A one-to-one map between $M_{ntimes n}$ and the $mathbb{R}^{n^{2}}$?












-1












$begingroup$


I've been trying to think in something that can make this happen, but i'm not get anywhere. Plus, i have to show something through this map that can make this space($M_{ntimes n}$) a metric space, so a think to use the trace of a matrix. Is that correct?






linear-algebra general-topology matrices metric-spaces







share|cite|improve this question











$endgroup$

















    -1












    $begingroup$


    I've been trying to think in something that can make this happen, but i'm not get anywhere. Plus, i have to show something through this map that can make this space($M_{ntimes n}$) a metric space, so a think to use the trace of a matrix. Is that correct?






    linear-algebra general-topology matrices metric-spaces







    share|cite|improve this question











    $endgroup$















      -1












      -1








      -1





      $begingroup$


      I've been trying to think in something that can make this happen, but i'm not get anywhere. Plus, i have to show something through this map that can make this space($M_{ntimes n}$) a metric space, so a think to use the trace of a matrix. Is that correct?






      linear-algebra general-topology matrices metric-spaces







      share|cite|improve this question











      $endgroup$




      I've been trying to think in something that can make this happen, but i'm not get anywhere. Plus, i have to show something through this map that can make this space($M_{ntimes n}$) a metric space, so a think to use the trace of a matrix. Is that correct?






      linear-algebra general-topology matrices metric-spaces




      linear-algebra general-topology matrices metric-spaces






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 4 '18 at 20:02









      J.G.

      24.6k22539




      24.6k22539










      asked Dec 4 '18 at 19:51









      gbcostagbcosta

      64




      64






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          The trace of matrix is invalid since it reduces the dimension from $n^2$ to $n$. You can define the set of $ntimes n$ matrices with Frobenius norm and reorder the rows of the matrices back to back and define the Euclidean $2$-norm on $Bbb R^{n^2}$. In this manner the two metric spaces would become homeomorphic.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Actually, $mathrm{tr}:M_{ntimes n}tomathbb R$ reduces the dimension to just $1$.
            $endgroup$
            – Federico
            Dec 4 '18 at 20:05








          • 1




            $begingroup$
            I mean the degree of freedom reduces to $n$ since only diagonal entries are important.
            $endgroup$
            – Mostafa Ayaz
            Dec 4 '18 at 20:05










          • $begingroup$
            Yeah, i saw my mistake about the trace. I'm gonna look the sugestion of Frobenius norm here, thanks a lot.
            $endgroup$
            – gbcosta
            Dec 4 '18 at 20:33










          • $begingroup$
            You're welcome. I also can help you further if you needed....
            $endgroup$
            – Mostafa Ayaz
            Dec 4 '18 at 20:40










          • $begingroup$
            @Federico actually, any metric $m: X times X to mathbb{R}$ has dimension 1 codomain, so I don't think that affects anything...
            $endgroup$
            – Monstrous Moonshiner
            Dec 4 '18 at 21:10





















          1












          $begingroup$

          Just take
          $$
          f:Ato (a_{1,1},a_{1,2},dots,a_{1,n},a_{2,1},dots a_{2,n},a_{3,1}dots,a_{n,1},a_{n,2},dots, a_{n,n})
          $$
          i.e just rearrange the entries of $A$ into a vector in $mathbb{R}^{n^2}$.
          Then define $d(A,B):=||f(A)-f(B)||$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            But this map is one-to-one? I think that $$f:Arightarrow (a_{1,1},..., a_{1,n}, a_{2,1}...,a{2,n},...,a_{n,1},...,a_{n,n})$$ answer the question, but i'm not sure of that.
            $endgroup$
            – gbcosta
            Dec 4 '18 at 20:40












          • $begingroup$
            Take n=3. If $A=begin{array}{lcr} a & b & c \ d & e & f \ g & h & j end{array}$ and $B = begin{array}{lcr} a & b & c \ k & l & m \ g & h & j end{array}$. So $Aneq B$ but f(A)=f(B)
            $endgroup$
            – gbcosta
            Dec 4 '18 at 20:57












          • $begingroup$
            No problem, anyway you helped me a lot, thank you so much.
            $endgroup$
            – gbcosta
            Dec 4 '18 at 21:04










          • $begingroup$
            @gbcosta no problem
            $endgroup$
            – qbert
            Dec 4 '18 at 21:04






