Proof verification for Identity matrices











up vote
7
down vote

favorite












So I have the following question:



Analyze the following 'Claim' (which may or may not be true) and the corresponding 'Proof', by writing 'TRUE' or 'FALSE' (together with the reason) for each step. [Note: $I_n$ is the $n times n$ identity matrix.]



Claim: Let $A$ be any $n times n$ matrix satisfying $A^2=I_n$. Then either $A=I_n$ or $A=-I_n$.



'Proof'.



Step 1: $A$ satisfies $A^2-I_n = 0$ (True or False)



True.



My reasoning: Clearly, this is true. $A^2=I_n$ is not always true, but because it is true, I should have no problem moving the Identity matrix the the LHS.



Step 2: So $(A+I_n)(A-I_n)=0$ (True or false)



True.



My reasoning: Because $I_n$ is the identity matrix, there should be no issues with factoring just like normal algebra.



Step 3: $A+I_n=0$ or $A-I_n=0$



I'm not sure about this part. I'm very tempted to say this is fine but I am not sure how I can justify this, if I even can.



Therefore $A=-I_n$ or $A=I_n$. (End of 'Proof'.)



Is what I am doing right so far or am I messing up somewhere?










share|cite|improve this question


















  • 1




    What's the square of $A = begin{pmatrix}1 & 0 \ 0 & -1end{pmatrix}$?
    – md2perpe
    Nov 30 at 17:55















up vote
7
down vote

favorite












So I have the following question:



Analyze the following 'Claim' (which may or may not be true) and the corresponding 'Proof', by writing 'TRUE' or 'FALSE' (together with the reason) for each step. [Note: $I_n$ is the $n times n$ identity matrix.]



Claim: Let $A$ be any $n times n$ matrix satisfying $A^2=I_n$. Then either $A=I_n$ or $A=-I_n$.



'Proof'.



Step 1: $A$ satisfies $A^2-I_n = 0$ (True or False)



True.



My reasoning: Clearly, this is true. $A^2=I_n$ is not always true, but because it is true, I should have no problem moving the Identity matrix the the LHS.



Step 2: So $(A+I_n)(A-I_n)=0$ (True or false)



True.



My reasoning: Because $I_n$ is the identity matrix, there should be no issues with factoring just like normal algebra.



Step 3: $A+I_n=0$ or $A-I_n=0$



I'm not sure about this part. I'm very tempted to say this is fine but I am not sure how I can justify this, if I even can.



Therefore $A=-I_n$ or $A=I_n$. (End of 'Proof'.)



Is what I am doing right so far or am I messing up somewhere?










share|cite|improve this question


















  • 1




    What's the square of $A = begin{pmatrix}1 & 0 \ 0 & -1end{pmatrix}$?
    – md2perpe
    Nov 30 at 17:55













up vote
7
down vote

favorite









up vote
7
down vote

favorite











So I have the following question:



Analyze the following 'Claim' (which may or may not be true) and the corresponding 'Proof', by writing 'TRUE' or 'FALSE' (together with the reason) for each step. [Note: $I_n$ is the $n times n$ identity matrix.]



Claim: Let $A$ be any $n times n$ matrix satisfying $A^2=I_n$. Then either $A=I_n$ or $A=-I_n$.



'Proof'.



Step 1: $A$ satisfies $A^2-I_n = 0$ (True or False)



True.



My reasoning: Clearly, this is true. $A^2=I_n$ is not always true, but because it is true, I should have no problem moving the Identity matrix the the LHS.



Step 2: So $(A+I_n)(A-I_n)=0$ (True or false)



True.



My reasoning: Because $I_n$ is the identity matrix, there should be no issues with factoring just like normal algebra.



Step 3: $A+I_n=0$ or $A-I_n=0$



I'm not sure about this part. I'm very tempted to say this is fine but I am not sure how I can justify this, if I even can.



Therefore $A=-I_n$ or $A=I_n$. (End of 'Proof'.)



Is what I am doing right so far or am I messing up somewhere?










share|cite|improve this question













So I have the following question:



Analyze the following 'Claim' (which may or may not be true) and the corresponding 'Proof', by writing 'TRUE' or 'FALSE' (together with the reason) for each step. [Note: $I_n$ is the $n times n$ identity matrix.]



Claim: Let $A$ be any $n times n$ matrix satisfying $A^2=I_n$. Then either $A=I_n$ or $A=-I_n$.



'Proof'.



