injectivity and bijectivity of linear maps












0












$begingroup$


I have got in trouble with one of the problems I'm solving for my university studies. This is the Problem:




Let $A in Bbb R^{n times n}$ and $Uin Bbb R^{n times n}$ such that $ker U = {0}$.
Define $B := UAU^{-1}$. Show that $f_A$ is injective if and only if $f_B$ is injective. What about surjectivity?




This is how far I have come:



$ f_U $ is injective and surjective and therefor bijective since $ ker A = {0} land operatorname{ran} U = Bbb R^{n} $.
And I have got to show that $ ker A = {0}$ iff $ker B = {0}$. That means I have to show that $ker A = {o} Rightarrow ker B = {o} land ker B = {o} Rightarrow ker A = {o}$.



Some help of you guys would come in handy.






linear-algebra proof-writing







share|cite|improve this question









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    0












    $begingroup$


    I have got in trouble with one of the problems I'm solving for my university studies. This is the Problem:




    Let $A in Bbb R^{n times n}$ and $Uin Bbb R^{n times n}$ such that $ker U = {0}$.
    Define $B := UAU^{-1}$. Show that $f_A$ is injective if and only if $f_B$ is injective. What about surjectivity?




    This is how far I have come:



    $ f_U $ is injective and surjective and therefor bijective since $ ker A = {0} land operatorname{ran} U = Bbb R^{n} $.
    And I have got to show that $ ker A = {0}$ iff $ker B = {0}$. That means I have to show that $ker A = {o} Rightarrow ker B = {o} land ker B = {o} Rightarrow ker A = {o}$.



    Some help of you guys would come in handy.






    linear-algebra proof-writing







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I have got in trouble with one of the problems I'm solving for my university studies. This is the Problem:




      Let $A in Bbb R^{n times n}$ and $Uin Bbb R^{n times n}$ such that $ker U = {0}$.
      Define $B := UAU^{-1}$. Show that $f_A$ is injective if and only if $f_B$ is injective. What about surjectivity?




      This is how far I have come:



      $ f_U $ is injective and surjective and therefor bijective since $ ker A = {0} land operatorname{ran} U = Bbb R^{n} $.
      And I have got to show that $ ker A = {0}$ iff $ker B = {0}$. That means I have to show that $ker A = {o} Rightarrow ker B = {o} land ker B = {o} Rightarrow ker A = {o}$.



      Some help of you guys would come in handy.






      linear-algebra proof-writing







      share|cite|improve this question









      $endgroup$




      I have got in trouble with one of the problems I'm solving for my university studies. This is the Problem:




      Let $A in Bbb R^{n times n}$ and $Uin Bbb R^{n times n}$ such that $ker U = {0}$.
      Define $B := UAU^{-1}$. Show that $f_A$ is injective if and only if $f_B$ is injective. What about surjectivity?




      This is how far I have come:



      $ f_U $ is injective and surjective and therefor bijective since $ ker A = {0} land operatorname{ran} U = Bbb R^{n} $.
      And I have got to show that $ ker A = {0}$ iff $ker B = {0}$. That means I have to show that $ker A = {o} Rightarrow ker B = {o} land ker B = {o} Rightarrow ker A = {o}$.



      Some help of you guys would come in handy.






      linear-algebra proof-writing




      linear-algebra proof-writing






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 2 '18 at 15:28









      MathmeeeeenMathmeeeeen

      133




      133






















          1 Answer
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          $begingroup$

          More generally let $f:Xto X$ and $g:Yto Y$ be functions, and let
          $h:Yto X$ be an invertible function with $g=hcirc fcirc h^{-1}$.



          Then $h$ and $h^{-1}$ are both bijective.



          So if $f$ is injective then $g$ is a composition of injective functions
          and if $f$ is surjective then $g$ is a composition of surjective
          functions.




          So we can apply the following lemma's:



          1) If $u:Ato B$ and $v:Bto C$ are injective then so is $vcirc u:Ato C$



          Proof: $vcirc uleft(aright)=vcirc uleft(a'right)stackrel{vtext{ injective}}{implies}uleft(aright)=uleft(a'right)stackrel{utext{ injective}}{implies}a=a'$



          2) If $u:Ato B$ and $v:Bto C$ are surjective then so is $vcirc u:Ato C$



          Proof: Let $cin C$. Then $c=vleft(bright)$ for some $bin B$
          and $b=uleft(aright)$ for some $ain A$. So $c=ucirc vleft(aright)$.




          Also we have $f=h^{-1}circ gcirc h$ so in both cases also the converse is true. That means that we end up with:$$ftext{ is injective }iff gtext{ is injective }$$and:$$ftext{ is surjective }iff gtext{ is surjective }$$






          share|cite|improve this answer









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            1 Answer
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            active

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            0












            $begingroup$

            More generally let $f:Xto X$ and $g:Yto Y$ be functions, and let
            $h:Yto X$ be an invertible function with $g=hcirc fcirc h^{-1}$.



            Then $h$ and $h^{-1}$ are both bijective.



