$T^2=I$ implies $T$ is diagonalizable












3












$begingroup$


Suppose $T:Vrightarrow V$ is linear and $T^2=I$. Prove that $T$ is diagonalizable.



First, I know that $T$ has only eigenvalues 1 or -1. Also I observed that $(T-I)(T+I)=0$, does this fact help to show that $T$ is diagonalizable?






linear-algebra







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Yes it does. Do you know about minimal polynomials?
    $endgroup$
    – Omnomnomnom
    Jun 8 '15 at 20:02






  • 1




    $begingroup$
    @Omnomnomnom: Not really. I will try john's method and see if it works.
    $endgroup$
    – user112358
    Jun 8 '15 at 20:11










  • $begingroup$
    From Jordan form it follows immediately $ PJP^{-1} PJP^{-1}=I $,Squarred Jordan block is identity only if it is diagonal.
    $endgroup$
    – Widawensen
    Dec 13 '18 at 9:10


















3












$begingroup$


Suppose $T:Vrightarrow V$ is linear and $T^2=I$. Prove that $T$ is diagonalizable.



First, I know that $T$ has only eigenvalues 1 or -1. Also I observed that $(T-I)(T+I)=0$, does this fact help to show that $T$ is diagonalizable?






linear-algebra







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Yes it does. Do you know about minimal polynomials?
    $endgroup$
    – Omnomnomnom
    Jun 8 '15 at 20:02






  • 1




    $begingroup$
    @Omnomnomnom: Not really. I will try john's method and see if it works.
    $endgroup$
    – user112358
    Jun 8 '15 at 20:11










  • $begingroup$
    From Jordan form it follows immediately $ PJP^{-1} PJP^{-1}=I $,Squarred Jordan block is identity only if it is diagonal.
    $endgroup$
    – Widawensen
    Dec 13 '18 at 9:10
















3












3








3





$begingroup$


Suppose $T:Vrightarrow V$ is linear and $T^2=I$. Prove that $T$ is diagonalizable.



First, I know that $T$ has only eigenvalues 1 or -1. Also I observed that $(T-I)(T+I)=0$, does this fact help to show that $T$ is diagonalizable?






linear-algebra







share|cite|improve this question











$endgroup$




Suppose $T:Vrightarrow V$ is linear and $T^2=I$. Prove that $T$ is diagonalizable.



First, I know that $T$ has only eigenvalues 1 or -1. Also I observed that $(T-I)(T+I)=0$, does this fact help to show that $T$ is diagonalizable?






linear-algebra




linear-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 7 '16 at 5:14







user112358

















asked Jun 8 '15 at 19:59









user112358user112358

878518




878518








  • 1




    $begingroup$
    Yes it does. Do you know about minimal polynomials?
    $endgroup$
    – Omnomnomnom
    Jun 8 '15 at 20:02






  • 1




    $begingroup$
    @Omnomnomnom: Not really. I will try john's method and see if it works.
    $endgroup$
    – user112358
    Jun 8 '15 at 20:11










  • $begingroup$
    From Jordan form it follows immediately $ PJP^{-1} PJP^{-1}=I $,Squarred Jordan block is identity only if it is diagonal.
    $endgroup$
    – Widawensen
    Dec 13 '18 at 9:10
















  • 1




    $begingroup$
    Yes it does. Do you know about minimal polynomials?
    $endgroup$
    – Omnomnomnom
    Jun 8 '15 at 20:02






  • 1




    $begingroup$
    @Omnomnomnom: Not really. I will try john's method and see if it works.
    $endgroup$
    – user112358
    Jun 8 '15 at 20:11










  • $begingroup$
    From Jordan form it follows immediately $ PJP^{-1} PJP^{-1}=I $,Squarred Jordan block is identity only if it is diagonal.
    $endgroup$
    – Widawensen
    Dec 13 '18 at 9:10










1




1




$begingroup$
Yes it does. Do you know about minimal polynomials?
$endgroup$
– Omnomnomnom
Jun 8 '15 at 20:02




$begingroup$
Yes it does. Do you know about minimal polynomials?
$endgroup$
– Omnomnomnom
Jun 8 '15 at 20:02




