4)=k(a+d)=kcdottext{tr}(A)$$
    as we want. Hence it is enough to show that $g$ satisfies $(*)$.



    Note that by the condition $g(AB)=g(BA)$ we have
    $$g(E_2)=g(E_1E_2)=g(E_2E_1)=g(0)=0$$
    and
    $$g(E_3)=g(E_4E_3)=g(E_3E_4)=g(0)=0 $$
    Now
    $(E_1+E_3)(E_1+E_2)=E_1+E_2+E_3+E_4$ and $(E_1+E_2)(E_1+E_3)=2E_1$. Then
    $$g(2E_1)=g(E_1+E_2+E_3+E_4)=g(E_1)+g(E_4)$$ which implies that
    $$g(E_1)=g(E_4)=k text{ for some } kin mathbb{R}.$$
    Hence we are done.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      So I just checked again and apparently the original problem was stated for $M_n$, I just wrote it $2$ because that's what I wrote on my paper when solving the question and translate it incorrectly.
      $endgroup$
      – Hawk
      Dec 12 '18 at 11:50










    • $begingroup$
      Oh no. Maybe we can use the same idea?
      $endgroup$
      – user9077
      Dec 12 '18 at 12:04










    • $begingroup$
      Well i ruled that out since the columns and rows can get large. For $n =4$ you have to do 16 cases...for case $n$ it is general to talk about, but I found this problem in an exam paper...so something tells me our method here won't be applicable.
      $endgroup$
      – Hawk
      Dec 12 '18 at 12:25










    • $begingroup$
      Well, it can be done, I just tried it. We just need to choose $A,B$ wisely when using $g(AB)=g(BA)$.
      $endgroup$
      – user9077
      Dec 12 '18 at 12:53
















    1












    $begingroup$

    Let $E_1=begin{pmatrix}1&0\0&0end{pmatrix},E_2=begin{pmatrix}0&1\0&0end{pmatrix},E_3=begin{pmatrix}0&0\1&0end{pmatrix}$ and $E_4=begin{pmatrix}0&0\0&1end{pmatrix}$.



    If we can show that $$g(E_1)=g(E_4)=k text{ and } g(E_2)=g(E_3)=0,quad (*)$$ then
    $$gbegin{pmatrix}a&b\c&dend{pmatrix}=ag(E_1)+bg(E_2)+cg(E_3)+dg(E_4)=k(a+d)=kcdottext{tr}(A)$$
    as we want. Hence it is enough to show that $g$ satisfies $(*)$.



    Note that by the condition $g(AB)=g(BA)$ we have
    $$g(E_2)=g(E_1E_2)=g(E_2E_1)=g(0)=0$$
    and
    $$g(E_3)=g(E_4E_3)=g(E_3E_4)=g(0)=0 $$
    Now
    $(E_1+E_3)(E_1+E_2)=E_1+E_2+E_3+E_4$ and $(E_1+E_2)(E_1+E_3)=2E_1$. Then
    $$g(2E_1)=g(E_1+E_2+E_3+E_4)=g(E_1)+g(E_4)$$ which implies that
    $$g(E_1)=g(E_4)=k text{ for some } kin mathbb{R}.$$
    Hence we are done.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      So I just checked again and apparently the original problem was stated for $M_n$, I just wrote it $2$ because that's what I wrote on my paper when solving the question and translate it incorrectly.
      $endgroup$
      – Hawk
      Dec 12 '18 at 11:50










    • $begingroup$
      Oh no. Maybe we can use the same idea?
      $endgroup$
      – user9077
      Dec 12 '18 at 12:04










    • $begingroup$
      Well i ruled that out since the columns and rows can get large. For $n =4$ you have to do 16 cases...for case $n$ it is general to talk about, but I found this problem in an exam paper...so something tells me our method here won't be applicable.
      $endgroup$
      – Hawk
      Dec 12 '18 at 12:25










    • $begingroup$
      Well, it can be done, I just tried it. We just need to choose $A,B$ wisely when using $g(AB)=g(BA)$.
      $endgroup$
      – user9077
      Dec 12 '18 at 12:53














    1












    1








    1





    $begingroup$

    Let $E_1=begin{pmatrix}1&0\0&0end{pmatrix},E_2=begin{pmatrix}0&1\0&0end{pmatrix},E_3=begin{pmatrix}0&0\1&0end{pmatrix}$ and $E_4=begin{pmatrix}0&0\0&1end{pmatrix}$.