Step 1: $A$ satisfies $A^2-I_n = 0$ (True or False)



True.



My reasoning: Clearly, this is true. $A^2=I_n$ is not always true, but because it is true, I should have no problem moving the Identity matrix the the LHS.



Step 2: So $(A+I_n)(A-I_n)=0$ (True or false)



True.



My reasoning: Because $I_n$ is the identity matrix, there should be no issues with factoring just like normal algebra.



Step 3: $A+I_n=0$ or $A-I_n=0$



I'm not sure about this part. I'm very tempted to say this is fine but I am not sure how I can justify this, if I even can.



Therefore $A=-I_n$ or $A=I_n$. (End of 'Proof'.)



Is what I am doing right so far or am I messing up somewhere?







linear-algebra matrices proof-verification






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 30 at 7:51









Future Math person

857717




857717








  • 1




    What's the square of $A = begin{pmatrix}1 & 0 \ 0 & -1end{pmatrix}$?
    – md2perpe
    Nov 30 at 17:55














  • 1




    What's the square of $A = begin{pmatrix}1 & 0 \ 0 & -1end{pmatrix}$?
    – md2perpe
    Nov 30 at 17:55








1




1




What's the square of $A = begin{pmatrix}1 & 0 \ 0 & -1end{pmatrix}$?
– md2perpe
Nov 30 at 17:55




What's the square of $A = begin{pmatrix}1 & 0 \ 0 & -1end{pmatrix}$?
– md2perpe
Nov 30 at 17:55










3 Answers
3






active

oldest

votes

















up vote
12
down vote



accepted











  • Rather than saying that moving the identiy to the LHS, it is due to we add $-I$ to both sides.


  • We have $A^2-I=(A-I)(A+I)$, we just have to expand the right hand side to verify that.


  • In matrices, $AB=0$ doesn't imply that $A=0$ or $B=0$. For example $$begin{bmatrix} 2 & 0 \ 0 & 0end{bmatrix}begin{bmatrix} 0 & 0 \ 0 & -2end{bmatrix}= begin{bmatrix} 0 & 0 \ 0 & 0end{bmatrix}$$


  • In particular,



$$left(begin{bmatrix} 1 & 0 \ 0 & -1end{bmatrix}+begin{bmatrix} 1 & 0 \ 0 & 1end{bmatrix}right)left(begin{bmatrix} 1 & 0 \ 0 & -1end{bmatrix}-begin{bmatrix} 1 & 0 \ 0 & 1end{bmatrix}right)= begin{bmatrix} 0 & 0 \ 0 & 0end{bmatrix}$$



that is we cant' conclude that $(A+I)(A-I)=0$ implies $A+I=0$ or $A-I=0$ as well.






share|cite|improve this answer























  • Aha. I knew something looked fishy with that last part. Thanks!
    – Future Math person
    Nov 30 at 7:59






  • 2




    When expanding the right hand side, be sure to respect the non-commutativity of matrix multiplication. $$(A+I)(A-I) = (A+I)A - (A+I)I$$ $$qquad = A^2 + IA - AI -I^2 = A^2 - I^2.$$ Although the difference in this particular problem is negligible, it is not generally so.
    – Eric Towers
    Nov 30 at 13:36








  • 1




    In the third point it should be "that $A=0$ or $B=0$".
    – Lonidard
    Nov 30 at 14:29












  • thanks for pointing that out.
    – Siong Thye Goh
    Nov 30 at 14:33






  • 1




    I would like this answer better if $A$ and $B$ were of the form $Apm I$ so that it matches OP's proof. Using the other answer, you could take $A=diag(2,0)$ and $B=diag(0,-2)$, so that they are $diag(1,-1)pm I$.
    – Teepeemm
    Nov 30 at 18:45


















up vote
9
down vote













Consider any diagonal matrix with diagonal elements $pm 1$. Show that$A^{2}=I_n$. you get $2^{n}$ matrices whose square is $I_n$.






share|cite|improve this answer




























    up vote
    2
    down vote













    Furthermore if you want a concrete example of a matrix whose square is the identity but not itself a simple matrix consider for example this one:



    $$begin{bmatrix}frac{1}{2} & frac{3}{4} \ 1 & -frac{1}{2}end{bmatrix}$$



    These matrices are called involuntory






    share|cite|improve this answer





















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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      12
      down vote



      accepted











      • Rather than saying that moving the identiy to the LHS, it is due to we add $-I$ to both sides.