1




1




$begingroup$
@Omnomnomnom: Not really. I will try john's method and see if it works.
$endgroup$
– user112358
Jun 8 '15 at 20:11




$begingroup$
@Omnomnomnom: Not really. I will try john's method and see if it works.
$endgroup$
– user112358
Jun 8 '15 at 20:11












$begingroup$
From Jordan form it follows immediately $ PJP^{-1} PJP^{-1}=I $,Squarred Jordan block is identity only if it is diagonal.
$endgroup$
– Widawensen
Dec 13 '18 at 9:10






$begingroup$
From Jordan form it follows immediately $ PJP^{-1} PJP^{-1}=I $,Squarred Jordan block is identity only if it is diagonal.
$endgroup$
– Widawensen
Dec 13 '18 at 9:10












4 Answers
4






active

oldest

votes


















5












$begingroup$

In this case you can split $V = V_1 oplus V_{-1}$ where $T$ acts as either $1$ or $-1$. Then take the direct sum over each subspace. $T = T_1 oplus T_{-1}$




Since $T^2 v = v$ for all $v in mathbb{V}$ (an involution), we can decompose every vector as:



$$ v = underbrace{tfrac{1}{2}(v + Tv)}_{E_1} +
underbrace{tfrac{1}{2}(v - Tv)}_{E_{-1}},$$



where $E_lambda$ is an eigenspace with eigenvalue $lambda$, so this eigenspace decomposition exists for all vector.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    how do I know that $V$ is the sum of the eigenspaces? Thanks.
    $endgroup$
    – user112358
    Jun 8 '15 at 20:16



















1












$begingroup$

Since the minimum polynomial factors over the field into distinct linear factors this implies that $T$ is diagonalisable.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    According to this answer, the Jordan Canonical Form of $A$ is a diagonal matrix with only $+1$ or $-1$ along the diagonal. That is, $A=SDS^{-1}$ where $D$ a diagonal matrix where each element on the diagonal is $+1$ or $-1$. Thus, $A$ is diagonalizable.






    share|cite|improve this answer











    $endgroup$





















      0












      $begingroup$

      Your proposition is in general false unless you specify which elements the matrices are allowed to take.



      The matrix $T = left(begin{array}{cc} 0&1\1&0end{array}right)$ with elements in non-negative reals or integers has $T^2 = I$ without being diagonalizable.



      This particular T gives $aT$ for positive real a is a representation for negative reals if $aI$ represent positive a. Furthermore if we define addition as the subspace spanned by all matrices except $left(begin{array}{cc} 1&1\1&1end{array}right)$ we can perform both addition and multiplication - we have constructed new numbers!



      Example, add 2 and -1: Just ordinary matrix addition: $left(begin{array}{cc} 2&1\1&2end{array}right)$, but after stripping the projection on $left(begin{array}{cc} 1&1\1&1end{array}right)$, we are left with $left(begin{array}{cc} 1&0\0&1end{array}right)$ which represents 1.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Aren't ${(1,1), (1,-1)}$ eigenvectors?
        $endgroup$
        – cactus314
        Jun 8 '15 at 21:00












      • $begingroup$
        Not if -1 is not allowed as an element.
        $endgroup$
        – mathreadler
        Jun 8 '15 at 21:02












      • $begingroup$
        Which field? Non-negative reals $mathbb{R}_{geq 0}$ are not a field, since $1$ does not have an additive inverse. $mathbb{Z}$ is not a field since $frac{1}{2} notin mathbb{Z}$.
        $endgroup$
        – cactus314
        Jun 8 '15 at 21:03












      • $begingroup$
        Yes you are right, it's not a field. Let's just say the set of elements is limited to positive reals.
        $endgroup$
        – mathreadler
        Jun 8 '15 at 21:05










      • $begingroup$
        If we use $mathbb{Z}$ instead of $mathbb{R}$ we get a $mathbb{Z}$-module. In the other case, $mathbb{R}_{geq 0}$ is not even a group, so non-negative matrices are a world of their own.
        $endgroup$
        – cactus314
        Jun 8 '15 at 21:09













      Your Answer





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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5












      $begingroup$

      In this case you can split $V = V_1 oplus V_{-1}$ where $T$ acts as either $1$ or $-1$. Then take the direct sum over each subspace. $T = T_1 oplus T_{-1}$