    If we can show that $$g(E_1)=g(E_4)=k text{ and } g(E_2)=g(E_3)=0,quad (*)$$ then
    $$gbegin{pmatrix}a&b\c&dend{pmatrix}=ag(E_1)+bg(E_2)+cg(E_3)+dg(E_4)=k(a+d)=kcdottext{tr}(A)$$
    as we want. Hence it is enough to show that $g$ satisfies $(*)$.



    Note that by the condition $g(AB)=g(BA)$ we have
    $$g(E_2)=g(E_1E_2)=g(E_2E_1)=g(0)=0$$
    and
    $$g(E_3)=g(E_4E_3)=g(E_3E_4)=g(0)=0 $$
    Now
    $(E_1+E_3)(E_1+E_2)=E_1+E_2+E_3+E_4$ and $(E_1+E_2)(E_1+E_3)=2E_1$. Then
    $$g(2E_1)=g(E_1+E_2+E_3+E_4)=g(E_1)+g(E_4)$$ which implies that
    $$g(E_1)=g(E_4)=k text{ for some } kin mathbb{R}.$$
    Hence we are done.






    share|cite|improve this answer









    $endgroup$



    Let $E_1=begin{pmatrix}1&0\0&0end{pmatrix},E_2=begin{pmatrix}0&1\0&0end{pmatrix},E_3=begin{pmatrix}0&0\1&0end{pmatrix}$ and $E_4=begin{pmatrix}0&0\0&1end{pmatrix}$.



    If we can show that $$g(E_1)=g(E_4)=k text{ and } g(E_2)=g(E_3)=0,quad (*)$$ then
    $$gbegin{pmatrix}a&b\c&dend{pmatrix}=ag(E_1)+bg(E_2)+cg(E_3)+dg(E_4)=k(a+d)=kcdottext{tr}(A)$$
    as we want. Hence it is enough to show that $g$ satisfies $(*)$.



    Note that by the condition $g(AB)=g(BA)$ we have
    $$g(E_2)=g(E_1E_2)=g(E_2E_1)=g(0)=0$$
    and
    $$g(E_3)=g(E_4E_3)=g(E_3E_4)=g(0)=0 $$
    Now
    $(E_1+E_3)(E_1+E_2)=E_1+E_2+E_3+E_4$ and $(E_1+E_2)(E_1+E_3)=2E_1$. Then
    $$g(2E_1)=g(E_1+E_2+E_3+E_4)=g(E_1)+g(E_4)$$ which implies that
    $$g(E_1)=g(E_4)=k text{ for some } kin mathbb{R}.$$
    Hence we are done.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 12 '18 at 2:47









    user9077user9077

    1,239612




    1,239612












    • $begingroup$
      So I just checked again and apparently the original problem was stated for $M_n$, I just wrote it $2$ because that's what I wrote on my paper when solving the question and translate it incorrectly.
      $endgroup$
      – Hawk
      Dec 12 '18 at 11:50










    • $begingroup$
      Oh no. Maybe we can use the same idea?
      $endgroup$
      – user9077
      Dec 12 '18 at 12:04










    • $begingroup$
      Well i ruled that out since the columns and rows can get large. For $n =4$ you have to do 16 cases...for case $n$ it is general to talk about, but I found this problem in an exam paper...so something tells me our method here won't be applicable.
      $endgroup$
      – Hawk
      Dec 12 '18 at 12:25










    • $begingroup$
      Well, it can be done, I just tried it. We just need to choose $A,B$ wisely when using $g(AB)=g(BA)$.
      $endgroup$
      – user9077
      Dec 12 '18 at 12:53


















    • $begingroup$
      So I just checked again and apparently the original problem was stated for $M_n$, I just wrote it $2$ because that's what I wrote on my paper when solving the question and translate it incorrectly.
      $endgroup$
      – Hawk
      Dec 12 '18 at 11:50










    • $begingroup$
      Oh no. Maybe we can use the same idea?
      $endgroup$
      – user9077
      Dec 12 '18 at 12:04










    • $begingroup$
      Well i ruled that out since the columns and rows can get large. For $n =4$ you have to do 16 cases...for case $n$ it is general to talk about, but I found this problem in an exam paper...so something tells me our method here won't be applicable.
      $endgroup$
      – Hawk
      Dec 12 '18 at 12:25