      • We have $A^2-I=(A-I)(A+I)$, we just have to expand the right hand side to verify that.


      • In matrices, $AB=0$ doesn't imply that $A=0$ or $B=0$. For example $$begin{bmatrix} 2 & 0 \ 0 & 0end{bmatrix}begin{bmatrix} 0 & 0 \ 0 & -2end{bmatrix}= begin{bmatrix} 0 & 0 \ 0 & 0end{bmatrix}$$


      • In particular,



      $$left(begin{bmatrix} 1 & 0 \ 0 & -1end{bmatrix}+begin{bmatrix} 1 & 0 \ 0 & 1end{bmatrix}right)left(begin{bmatrix} 1 & 0 \ 0 & -1end{bmatrix}-begin{bmatrix} 1 & 0 \ 0 & 1end{bmatrix}right)= begin{bmatrix} 0 & 0 \ 0 & 0end{bmatrix}$$



      that is we cant' conclude that $(A+I)(A-I)=0$ implies $A+I=0$ or $A-I=0$ as well.






      share|cite|improve this answer























      • Aha. I knew something looked fishy with that last part. Thanks!
        – Future Math person
        Nov 30 at 7:59






      • 2




        When expanding the right hand side, be sure to respect the non-commutativity of matrix multiplication. $$(A+I)(A-I) = (A+I)A - (A+I)I$$ $$qquad = A^2 + IA - AI -I^2 = A^2 - I^2.$$ Although the difference in this particular problem is negligible, it is not generally so.
        – Eric Towers
        Nov 30 at 13:36








      • 1




        In the third point it should be "that $A=0$ or $B=0$".
        – Lonidard
        Nov 30 at 14:29












      • thanks for pointing that out.
        – Siong Thye Goh
        Nov 30 at 14:33






      • 1




        I would like this answer better if $A$ and $B$ were of the form $Apm I$ so that it matches OP's proof. Using the other answer, you could take $A=diag(2,0)$ and $B=diag(0,-2)$, so that they are $diag(1,-1)pm I$.
        – Teepeemm
        Nov 30 at 18:45















      up vote
      12
      down vote



      accepted











      • Rather than saying that moving the identiy to the LHS, it is due to we add $-I$ to both sides.


      • We have $A^2-I=(A-I)(A+I)$, we just have to expand the right hand side to verify that.


      • In matrices, $AB=0$ doesn't imply that $A=0$ or $B=0$. For example $$begin{bmatrix} 2 & 0 \ 0 & 0end{bmatrix}begin{bmatrix} 0 & 0 \ 0 & -2end{bmatrix}= begin{bmatrix} 0 & 0 \ 0 & 0end{bmatrix}$$


      • In particular,



      $$left(begin{bmatrix} 1 & 0 \ 0 & -1end{bmatrix}+begin{bmatrix} 1 & 0 \ 0 & 1end{bmatrix}right)left(begin{bmatrix} 1 & 0 \ 0 & -1end{bmatrix}-begin{bmatrix} 1 & 0 \ 0 & 1end{bmatrix}right)= begin{bmatrix} 0 & 0 \ 0 & 0end{bmatrix}$$



      that is we cant' conclude that $(A+I)(A-I)=0$ implies $A+I=0$ or $A-I=0$ as well.






      share|cite|improve this answer























      • Aha. I knew something looked fishy with that last part. Thanks!
        – Future Math person
        Nov 30 at 7:59






      • 2




        When expanding the right hand side, be sure to respect the non-commutativity of matrix multiplication. $$(A+I)(A-I) = (A+I)A - (A+I)I$$ $$qquad = A^2 + IA - AI -I^2 = A^2 - I^2.$$ Although the difference in this particular problem is negligible, it is not generally so.
        – Eric Towers
        Nov 30 at 13:36








      • 1




        In the third point it should be "that $A=0$ or $B=0$".
        – Lonidard
        Nov 30 at 14:29












      • thanks for pointing that out.
        – Siong Thye Goh
        Nov 30 at 14:33






      • 1




        I would like this answer better if $A$ and $B$ were of the form $Apm I$ so that it matches OP's proof. Using the other answer, you could take $A=diag(2,0)$ and $B=diag(0,-2)$, so that they are $diag(1,-1)pm I$.
        – Teepeemm
        Nov 30 at 18:45













      up vote
      12
      down vote



      accepted







      up vote
      12
      down vote



      accepted







      • Rather than saying that moving the identiy to the LHS, it is due to we add $-I$ to both sides.


