Prove that the range is a subspace












1












$begingroup$


In Linear Algebra Done Right, it said



If $T in mathcal{L}(V,W)$, then range $T$ is a subspace of $W$.



Proof:



Suppose $T in mathcal{L}(V,W)$. Then $T(0) = 0$, which implies that $0 in text{range } T$.



If $w_1,w_2 in text{range } T$, then there exist $v_1,v_2 in V$ such that $Tv_1 = w_1$ and $Tv_2 = w_2$. Thus
$$T(v_1+v_2) = Tv_1 + Tv_2 = w_1 + w_2$$
Hence $w_1+w_2 in text{range } T$



Why $w_1+w_2 in text{range } T$ ? I can follow the steps but don't understand why it is in the range



Similarly, I don't understand why $lambda w$ is in the range of $T$






linear-algebra







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    In Linear Algebra Done Right, it said



    If $T in mathcal{L}(V,W)$, then range $T$ is a subspace of $W$.



    Proof:



    Suppose $T in mathcal{L}(V,W)$. Then $T(0) = 0$, which implies that $0 in text{range } T$.



    If $w_1,w_2 in text{range } T$, then there exist $v_1,v_2 in V$ such that $Tv_1 = w_1$ and $Tv_2 = w_2$. Thus
    $$T(v_1+v_2) = Tv_1 + Tv_2 = w_1 + w_2$$
    Hence $w_1+w_2 in text{range } T$



    Why $w_1+w_2 in text{range } T$ ? I can follow the steps but don't understand why it is in the range



    Similarly, I don't understand why $lambda w$ is in the range of $T$






    linear-algebra







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      In Linear Algebra Done Right, it said



      If $T in mathcal{L}(V,W)$, then range $T$ is a subspace of $W$.



      Proof:



      Suppose $T in mathcal{L}(V,W)$. Then $T(0) = 0$, which implies that $0 in text{range } T$.



      If $w_1,w_2 in text{range } T$, then there exist $v_1,v_2 in V$ such that $Tv_1 = w_1$ and $Tv_2 = w_2$. Thus
      $$T(v_1+v_2) = Tv_1 + Tv_2 = w_1 + w_2$$
      Hence $w_1+w_2 in text{range } T$



      Why $w_1+w_2 in text{range } T$ ? I can follow the steps but don't understand why it is in the range



      Similarly, I don't understand why $lambda w$ is in the range of $T$






      linear-algebra







      share|cite|improve this question









      $endgroup$




      In Linear Algebra Done Right, it said



      If $T in mathcal{L}(V,W)$, then range $T$ is a subspace of $W$.



      Proof:



      Suppose $T in mathcal{L}(V,W)$. Then $T(0) = 0$, which implies that $0 in text{range } T$.



      If $w_1,w_2 in text{range } T$, then there exist $v_1,v_2 in V$ such that $Tv_1 = w_1$ and $Tv_2 = w_2$. Thus
      $$T(v_1+v_2) = Tv_1 + Tv_2 = w_1 + w_2$$
      Hence $w_1+w_2 in text{range } T$



      Why $w_1+w_2 in text{range } T$ ? I can follow the steps but don't understand why it is in the range



      Similarly, I don't understand why $lambda w$ is in the range of $T$






      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 Jan 16 at 23:05









      JOHN JOHN

      3268




      3268






















          3 Answers
          3






          active

          oldest

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          1












          $begingroup$

          Here $y$ is in the range of $T$ if $y = T(x)$ for some $x in V$. Note that $v_1+v_2 in V$ since $V$ is a vector space, and $w_1+w_2 = T(v_1+v_2)$, then $w_1+w_2$ is in the range of $T$. Likewise, $lambda w = T(lambda v)$, implying $lambda w$ is in the range of $T$. Thus The range of $T$ is a subspace of $W$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            $v_1 + v_2 in V$ helps me understand why $w_1 + w_2 = T(v_1 + v_2)$, then $w_1 + w_2$ is in the range of $T$.
            $endgroup$
            – JOHN
            Jan 17 at 7:31



















          2












          $begingroup$

          Range of $T$ consists all vectors of the form $T(v)$ where $v in V$. $w_1+w_2=T(v_1+v_2)$. Hence $w_1+w_2$ belongs to range of $T$. If $w=T(v)$ then $lambda w=T(lambda v)$ so $lambda w$ is in the range of $T$.






          share|cite|improve this answer









          $endgroup$





















            -1












            $begingroup$

            Note the previous line: $T(text{something}) = w_1 + w_2$. Thus, there is something that, when we hit it with $t$, we get w_1 + w_2. The proof for $lambda w$ will be similar: since $T$ is linear, $T(lambda v) = lambda T(v) = lambda w$, so there is something that, when we hit it with $T$, we get $lambda w$, so $lambda w$ is in the range of $T$.






            share|cite|improve this answer









            $endgroup$













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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              1












              $begingroup$

              Here $y$ is in the range of $T$ if $y = T(x)$ for some $x in V$. Note that $v_1+v_2 in V$ since $V$ is a vector space, and $w_1+w_2 = T(v_1+v_2)$, then $w_1+w_2$ is in the range of $T$. Likewise, $lambda w = T(lambda v)$, implying $lambda w$ is in the range of $T$. Thus The range of $T$ is a subspace of $W$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                $v_1 + v_2 in V$ helps me understand why $w_1 + w_2 = T(v_1 + v_2)$, then $w_1 + w_2$ is in the range of $T$.
                $endgroup$
                – JOHN
                Jan 17 at 7:31
