      Since $T^2 v = v$ for all $v in mathbb{V}$ (an involution), we can decompose every vector as:



      $$ v = underbrace{tfrac{1}{2}(v + Tv)}_{E_1} +
      underbrace{tfrac{1}{2}(v - Tv)}_{E_{-1}},$$



      where $E_lambda$ is an eigenspace with eigenvalue $lambda$, so this eigenspace decomposition exists for all vector.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        how do I know that $V$ is the sum of the eigenspaces? Thanks.
        $endgroup$
        – user112358
        Jun 8 '15 at 20:16
















      5












      $begingroup$

      In this case you can split $V = V_1 oplus V_{-1}$ where $T$ acts as either $1$ or $-1$. Then take the direct sum over each subspace. $T = T_1 oplus T_{-1}$




      Since $T^2 v = v$ for all $v in mathbb{V}$ (an involution), we can decompose every vector as:



      $$ v = underbrace{tfrac{1}{2}(v + Tv)}_{E_1} +
      underbrace{tfrac{1}{2}(v - Tv)}_{E_{-1}},$$



      where $E_lambda$ is an eigenspace with eigenvalue $lambda$, so this eigenspace decomposition exists for all vector.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        how do I know that $V$ is the sum of the eigenspaces? Thanks.
        $endgroup$
        – user112358
        Jun 8 '15 at 20:16














      5












      5








      5





      $begingroup$

      In this case you can split $V = V_1 oplus V_{-1}$ where $T$ acts as either $1$ or $-1$. Then take the direct sum over each subspace. $T = T_1 oplus T_{-1}$




      Since $T^2 v = v$ for all $v in mathbb{V}$ (an involution), we can decompose every vector as:



      $$ v = underbrace{tfrac{1}{2}(v + Tv)}_{E_1} +
      underbrace{tfrac{1}{2}(v - Tv)}_{E_{-1}},$$



      where $E_lambda$ is an eigenspace with eigenvalue $lambda$, so this eigenspace decomposition exists for all vector.






      share|cite|improve this answer











      $endgroup$



      In this case you can split $V = V_1 oplus V_{-1}$ where $T$ acts as either $1$ or $-1$. Then take the direct sum over each subspace. $T = T_1 oplus T_{-1}$




      Since $T^2 v = v$ for all $v in mathbb{V}$ (an involution), we can decompose every vector as:



      $$ v = underbrace{tfrac{1}{2}(v + Tv)}_{E_1} +
      underbrace{tfrac{1}{2}(v - Tv)}_{E_{-1}},$$



      where $E_lambda$ is an eigenspace with eigenvalue $lambda$, so this eigenspace decomposition exists for all vector.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Dec 13 '18 at 1:10









      Zack Ni

      3,480729




      3,480729










      answered Jun 8 '15 at 20:04









      cactus314cactus314

      15.4k42269




      15.4k42269












      • $begingroup$
        how do I know that $V$ is the sum of the eigenspaces? Thanks.
        $endgroup$
        – user112358
        Jun 8 '15 at 20:16


















      • $begingroup$
        how do I know that $V$ is the sum of the eigenspaces? Thanks.
        $endgroup$
        – user112358
        Jun 8 '15 at 20:16
















      $begingroup$
      how do I know that $V$ is the sum of the eigenspaces? Thanks.
      $endgroup$
      – user112358
      Jun 8 '15 at 20:16




      $begingroup$
      how do I know that $V$ is the sum of the eigenspaces? Thanks.
      $endgroup$
      – user112358
      Jun 8 '15 at 20:16











      1












      $begingroup$

      Since the minimum polynomial factors over the field into distinct linear factors this implies that $T$ is diagonalisable.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Since the minimum polynomial factors over the field into distinct linear factors this implies that $T$ is diagonalisable.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Since the minimum polynomial factors over the field into distinct linear factors this implies that $T$ is diagonalisable.






          share|cite|improve this answer









          $endgroup$



          Since the minimum polynomial factors over the field into distinct linear factors this implies that $T$ is diagonalisable.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jun 8 '15 at 20:02









          Rogelio MolinaRogelio Molina

          2,578818




          2,578818























              1












              $begingroup$

              According to this answer, the Jordan Canonical Form of $A$ is a diagonal matrix with only $+1$ or $-1$ along the diagonal. That is, $A=SDS^{-1}$ where $D$ a diagonal matrix where each element on the diagonal is $+1$ or $-1$. Thus, $A$ is diagonalizable.