    • $begingroup$
      Well, it can be done, I just tried it. We just need to choose $A,B$ wisely when using $g(AB)=g(BA)$.
      $endgroup$
      – user9077
      Dec 12 '18 at 12:53
















    $begingroup$
    So I just checked again and apparently the original problem was stated for $M_n$, I just wrote it $2$ because that's what I wrote on my paper when solving the question and translate it incorrectly.
    $endgroup$
    – Hawk
    Dec 12 '18 at 11:50




    $begingroup$
    So I just checked again and apparently the original problem was stated for $M_n$, I just wrote it $2$ because that's what I wrote on my paper when solving the question and translate it incorrectly.
    $endgroup$
    – Hawk
    Dec 12 '18 at 11:50












    $begingroup$
    Oh no. Maybe we can use the same idea?
    $endgroup$
    – user9077
    Dec 12 '18 at 12:04




    $begingroup$
    Oh no. Maybe we can use the same idea?
    $endgroup$
    – user9077
    Dec 12 '18 at 12:04












    $begingroup$
    Well i ruled that out since the columns and rows can get large. For $n =4$ you have to do 16 cases...for case $n$ it is general to talk about, but I found this problem in an exam paper...so something tells me our method here won't be applicable.
    $endgroup$
    – Hawk
    Dec 12 '18 at 12:25




    $begingroup$
    Well i ruled that out since the columns and rows can get large. For $n =4$ you have to do 16 cases...for case $n$ it is general to talk about, but I found this problem in an exam paper...so something tells me our method here won't be applicable.
    $endgroup$
    – Hawk
    Dec 12 '18 at 12:25












    $begingroup$
    Well, it can be done, I just tried it. We just need to choose $A,B$ wisely when using $g(AB)=g(BA)$.
    $endgroup$
    – user9077
    Dec 12 '18 at 12:53




    $begingroup$
    Well, it can be done, I just tried it. We just need to choose $A,B$ wisely when using $g(AB)=g(BA)$.
    $endgroup$
    – user9077
    Dec 12 '18 at 12:53











    0












    $begingroup$

    We can use the same idea if $g:M_n(mathbb{R})tomathbb{R}$. The idea is that if $B$ is a basis of $M_n$, also $g$ and $text{tr}$ agree on $B$ up to some constant multiple $alpha$, then $$g(A)=alphacdot text{tr}(A) $$ for all $Ain M_n$.



    Let $E_{i,j}$ be the matrix where all the entries are zero except it is 1 at the position $(i,j)$. Let $T_k=text{diag}(underbrace{1,1,ldots,1}_k,0,ldots,0)$. Notice that for $ineq j$ we have $E_{i,j}T_j=E_{i,j}$ while $T_jE_{i,j}=0$. Therefore for all $ineq j$ we have $g(E_{i,j})=0$.



    Now let $J_k$ be a $ktimes n$ matrix where all the entries are 1. Define an $ntimes n$ matrix $S_k$ as a block matrix $$S_k:=begin{pmatrix}J_k\0end{pmatrix}$$.



    Note that since $g(E_{i,j})=0$ for $ineq j$, we have $g(S_k)=g(T_k)$ for all $k$.



    Now
    begin{align*}
    S_kS_1&=S_k \ S_1S_k&=kS_1
    end{align*}



    Hence $g(T_k)=g(S_k)=g(kS_1)=kg(S_1)=kg(T_1)$. Notice that the trace map also satisfies $text{tr}(T_k)=k$. By taking $alpha=g(T_1)$ then $g$ and $alpha cdot text{tr}$ are two linear maps that agree on $B={E_{i,j}mid ineq j}cup {T_kmid k=1,..,n}$ which is a basis of $M_n$. Hence $g=alphacdot text{tr}.$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      We can use the same idea if $g:M_n(mathbb{R})tomathbb{R}$. The idea is that if $B$ is a basis of $M_n$, also $g$ and $text{tr}$ agree on $B$ up to some constant multiple $alpha$, then $$g(A)=alphacdot text{tr}(A) $$ for all $Ain M_n$.



      Let $E_{i,j}$ be the matrix where all the entries are zero except it is 1 at the position $(i,j)$. Let $T_k=text{diag}(underbrace{1,1,ldots,1}_k,0,ldots,0)$. Notice that for $ineq j$ we have $E_{i,j}T_j=E_{i,j}$ while $T_jE_{i,j}=0$. Therefore for all $ineq j$ we have $g(E_{i,j})=0$.