          • 2




            $begingroup$
            Your objection : "Take n=3..." is not valid : we order the entries f(A)=(a, b, c, d, e, f, g, h, i) is different from f(B)=(a, b, c, k, l, m, g, h, i) !
            $endgroup$
            – Jean Marie
            Dec 4 '18 at 21:12











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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          The trace of matrix is invalid since it reduces the dimension from $n^2$ to $n$. You can define the set of $ntimes n$ matrices with Frobenius norm and reorder the rows of the matrices back to back and define the Euclidean $2$-norm on $Bbb R^{n^2}$. In this manner the two metric spaces would become homeomorphic.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Actually, $mathrm{tr}:M_{ntimes n}tomathbb R$ reduces the dimension to just $1$.
            $endgroup$
            – Federico
            Dec 4 '18 at 20:05








          • 1




            $begingroup$
            I mean the degree of freedom reduces to $n$ since only diagonal entries are important.
            $endgroup$
            – Mostafa Ayaz
            Dec 4 '18 at 20:05










          • $begingroup$
            Yeah, i saw my mistake about the trace. I'm gonna look the sugestion of Frobenius norm here, thanks a lot.
            $endgroup$
            – gbcosta
            Dec 4 '18 at 20:33










          • $begingroup$
            You're welcome. I also can help you further if you needed....
            $endgroup$
            – Mostafa Ayaz
            Dec 4 '18 at 20:40










          • $begingroup$
            @Federico actually, any metric $m: X times X to mathbb{R}$ has dimension 1 codomain, so I don't think that affects anything...
            $endgroup$
            – Monstrous Moonshiner
            Dec 4 '18 at 21:10


















          1












          $begingroup$

          The trace of matrix is invalid since it reduces the dimension from $n^2$ to $n$. You can define the set of $ntimes n$ matrices with Frobenius norm and reorder the rows of the matrices back to back and define the Euclidean $2$-norm on $Bbb R^{n^2}$. In this manner the two metric spaces would become homeomorphic.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Actually, $mathrm{tr}:M_{ntimes n}tomathbb R$ reduces the dimension to just $1$.
            $endgroup$
            – Federico
            Dec 4 '18 at 20:05








          • 1




            $begingroup$
            I mean the degree of freedom reduces to $n$ since only diagonal entries are important.
            $endgroup$
            – Mostafa Ayaz
            Dec 4 '18 at 20:05










          • $begingroup$
            Yeah, i saw my mistake about the trace. I'm gonna look the sugestion of Frobenius norm here, thanks a lot.
            $endgroup$
            – gbcosta
            Dec 4 '18 at 20:33










          • $begingroup$
            You're welcome. I also can help you further if you needed....
            $endgroup$
            – Mostafa Ayaz
            Dec 4 '18 at 20:40










          • $begingroup$
            @Federico actually, any metric $m: X times X to mathbb{R}$ has dimension 1 codomain, so I don't think that affects anything...
            $endgroup$
            – Monstrous Moonshiner
            Dec 4 '18 at 21:10
















          1












          1








          1





          $begingroup$

          The trace of matrix is invalid since it reduces the dimension from $n^2$ to $n$. You can define the set of $ntimes n$ matrices with Frobenius norm and reorder the rows of the matrices back to back and define the Euclidean $2$-norm on $Bbb R^{n^2}$. In this manner the two metric spaces would become homeomorphic.






          share|cite|improve this answer









          $endgroup$



          The trace of matrix is invalid since it reduces the dimension from $n^2$ to $n$. You can define the set of $ntimes n$ matrices with Frobenius norm and reorder the rows of the matrices back to back and define the Euclidean $2$-norm on $Bbb R^{n^2}$. In this manner the two metric spaces would become homeomorphic.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 4 '18 at 19:57









          Mostafa AyazMostafa Ayaz

          15.4k3939




          15.4k3939












          • $begingroup$
            Actually, $mathrm{tr}:M_{ntimes n}tomathbb R$ reduces the dimension to just $1$.
            $endgroup$
            – Federico
            Dec 4 '18 at 20:05








          • 1




            $begingroup$
            I mean the degree of freedom reduces to $n$ since only diagonal entries are important.
            $endgroup$
            – Mostafa Ayaz
            Dec 4 '18 at 20:05










          • $begingroup$
            Yeah, i saw my mistake about the trace. I'm gonna look the sugestion of Frobenius norm here, thanks a lot.
            $endgroup$
            – gbcosta
            Dec 4 '18 at 20:33