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                According to this answer, the Jordan Canonical Form of $A$ is a diagonal matrix with only $+1$ or $-1$ along the diagonal. That is, $A=SDS^{-1}$ where $D$ a diagonal matrix where each element on the diagonal is $+1$ or $-1$. Thus, $A$ is diagonalizable.






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  According to this answer, the Jordan Canonical Form of $A$ is a diagonal matrix with only $+1$ or $-1$ along the diagonal. That is, $A=SDS^{-1}$ where $D$ a diagonal matrix where each element on the diagonal is $+1$ or $-1$. Thus, $A$ is diagonalizable.






                  share|cite|improve this answer











                  $endgroup$



                  According to this answer, the Jordan Canonical Form of $A$ is a diagonal matrix with only $+1$ or $-1$ along the diagonal. That is, $A=SDS^{-1}$ where $D$ a diagonal matrix where each element on the diagonal is $+1$ or $-1$. Thus, $A$ is diagonalizable.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Apr 13 '17 at 12:20









                  Community♦

                  1




                  1










                  answered Jun 9 '15 at 21:15









                  robjohn♦robjohn

                  268k27308634




                  268k27308634























                      0












                      $begingroup$

                      Your proposition is in general false unless you specify which elements the matrices are allowed to take.



                      The matrix $T = left(begin{array}{cc} 0&1\1&0end{array}right)$ with elements in non-negative reals or integers has $T^2 = I$ without being diagonalizable.



                      This particular T gives $aT$ for positive real a is a representation for negative reals if $aI$ represent positive a. Furthermore if we define addition as the subspace spanned by all matrices except $left(begin{array}{cc} 1&1\1&1end{array}right)$ we can perform both addition and multiplication - we have constructed new numbers!



                      Example, add 2 and -1: Just ordinary matrix addition: $left(begin{array}{cc} 2&1\1&2end{array}right)$, but after stripping the projection on $left(begin{array}{cc} 1&1\1&1end{array}right)$, we are left with $left(begin{array}{cc} 1&0\0&1end{array}right)$ which represents 1.






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        Aren't ${(1,1), (1,-1)}$ eigenvectors?
                        $endgroup$
                        – cactus314
                        Jun 8 '15 at 21:00












                      • $begingroup$
                        Not if -1 is not allowed as an element.
                        $endgroup$
                        – mathreadler
                        Jun 8 '15 at 21:02












                      • $begingroup$
                        Which field? Non-negative reals $mathbb{R}_{geq 0}$ are not a field, since $1$ does not have an additive inverse. $mathbb{Z}$ is not a field since $frac{1}{2} notin mathbb{Z}$.
                        $endgroup$
                        – cactus314
                        Jun 8 '15 at 21:03












                      • $begingroup$
                        Yes you are right, it's not a field. Let's just say the set of elements is limited to positive reals.
                        $endgroup$
                        – mathreadler
                        Jun 8 '15 at 21:05










                      • $begingroup$
                        If we use $mathbb{Z}$ instead of $mathbb{R}$ we get a $mathbb{Z}$-module. In the other case, $mathbb{R}_{geq 0}$ is not even a group, so non-negative matrices are a world of their own.
                        $endgroup$
                        – cactus314
                        Jun 8 '15 at 21:09


















                      0












                      $begingroup$

                      Your proposition is in general false unless you specify which elements the matrices are allowed to take.



                      The matrix $T = left(begin{array}{cc} 0&1\1&0end{array}right)$ with elements in non-negative reals or integers has $T^2 = I$ without being diagonalizable.



                      This particular T gives $aT$ for positive real a is a representation for negative reals if $aI$ represent positive a. Furthermore if we define addition as the subspace spanned by all matrices except $left(begin{array}{cc} 1&1\1&1end{array}right)$ we can perform both addition and multiplication - we have constructed new numbers!