      Now let $J_k$ be a $ktimes n$ matrix where all the entries are 1. Define an $ntimes n$ matrix $S_k$ as a block matrix $$S_k:=begin{pmatrix}J_k\0end{pmatrix}$$.



      Note that since $g(E_{i,j})=0$ for $ineq j$, we have $g(S_k)=g(T_k)$ for all $k$.



      Now
      begin{align*}
      S_kS_1&=S_k \ S_1S_k&=kS_1
      end{align*}



      Hence $g(T_k)=g(S_k)=g(kS_1)=kg(S_1)=kg(T_1)$. Notice that the trace map also satisfies $text{tr}(T_k)=k$. By taking $alpha=g(T_1)$ then $g$ and $alpha cdot text{tr}$ are two linear maps that agree on $B={E_{i,j}mid ineq j}cup {T_kmid k=1,..,n}$ which is a basis of $M_n$. Hence $g=alphacdot text{tr}.$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        We can use the same idea if $g:M_n(mathbb{R})tomathbb{R}$. The idea is that if $B$ is a basis of $M_n$, also $g$ and $text{tr}$ agree on $B$ up to some constant multiple $alpha$, then $$g(A)=alphacdot text{tr}(A) $$ for all $Ain M_n$.



        Let $E_{i,j}$ be the matrix where all the entries are zero except it is 1 at the position $(i,j)$. Let $T_k=text{diag}(underbrace{1,1,ldots,1}_k,0,ldots,0)$. Notice that for $ineq j$ we have $E_{i,j}T_j=E_{i,j}$ while $T_jE_{i,j}=0$. Therefore for all $ineq j$ we have $g(E_{i,j})=0$.



        Now let $J_k$ be a $ktimes n$ matrix where all the entries are 1. Define an $ntimes n$ matrix $S_k$ as a block matrix $$S_k:=begin{pmatrix}J_k\0end{pmatrix}$$.



        Note that since $g(E_{i,j})=0$ for $ineq j$, we have $g(S_k)=g(T_k)$ for all $k$.



        Now
        begin{align*}
        S_kS_1&=S_k \ S_1S_k&=kS_1
        end{align*}



        Hence $g(T_k)=g(S_k)=g(kS_1)=kg(S_1)=kg(T_1)$. Notice that the trace map also satisfies $text{tr}(T_k)=k$. By taking $alpha=g(T_1)$ then $g$ and $alpha cdot text{tr}$ are two linear maps that agree on $B={E_{i,j}mid ineq j}cup {T_kmid k=1,..,n}$ which is a basis of $M_n$. Hence $g=alphacdot text{tr}.$






        share|cite|improve this answer









        $endgroup$



        We can use the same idea if $g:M_n(mathbb{R})tomathbb{R}$. The idea is that if $B$ is a basis of $M_n$, also $g$ and $text{tr}$ agree on $B$ up to some constant multiple $alpha$, then $$g(A)=alphacdot text{tr}(A) $$ for all $Ain M_n$.



        Let $E_{i,j}$ be the matrix where all the entries are zero except it is 1 at the position $(i,j)$. Let $T_k=text{diag}(underbrace{1,1,ldots,1}_k,0,ldots,0)$. Notice that for $ineq j$ we have $E_{i,j}T_j=E_{i,j}$ while $T_jE_{i,j}=0$. Therefore for all $ineq j$ we have $g(E_{i,j})=0$.



        Now let $J_k$ be a $ktimes n$ matrix where all the entries are 1. Define an $ntimes n$ matrix $S_k$ as a block matrix $$S_k:=begin{pmatrix}J_k\0end{pmatrix}$$.



        Note that since $g(E_{i,j})=0$ for $ineq j$, we have $g(S_k)=g(T_k)$ for all $k$.



        Now
        begin{align*}
        S_kS_1&=S_k \ S_1S_k&=kS_1
        end{align*}



        Hence $g(T_k)=g(S_k)=g(kS_1)=kg(S_1)=kg(T_1)$. Notice that the trace map also satisfies $text{tr}(T_k)=k$. By taking $alpha=g(T_1)$ then $g$ and $alpha cdot text{tr}$ are two linear maps that agree on $B={E_{i,j}mid ineq j}cup {T_kmid k=1,..,n}$ which is a basis of $M_n$. Hence $g=alphacdot text{tr}.$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 12 '18 at 13:28









        user9077user9077

        1,239612




        1,239612






























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