          • $begingroup$
            You're welcome. I also can help you further if you needed....
            $endgroup$
            – Mostafa Ayaz
            Dec 4 '18 at 20:40










          • $begingroup$
            @Federico actually, any metric $m: X times X to mathbb{R}$ has dimension 1 codomain, so I don't think that affects anything...
            $endgroup$
            – Monstrous Moonshiner
            Dec 4 '18 at 21:10




















          • $begingroup$
            Actually, $mathrm{tr}:M_{ntimes n}tomathbb R$ reduces the dimension to just $1$.
            $endgroup$
            – Federico
            Dec 4 '18 at 20:05








          • 1




            $begingroup$
            I mean the degree of freedom reduces to $n$ since only diagonal entries are important.
            $endgroup$
            – Mostafa Ayaz
            Dec 4 '18 at 20:05










          • $begingroup$
            Yeah, i saw my mistake about the trace. I'm gonna look the sugestion of Frobenius norm here, thanks a lot.
            $endgroup$
            – gbcosta
            Dec 4 '18 at 20:33










          • $begingroup$
            You're welcome. I also can help you further if you needed....
            $endgroup$
            – Mostafa Ayaz
            Dec 4 '18 at 20:40










          • $begingroup$
            @Federico actually, any metric $m: X times X to mathbb{R}$ has dimension 1 codomain, so I don't think that affects anything...
            $endgroup$
            – Monstrous Moonshiner
            Dec 4 '18 at 21:10


















          $begingroup$
          Actually, $mathrm{tr}:M_{ntimes n}tomathbb R$ reduces the dimension to just $1$.
          $endgroup$
          – Federico
          Dec 4 '18 at 20:05






          $begingroup$
          Actually, $mathrm{tr}:M_{ntimes n}tomathbb R$ reduces the dimension to just $1$.
          $endgroup$
          – Federico
          Dec 4 '18 at 20:05






          1




          1




          $begingroup$
          I mean the degree of freedom reduces to $n$ since only diagonal entries are important.
          $endgroup$
          – Mostafa Ayaz
          Dec 4 '18 at 20:05




          $begingroup$
          I mean the degree of freedom reduces to $n$ since only diagonal entries are important.
          $endgroup$
          – Mostafa Ayaz
          Dec 4 '18 at 20:05












          $begingroup$
          Yeah, i saw my mistake about the trace. I'm gonna look the sugestion of Frobenius norm here, thanks a lot.
          $endgroup$
          – gbcosta
          Dec 4 '18 at 20:33




          $begingroup$
          Yeah, i saw my mistake about the trace. I'm gonna look the sugestion of Frobenius norm here, thanks a lot.
          $endgroup$
          – gbcosta
          Dec 4 '18 at 20:33












          $begingroup$
          You're welcome. I also can help you further if you needed....
          $endgroup$
          – Mostafa Ayaz
          Dec 4 '18 at 20:40




          $begingroup$
          You're welcome. I also can help you further if you needed....
          $endgroup$
          – Mostafa Ayaz
          Dec 4 '18 at 20:40












          $begingroup$
          @Federico actually, any metric $m: X times X to mathbb{R}$ has dimension 1 codomain, so I don't think that affects anything...
          $endgroup$
          – Monstrous Moonshiner
          Dec 4 '18 at 21:10






          $begingroup$
          @Federico actually, any metric $m: X times X to mathbb{R}$ has dimension 1 codomain, so I don't think that affects anything...
          $endgroup$
          – Monstrous Moonshiner
          Dec 4 '18 at 21:10













          1












          $begingroup$

          Just take
          $$
          f:Ato (a_{1,1},a_{1,2},dots,a_{1,n},a_{2,1},dots a_{2,n},a_{3,1}dots,a_{n,1},a_{n,2},dots, a_{n,n})
          $$
          i.e just rearrange the entries of $A$ into a vector in $mathbb{R}^{n^2}$.
          Then define $d(A,B):=||f(A)-f(B)||$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            But this map is one-to-one? I think that $$f:Arightarrow (a_{1,1},..., a_{1,n}, a_{2,1}...,a{2,n},...,a_{n,1},...,a_{n,n})$$ answer the question, but i'm not sure of that.
            $endgroup$
            – gbcosta
            Dec 4 '18 at 20:40