      • We have $A^2-I=(A-I)(A+I)$, we just have to expand the right hand side to verify that.


      • In matrices, $AB=0$ doesn't imply that $A=0$ or $B=0$. For example $$begin{bmatrix} 2 & 0 \ 0 & 0end{bmatrix}begin{bmatrix} 0 & 0 \ 0 & -2end{bmatrix}= begin{bmatrix} 0 & 0 \ 0 & 0end{bmatrix}$$


      • In particular,



      $$left(begin{bmatrix} 1 & 0 \ 0 & -1end{bmatrix}+begin{bmatrix} 1 & 0 \ 0 & 1end{bmatrix}right)left(begin{bmatrix} 1 & 0 \ 0 & -1end{bmatrix}-begin{bmatrix} 1 & 0 \ 0 & 1end{bmatrix}right)= begin{bmatrix} 0 & 0 \ 0 & 0end{bmatrix}$$



      that is we cant' conclude that $(A+I)(A-I)=0$ implies $A+I=0$ or $A-I=0$ as well.






      share|cite|improve this answer















      • Rather than saying that moving the identiy to the LHS, it is due to we add $-I$ to both sides.


      • We have $A^2-I=(A-I)(A+I)$, we just have to expand the right hand side to verify that.


      • In matrices, $AB=0$ doesn't imply that $A=0$ or $B=0$. For example $$begin{bmatrix} 2 & 0 \ 0 & 0end{bmatrix}begin{bmatrix} 0 & 0 \ 0 & -2end{bmatrix}= begin{bmatrix} 0 & 0 \ 0 & 0end{bmatrix}$$


      • In particular,



      $$left(begin{bmatrix} 1 & 0 \ 0 & -1end{bmatrix}+begin{bmatrix} 1 & 0 \ 0 & 1end{bmatrix}right)left(begin{bmatrix} 1 & 0 \ 0 & -1end{bmatrix}-begin{bmatrix} 1 & 0 \ 0 & 1end{bmatrix}right)= begin{bmatrix} 0 & 0 \ 0 & 0end{bmatrix}$$



      that is we cant' conclude that $(A+I)(A-I)=0$ implies $A+I=0$ or $A-I=0$ as well.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Dec 1 at 5:08

























      answered Nov 30 at 7:57









      Siong Thye Goh

      96.4k1462116




      96.4k1462116












      • Aha. I knew something looked fishy with that last part. Thanks!
        – Future Math person
        Nov 30 at 7:59






      • 2




        When expanding the right hand side, be sure to respect the non-commutativity of matrix multiplication. $$(A+I)(A-I) = (A+I)A - (A+I)I$$ $$qquad = A^2 + IA - AI -I^2 = A^2 - I^2.$$ Although the difference in this particular problem is negligible, it is not generally so.
        – Eric Towers
        Nov 30 at 13:36








      • 1




        In the third point it should be "that $A=0$ or $B=0$".
        – Lonidard
        Nov 30 at 14:29












      • thanks for pointing that out.
        – Siong Thye Goh
        Nov 30 at 14:33






      • 1




        I would like this answer better if $A$ and $B$ were of the form $Apm I$ so that it matches OP's proof. Using the other answer, you could take $A=diag(2,0)$ and $B=diag(0,-2)$, so that they are $diag(1,-1)pm I$.
        – Teepeemm
        Nov 30 at 18:45


















      • Aha. I knew something looked fishy with that last part. Thanks!
        – Future Math person
        Nov 30 at 7:59






      • 2




        When expanding the right hand side, be sure to respect the non-commutativity of matrix multiplication. $$(A+I)(A-I) = (A+I)A - (A+I)I$$ $$qquad = A^2 + IA - AI -I^2 = A^2 - I^2.$$ Although the difference in this particular problem is negligible, it is not generally so.
        – Eric Towers
        Nov 30 at 13:36








      • 1




        In the third point it should be "that $A=0$ or $B=0$".
        – Lonidard
        Nov 30 at 14:29












      • thanks for pointing that out.
        – Siong Thye Goh
        Nov 30 at 14:33






      • 1




        I would like this answer better if $A$ and $B$ were of the form $Apm I$ so that it matches OP's proof. Using the other answer, you could take $A=diag(2,0)$ and $B=diag(0,-2)$, so that they are $diag(1,-1)pm I$.
        – Teepeemm
        Nov 30 at 18:45
