                      Example, add 2 and -1: Just ordinary matrix addition: $left(begin{array}{cc} 2&1\1&2end{array}right)$, but after stripping the projection on $left(begin{array}{cc} 1&1\1&1end{array}right)$, we are left with $left(begin{array}{cc} 1&0\0&1end{array}right)$ which represents 1.






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        Aren't ${(1,1), (1,-1)}$ eigenvectors?
                        $endgroup$
                        – cactus314
                        Jun 8 '15 at 21:00












                      • $begingroup$
                        Not if -1 is not allowed as an element.
                        $endgroup$
                        – mathreadler
                        Jun 8 '15 at 21:02












                      • $begingroup$
                        Which field? Non-negative reals $mathbb{R}_{geq 0}$ are not a field, since $1$ does not have an additive inverse. $mathbb{Z}$ is not a field since $frac{1}{2} notin mathbb{Z}$.
                        $endgroup$
                        – cactus314
                        Jun 8 '15 at 21:03












                      • $begingroup$
                        Yes you are right, it's not a field. Let's just say the set of elements is limited to positive reals.
                        $endgroup$
                        – mathreadler
                        Jun 8 '15 at 21:05










                      • $begingroup$
                        If we use $mathbb{Z}$ instead of $mathbb{R}$ we get a $mathbb{Z}$-module. In the other case, $mathbb{R}_{geq 0}$ is not even a group, so non-negative matrices are a world of their own.
                        $endgroup$
                        – cactus314
                        Jun 8 '15 at 21:09
















                      0












                      0








                      0





                      $begingroup$

                      Your proposition is in general false unless you specify which elements the matrices are allowed to take.



                      The matrix $T = left(begin{array}{cc} 0&1\1&0end{array}right)$ with elements in non-negative reals or integers has $T^2 = I$ without being diagonalizable.



                      This particular T gives $aT$ for positive real a is a representation for negative reals if $aI$ represent positive a. Furthermore if we define addition as the subspace spanned by all matrices except $left(begin{array}{cc} 1&1\1&1end{array}right)$ we can perform both addition and multiplication - we have constructed new numbers!



                      Example, add 2 and -1: Just ordinary matrix addition: $left(begin{array}{cc} 2&1\1&2end{array}right)$, but after stripping the projection on $left(begin{array}{cc} 1&1\1&1end{array}right)$, we are left with $left(begin{array}{cc} 1&0\0&1end{array}right)$ which represents 1.






                      share|cite|improve this answer











                      $endgroup$



                      Your proposition is in general false unless you specify which elements the matrices are allowed to take.



                      The matrix $T = left(begin{array}{cc} 0&1\1&0end{array}right)$ with elements in non-negative reals or integers has $T^2 = I$ without being diagonalizable.



                      This particular T gives $aT$ for positive real a is a representation for negative reals if $aI$ represent positive a. Furthermore if we define addition as the subspace spanned by all matrices except $left(begin{array}{cc} 1&1\1&1end{array}right)$ we can perform both addition and multiplication - we have constructed new numbers!



                      Example, add 2 and -1: Just ordinary matrix addition: $left(begin{array}{cc} 2&1\1&2end{array}right)$, but after stripping the projection on $left(begin{array}{cc} 1&1\1&1end{array}right)$, we are left with $left(begin{array}{cc} 1&0\0&1end{array}right)$ which represents 1.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Jun 8 '15 at 21:23

























                      answered Jun 8 '15 at 20:54









                      mathreadlermathreadler

                      15k72263




                      15k72263












                      • $begingroup$
                        Aren't ${(1,1), (1,-1)}$ eigenvectors?
                        $endgroup$
                        – cactus314
                        Jun 8 '15 at 21:00












                      • $begingroup$
                        Not if -1 is not allowed as an element.
                        $endgroup$
                        – mathreadler
                        Jun 8 '15 at 21:02












                      • $begingroup$
                        Which field? Non-negative reals $mathbb{R}_{geq 0}$ are not a field, since $1$ does not have an additive inverse. $mathbb{Z}$ is not a field since $frac{1}{2} notin mathbb{Z}$.
                        $endgroup$
                        – cactus314
                        Jun 8 '15 at 21:03