          • $begingroup$
            Take n=3. If $A=begin{array}{lcr} a & b & c \ d & e & f \ g & h & j end{array}$ and $B = begin{array}{lcr} a & b & c \ k & l & m \ g & h & j end{array}$. So $Aneq B$ but f(A)=f(B)
            $endgroup$
            – gbcosta
            Dec 4 '18 at 20:57












          • $begingroup$
            No problem, anyway you helped me a lot, thank you so much.
            $endgroup$
            – gbcosta
            Dec 4 '18 at 21:04










          • $begingroup$
            @gbcosta no problem
            $endgroup$
            – qbert
            Dec 4 '18 at 21:04






          • 2




            $begingroup$
            Your objection : "Take n=3..." is not valid : we order the entries f(A)=(a, b, c, d, e, f, g, h, i) is different from f(B)=(a, b, c, k, l, m, g, h, i) !
            $endgroup$
            – Jean Marie
            Dec 4 '18 at 21:12
















          1












          $begingroup$

          Just take
          $$
          f:Ato (a_{1,1},a_{1,2},dots,a_{1,n},a_{2,1},dots a_{2,n},a_{3,1}dots,a_{n,1},a_{n,2},dots, a_{n,n})
          $$
          i.e just rearrange the entries of $A$ into a vector in $mathbb{R}^{n^2}$.
          Then define $d(A,B):=||f(A)-f(B)||$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            But this map is one-to-one? I think that $$f:Arightarrow (a_{1,1},..., a_{1,n}, a_{2,1}...,a{2,n},...,a_{n,1},...,a_{n,n})$$ answer the question, but i'm not sure of that.
            $endgroup$
            – gbcosta
            Dec 4 '18 at 20:40












          • $begingroup$
            Take n=3. If $A=begin{array}{lcr} a & b & c \ d & e & f \ g & h & j end{array}$ and $B = begin{array}{lcr} a & b & c \ k & l & m \ g & h & j end{array}$. So $Aneq B$ but f(A)=f(B)
            $endgroup$
            – gbcosta
            Dec 4 '18 at 20:57












          • $begingroup$
            No problem, anyway you helped me a lot, thank you so much.
            $endgroup$
            – gbcosta
            Dec 4 '18 at 21:04










          • $begingroup$
            @gbcosta no problem
            $endgroup$
            – qbert
            Dec 4 '18 at 21:04






          • 2




            $begingroup$
            Your objection : "Take n=3..." is not valid : we order the entries f(A)=(a, b, c, d, e, f, g, h, i) is different from f(B)=(a, b, c, k, l, m, g, h, i) !
            $endgroup$
            – Jean Marie
            Dec 4 '18 at 21:12














          1












          1








          1





          $begingroup$

          Just take
          $$
          f:Ato (a_{1,1},a_{1,2},dots,a_{1,n},a_{2,1},dots a_{2,n},a_{3,1}dots,a_{n,1},a_{n,2},dots, a_{n,n})
          $$
          i.e just rearrange the entries of $A$ into a vector in $mathbb{R}^{n^2}$.
          Then define $d(A,B):=||f(A)-f(B)||$.






          share|cite|improve this answer











          $endgroup$



          Just take
          $$
          f:Ato (a_{1,1},a_{1,2},dots,a_{1,n},a_{2,1},dots a_{2,n},a_{3,1}dots,a_{n,1},a_{n,2},dots, a_{n,n})
          $$
          i.e just rearrange the entries of $A$ into a vector in $mathbb{R}^{n^2}$.
          Then define $d(A,B):=||f(A)-f(B)||$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 4 '18 at 21:01

























          answered Dec 4 '18 at 20:08









          qbertqbert

          22.1k32460




          22.1k32460












          • $begingroup$
            But this map is one-to-one? I think that $$f:Arightarrow (a_{1,1},..., a_{1,n}, a_{2,1}...,a{2,n},...,a_{n,1},...,a_{n,n})$$ answer the question, but i'm not sure of that.
            $endgroup$
            – gbcosta
            Dec 4 '18 at 20:40












          • $begingroup$
            Take n=3. If $A=begin{array}{lcr} a & b & c \ d & e & f \ g & h & j end{array}$ and $B = begin{array}{lcr} a & b & c \ k & l & m \ g & h & j end{array}$. So $Aneq B$ but f(A)=f(B)
            $endgroup$
            – gbcosta
            Dec 4 '18 at 20:57