      Aha. I knew something looked fishy with that last part. Thanks!
      – Future Math person
      Nov 30 at 7:59




      Aha. I knew something looked fishy with that last part. Thanks!
      – Future Math person
      Nov 30 at 7:59




      2




      2




      When expanding the right hand side, be sure to respect the non-commutativity of matrix multiplication. $$(A+I)(A-I) = (A+I)A - (A+I)I$$ $$qquad = A^2 + IA - AI -I^2 = A^2 - I^2.$$ Although the difference in this particular problem is negligible, it is not generally so.
      – Eric Towers
      Nov 30 at 13:36






      When expanding the right hand side, be sure to respect the non-commutativity of matrix multiplication. $$(A+I)(A-I) = (A+I)A - (A+I)I$$ $$qquad = A^2 + IA - AI -I^2 = A^2 - I^2.$$ Although the difference in this particular problem is negligible, it is not generally so.
      – Eric Towers
      Nov 30 at 13:36






      1




      1




      In the third point it should be "that $A=0$ or $B=0$".
      – Lonidard
      Nov 30 at 14:29






      In the third point it should be "that $A=0$ or $B=0$".
      – Lonidard
      Nov 30 at 14:29














      thanks for pointing that out.
      – Siong Thye Goh
      Nov 30 at 14:33




      thanks for pointing that out.
      – Siong Thye Goh
      Nov 30 at 14:33




      1




      1




      I would like this answer better if $A$ and $B$ were of the form $Apm I$ so that it matches OP's proof. Using the other answer, you could take $A=diag(2,0)$ and $B=diag(0,-2)$, so that they are $diag(1,-1)pm I$.
      – Teepeemm
      Nov 30 at 18:45




      I would like this answer better if $A$ and $B$ were of the form $Apm I$ so that it matches OP's proof. Using the other answer, you could take $A=diag(2,0)$ and $B=diag(0,-2)$, so that they are $diag(1,-1)pm I$.
      – Teepeemm
      Nov 30 at 18:45










      up vote
      9
      down vote













      Consider any diagonal matrix with diagonal elements $pm 1$. Show that$A^{2}=I_n$. you get $2^{n}$ matrices whose square is $I_n$.






      share|cite|improve this answer

























        up vote
        9
        down vote













        Consider any diagonal matrix with diagonal elements $pm 1$. Show that$A^{2}=I_n$. you get $2^{n}$ matrices whose square is $I_n$.






        share|cite|improve this answer























          up vote
          9
          down vote










          up vote
          9
          down vote









          Consider any diagonal matrix with diagonal elements $pm 1$. Show that$A^{2}=I_n$. you get $2^{n}$ matrices whose square is $I_n$.






          share|cite|improve this answer












          Consider any diagonal matrix with diagonal elements $pm 1$. Show that$A^{2}=I_n$. you get $2^{n}$ matrices whose square is $I_n$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 30 at 7:53









          Kavi Rama Murthy

          45.1k31852




          45.1k31852






















              up vote
              2
              down vote













              Furthermore if you want a concrete example of a matrix whose square is the identity but not itself a simple matrix consider for example this one:



              $$begin{bmatrix}frac{1}{2} & frac{3}{4} \ 1 & -frac{1}{2}end{bmatrix}$$



              These matrices are called involuntory






              share|cite|improve this answer

























                up vote
                2
                down vote













                Furthermore if you want a concrete example of a matrix whose square is the identity but not itself a simple matrix consider for example this one:



                $$begin{bmatrix}frac{1}{2} & frac{3}{4} \ 1 & -frac{1}{2}end{bmatrix}$$



                These matrices are called involuntory






                share|cite|improve this answer























                  up vote
                  2
                  down vote










                  up vote
                  2
                  down vote









                  Furthermore if you want a concrete example of a matrix whose square is the identity but not itself a simple matrix consider for example this one:



                  $$begin{bmatrix}frac{1}{2} & frac{3}{4} \ 1 & -frac{1}{2}end{bmatrix}$$



                  These matrices are called involuntory






                  share|cite|improve this answer












                  Furthermore if you want a concrete example of a matrix whose square is the identity but not itself a simple matrix consider for example this one:



                  $$begin{bmatrix}frac{1}{2} & frac{3}{4} \ 1 & -frac{1}{2}end{bmatrix}$$



                  These matrices are called involuntory







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 30 at 15:43









                  gota

                  394315




                  394315






























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