                      • $begingroup$
                        Yes you are right, it's not a field. Let's just say the set of elements is limited to positive reals.
                        $endgroup$
                        – mathreadler
                        Jun 8 '15 at 21:05










                      • $begingroup$
                        If we use $mathbb{Z}$ instead of $mathbb{R}$ we get a $mathbb{Z}$-module. In the other case, $mathbb{R}_{geq 0}$ is not even a group, so non-negative matrices are a world of their own.
                        $endgroup$
                        – cactus314
                        Jun 8 '15 at 21:09




















                      • $begingroup$
                        Aren't ${(1,1), (1,-1)}$ eigenvectors?
                        $endgroup$
                        – cactus314
                        Jun 8 '15 at 21:00












                      • $begingroup$
                        Not if -1 is not allowed as an element.
                        $endgroup$
                        – mathreadler
                        Jun 8 '15 at 21:02












                      • $begingroup$
                        Which field? Non-negative reals $mathbb{R}_{geq 0}$ are not a field, since $1$ does not have an additive inverse. $mathbb{Z}$ is not a field since $frac{1}{2} notin mathbb{Z}$.
                        $endgroup$
                        – cactus314
                        Jun 8 '15 at 21:03












                      • $begingroup$
                        Yes you are right, it's not a field. Let's just say the set of elements is limited to positive reals.
                        $endgroup$
                        – mathreadler
                        Jun 8 '15 at 21:05










                      • $begingroup$
                        If we use $mathbb{Z}$ instead of $mathbb{R}$ we get a $mathbb{Z}$-module. In the other case, $mathbb{R}_{geq 0}$ is not even a group, so non-negative matrices are a world of their own.
                        $endgroup$
                        – cactus314
                        Jun 8 '15 at 21:09


















                      $begingroup$
                      Aren't ${(1,1), (1,-1)}$ eigenvectors?
                      $endgroup$
                      – cactus314
                      Jun 8 '15 at 21:00






                      $begingroup$
                      Aren't ${(1,1), (1,-1)}$ eigenvectors?
                      $endgroup$
                      – cactus314
                      Jun 8 '15 at 21:00














                      $begingroup$
                      Not if -1 is not allowed as an element.
                      $endgroup$
                      – mathreadler
                      Jun 8 '15 at 21:02






                      $begingroup$
                      Not if -1 is not allowed as an element.
                      $endgroup$
                      – mathreadler
                      Jun 8 '15 at 21:02














                      $begingroup$
                      Which field? Non-negative reals $mathbb{R}_{geq 0}$ are not a field, since $1$ does not have an additive inverse. $mathbb{Z}$ is not a field since $frac{1}{2} notin mathbb{Z}$.
                      $endgroup$
                      – cactus314
                      Jun 8 '15 at 21:03






                      $begingroup$
                      Which field? Non-negative reals $mathbb{R}_{geq 0}$ are not a field, since $1$ does not have an additive inverse. $mathbb{Z}$ is not a field since $frac{1}{2} notin mathbb{Z}$.
                      $endgroup$
                      – cactus314
                      Jun 8 '15 at 21:03














                      $begingroup$
                      Yes you are right, it's not a field. Let's just say the set of elements is limited to positive reals.
                      $endgroup$
                      – mathreadler
                      Jun 8 '15 at 21:05




                      $begingroup$
                      Yes you are right, it's not a field. Let's just say the set of elements is limited to positive reals.
                      $endgroup$
                      – mathreadler
                      Jun 8 '15 at 21:05












                      $begingroup$
                      If we use $mathbb{Z}$ instead of $mathbb{R}$ we get a $mathbb{Z}$-module. In the other case, $mathbb{R}_{geq 0}$ is not even a group, so non-negative matrices are a world of their own.
                      $endgroup$
                      – cactus314
                      Jun 8 '15 at 21:09






                      $begingroup$
                      If we use $mathbb{Z}$ instead of $mathbb{R}$ we get a $mathbb{Z}$-module. In the other case, $mathbb{R}_{geq 0}$ is not even a group, so non-negative matrices are a world of their own.
                      $endgroup$
                      – cactus314
                      Jun 8 '15 at 21:09




















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