          • $begingroup$
            No problem, anyway you helped me a lot, thank you so much.
            $endgroup$
            – gbcosta
            Dec 4 '18 at 21:04










          • $begingroup$
            @gbcosta no problem
            $endgroup$
            – qbert
            Dec 4 '18 at 21:04






          • 2




            $begingroup$
            Your objection : "Take n=3..." is not valid : we order the entries f(A)=(a, b, c, d, e, f, g, h, i) is different from f(B)=(a, b, c, k, l, m, g, h, i) !
            $endgroup$
            – Jean Marie
            Dec 4 '18 at 21:12


















          • $begingroup$
            But this map is one-to-one? I think that $$f:Arightarrow (a_{1,1},..., a_{1,n}, a_{2,1}...,a{2,n},...,a_{n,1},...,a_{n,n})$$ answer the question, but i'm not sure of that.
            $endgroup$
            – gbcosta
            Dec 4 '18 at 20:40












          • $begingroup$
            Take n=3. If $A=begin{array}{lcr} a & b & c \ d & e & f \ g & h & j end{array}$ and $B = begin{array}{lcr} a & b & c \ k & l & m \ g & h & j end{array}$. So $Aneq B$ but f(A)=f(B)
            $endgroup$
            – gbcosta
            Dec 4 '18 at 20:57












          • $begingroup$
            No problem, anyway you helped me a lot, thank you so much.
            $endgroup$
            – gbcosta
            Dec 4 '18 at 21:04










          • $begingroup$
            @gbcosta no problem
            $endgroup$
            – qbert
            Dec 4 '18 at 21:04






          • 2




            $begingroup$
            Your objection : "Take n=3..." is not valid : we order the entries f(A)=(a, b, c, d, e, f, g, h, i) is different from f(B)=(a, b, c, k, l, m, g, h, i) !
            $endgroup$
            – Jean Marie
            Dec 4 '18 at 21:12
















          $begingroup$
          But this map is one-to-one? I think that $$f:Arightarrow (a_{1,1},..., a_{1,n}, a_{2,1}...,a{2,n},...,a_{n,1},...,a_{n,n})$$ answer the question, but i'm not sure of that.
          $endgroup$
          – gbcosta
          Dec 4 '18 at 20:40






          $begingroup$
          But this map is one-to-one? I think that $$f:Arightarrow (a_{1,1},..., a_{1,n}, a_{2,1}...,a{2,n},...,a_{n,1},...,a_{n,n})$$ answer the question, but i'm not sure of that.
          $endgroup$
          – gbcosta
          Dec 4 '18 at 20:40














          $begingroup$
          Take n=3. If $A=begin{array}{lcr} a & b & c \ d & e & f \ g & h & j end{array}$ and $B = begin{array}{lcr} a & b & c \ k & l & m \ g & h & j end{array}$. So $Aneq B$ but f(A)=f(B)
          $endgroup$
          – gbcosta
          Dec 4 '18 at 20:57






          $begingroup$
          Take n=3. If $A=begin{array}{lcr} a & b & c \ d & e & f \ g & h & j end{array}$ and $B = begin{array}{lcr} a & b & c \ k & l & m \ g & h & j end{array}$. So $Aneq B$ but f(A)=f(B)
          $endgroup$
          – gbcosta
          Dec 4 '18 at 20:57














          $begingroup$
          No problem, anyway you helped me a lot, thank you so much.
          $endgroup$
          – gbcosta
          Dec 4 '18 at 21:04




          $begingroup$
          No problem, anyway you helped me a lot, thank you so much.
          $endgroup$
          – gbcosta
          Dec 4 '18 at 21:04












          $begingroup$
          @gbcosta no problem
          $endgroup$
          – qbert
          Dec 4 '18 at 21:04




          $begingroup$
          @gbcosta no problem
          $endgroup$
          – qbert
          Dec 4 '18 at 21:04




          2




          2




          $begingroup$
          Your objection : "Take n=3..." is not valid : we order the entries f(A)=(a, b, c, d, e, f, g, h, i) is different from f(B)=(a, b, c, k, l, m, g, h, i) !
          $endgroup$
          – Jean Marie
          Dec 4 '18 at 21:12




          $begingroup$
          Your objection : "Take n=3..." is not valid : we order the entries f(A)=(a, b, c, d, e, f, g, h, i) is different from f(B)=(a, b, c, k, l, m, g, h, i) !
          $endgroup$
          – Jean Marie
          Dec 4 '18 at 21:12


















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