              1












              $begingroup$

              Here $y$ is in the range of $T$ if $y = T(x)$ for some $x in V$. Note that $v_1+v_2 in V$ since $V$ is a vector space, and $w_1+w_2 = T(v_1+v_2)$, then $w_1+w_2$ is in the range of $T$. Likewise, $lambda w = T(lambda v)$, implying $lambda w$ is in the range of $T$. Thus The range of $T$ is a subspace of $W$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                $v_1 + v_2 in V$ helps me understand why $w_1 + w_2 = T(v_1 + v_2)$, then $w_1 + w_2$ is in the range of $T$.
                $endgroup$
                – JOHN
                Jan 17 at 7:31














              1












              1








              1





              $begingroup$

              Here $y$ is in the range of $T$ if $y = T(x)$ for some $x in V$. Note that $v_1+v_2 in V$ since $V$ is a vector space, and $w_1+w_2 = T(v_1+v_2)$, then $w_1+w_2$ is in the range of $T$. Likewise, $lambda w = T(lambda v)$, implying $lambda w$ is in the range of $T$. Thus The range of $T$ is a subspace of $W$.






              share|cite|improve this answer









              $endgroup$



              Here $y$ is in the range of $T$ if $y = T(x)$ for some $x in V$. Note that $v_1+v_2 in V$ since $V$ is a vector space, and $w_1+w_2 = T(v_1+v_2)$, then $w_1+w_2$ is in the range of $T$. Likewise, $lambda w = T(lambda v)$, implying $lambda w$ is in the range of $T$. Thus The range of $T$ is a subspace of $W$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 16 at 23:15









              DeepSeaDeepSea

              71.2k54487




              71.2k54487












              • $begingroup$
                $v_1 + v_2 in V$ helps me understand why $w_1 + w_2 = T(v_1 + v_2)$, then $w_1 + w_2$ is in the range of $T$.
                $endgroup$
                – JOHN
                Jan 17 at 7:31


















              • $begingroup$
                $v_1 + v_2 in V$ helps me understand why $w_1 + w_2 = T(v_1 + v_2)$, then $w_1 + w_2$ is in the range of $T$.
                $endgroup$
                – JOHN
                Jan 17 at 7:31
















              $begingroup$
              $v_1 + v_2 in V$ helps me understand why $w_1 + w_2 = T(v_1 + v_2)$, then $w_1 + w_2$ is in the range of $T$.
              $endgroup$
              – JOHN
              Jan 17 at 7:31




              $begingroup$
              $v_1 + v_2 in V$ helps me understand why $w_1 + w_2 = T(v_1 + v_2)$, then $w_1 + w_2$ is in the range of $T$.
              $endgroup$
              – JOHN
              Jan 17 at 7:31











              2












              $begingroup$

              Range of $T$ consists all vectors of the form $T(v)$ where $v in V$. $w_1+w_2=T(v_1+v_2)$. Hence $w_1+w_2$ belongs to range of $T$. If $w=T(v)$ then $lambda w=T(lambda v)$ so $lambda w$ is in the range of $T$.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Range of $T$ consists all vectors of the form $T(v)$ where $v in V$. $w_1+w_2=T(v_1+v_2)$. Hence $w_1+w_2$ belongs to range of $T$. If $w=T(v)$ then $lambda w=T(lambda v)$ so $lambda w$ is in the range of $T$.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Range of $T$ consists all vectors of the form $T(v)$ where $v in V$. $w_1+w_2=T(v_1+v_2)$. Hence $w_1+w_2$ belongs to range of $T$. If $w=T(v)$ then $lambda w=T(lambda v)$ so $lambda w$ is in the range of $T$.






                  share|cite|improve this answer









                  $endgroup$



                  Range of $T$ consists all vectors of the form $T(v)$ where $v in V$. $w_1+w_2=T(v_1+v_2)$. Hence $w_1+w_2$ belongs to range of $T$. If $w=T(v)$ then $lambda w=T(lambda v)$ so $lambda w$ is in the range of $T$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 16 at 23:13









                  Kavi Rama MurthyKavi Rama Murthy

                  59k42161




                  59k42161























                      -1












                      $begingroup$

                      Note the previous line: $T(text{something}) = w_1 + w_2$. Thus, there is something that, when we hit it with $t$, we get w_1 + w_2. The proof for $lambda w$ will be similar: since $T$ is linear, $T(lambda v) = lambda T(v) = lambda w$, so there is something that, when we hit it with $T$, we get $lambda w$, so $lambda w$ is in the range of $T$.






                      share|cite|improve this answer









                      $endgroup$


















                        -1












                        $begingroup$

                        Note the previous line: $T(text{something}) = w_1 + w_2$. Thus, there is something that, when we hit it with $t$, we get w_1 + w_2. The proof for $lambda w$ will be similar: since $T$ is linear, $T(lambda v) = lambda T(v) = lambda w$, so there is something that, when we hit it with $T$, we get $lambda w$, so $lambda w$ is in the range of $T$.






                        share|cite|improve this answer









                        $endgroup$
















                          -1












                          -1








                          -1





                          $begingroup$

                          Note the previous line: $T(text{something}) = w_1 + w_2$. Thus, there is something that, when we hit it with $t$, we get w_1 + w_2. The proof for $lambda w$ will be similar: since $T$ is linear, $T(lambda v) = lambda T(v) = lambda w$, so there is something that, when we hit it with $T$, we get $lambda w$, so $lambda w$ is in the range of $T$.






                          share|cite|improve this answer









                          $endgroup$



                          Note the previous line: $T(text{something}) = w_1 + w_2$. Thus, there is something that, when we hit it with $t$, we get w_1 + w_2. The proof for $lambda w$ will be similar: since $T$ is linear, $T(lambda v) = lambda T(v) = lambda w$, so there is something that, when we hit it with $T$, we get $lambda w$, so $lambda w$ is in the range of $T$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 16 at 23:14









                          user3482749user3482749

                          4,266919




                          4,266919






























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