            So if $f$ is injective then $g$ is a composition of injective functions
            and if $f$ is surjective then $g$ is a composition of surjective
            functions.




            So we can apply the following lemma's:



            1) If $u:Ato B$ and $v:Bto C$ are injective then so is $vcirc u:Ato C$



            Proof: $vcirc uleft(aright)=vcirc uleft(a'right)stackrel{vtext{ injective}}{implies}uleft(aright)=uleft(a'right)stackrel{utext{ injective}}{implies}a=a'$



            2) If $u:Ato B$ and $v:Bto C$ are surjective then so is $vcirc u:Ato C$



            Proof: Let $cin C$. Then $c=vleft(bright)$ for some $bin B$
            and $b=uleft(aright)$ for some $ain A$. So $c=ucirc vleft(aright)$.




            Also we have $f=h^{-1}circ gcirc h$ so in both cases also the converse is true. That means that we end up with:$$ftext{ is injective }iff gtext{ is injective }$$and:$$ftext{ is surjective }iff gtext{ is surjective }$$






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              More generally let $f:Xto X$ and $g:Yto Y$ be functions, and let
              $h:Yto X$ be an invertible function with $g=hcirc fcirc h^{-1}$.



              Then $h$ and $h^{-1}$ are both bijective.



              So if $f$ is injective then $g$ is a composition of injective functions
              and if $f$ is surjective then $g$ is a composition of surjective
              functions.




              So we can apply the following lemma's:



              1) If $u:Ato B$ and $v:Bto C$ are injective then so is $vcirc u:Ato C$



              Proof: $vcirc uleft(aright)=vcirc uleft(a'right)stackrel{vtext{ injective}}{implies}uleft(aright)=uleft(a'right)stackrel{utext{ injective}}{implies}a=a'$



              2) If $u:Ato B$ and $v:Bto C$ are surjective then so is $vcirc u:Ato C$



              Proof: Let $cin C$. Then $c=vleft(bright)$ for some $bin B$
              and $b=uleft(aright)$ for some $ain A$. So $c=ucirc vleft(aright)$.




              Also we have $f=h^{-1}circ gcirc h$ so in both cases also the converse is true. That means that we end up with:$$ftext{ is injective }iff gtext{ is injective }$$and:$$ftext{ is surjective }iff gtext{ is surjective }$$






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                More generally let $f:Xto X$ and $g:Yto Y$ be functions, and let
                $h:Yto X$ be an invertible function with $g=hcirc fcirc h^{-1}$.



                Then $h$ and $h^{-1}$ are both bijective.



                So if $f$ is injective then $g$ is a composition of injective functions
                and if $f$ is surjective then $g$ is a composition of surjective
                functions.




                So we can apply the following lemma's:



                1) If $u:Ato B$ and $v:Bto C$ are injective then so is $vcirc u:Ato C$



                Proof: $vcirc uleft(aright)=vcirc uleft(a'right)stackrel{vtext{ injective}}{implies}uleft(aright)=uleft(a'right)stackrel{utext{ injective}}{implies}a=a'$



                2) If $u:Ato B$ and $v:Bto C$ are surjective then so is $vcirc u:Ato C$



                Proof: Let $cin C$. Then $c=vleft(bright)$ for some $bin B$
                and $b=uleft(aright)$ for some $ain A$. So $c=ucirc vleft(aright)$.




                Also we have $f=h^{-1}circ gcirc h$ so in both cases also the converse is true. That means that we end up with:$$ftext{ is injective }iff gtext{ is injective }$$and:$$ftext{ is surjective }iff gtext{ is surjective }$$






                share|cite|improve this answer









                $endgroup$



                More generally let $f:Xto X$ and $g:Yto Y$ be functions, and let
                $h:Yto X$ be an invertible function with $g=hcirc fcirc h^{-1}$.



                Then $h$ and $h^{-1}$ are both bijective.



                So if $f$ is injective then $g$ is a composition of injective functions
                and if $f$ is surjective then $g$ is a composition of surjective
                functions.




                So we can apply the following lemma's:



                1) If $u:Ato B$ and $v:Bto C$ are injective then so is $vcirc u:Ato C$



                Proof: $vcirc uleft(aright)=vcirc uleft(a'right)stackrel{vtext{ injective}}{implies}uleft(aright)=uleft(a'right)stackrel{utext{ injective}}{implies}a=a'$



                2) If $u:Ato B$ and $v:Bto C$ are surjective then so is $vcirc u:Ato C$



                Proof: Let $cin C$. Then $c=vleft(bright)$ for some $bin B$
                and $b=uleft(aright)$ for some $ain A$. So $c=ucirc vleft(aright)$.




                Also we have $f=h^{-1}circ gcirc h$ so in both cases also the converse is true. That means that we end up with:$$ftext{ is injective }iff gtext{ is injective }$$and:$$ftext{ is surjective }iff gtext{ is surjective }$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 2 '18 at 15:36









                drhabdrhab

                98.9k544129




                98.9k544129






























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