Finding Range of a linear mapping












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In the following textbook example I understand how they get Null(L) but not how they get Range(L) which they say is clear to see. Can anyone elucidate the method of finding the Range of a linear mapping? Thank you!



enter image description here






linear-algebra linear-transformations







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  • $begingroup$
    Exactly where does the expression “clear to see” appear?
    $endgroup$
    – José Carlos Santos
    Jan 1 at 18:04
















0












$begingroup$


In the following textbook example I understand how they get Null(L) but not how they get Range(L) which they say is clear to see. Can anyone elucidate the method of finding the Range of a linear mapping? Thank you!



enter image description here






linear-algebra linear-transformations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Exactly where does the expression “clear to see” appear?
    $endgroup$
    – José Carlos Santos
    Jan 1 at 18:04














0












0








0





$begingroup$


In the following textbook example I understand how they get Null(L) but not how they get Range(L) which they say is clear to see. Can anyone elucidate the method of finding the Range of a linear mapping? Thank you!



enter image description here






linear-algebra linear-transformations







share|cite|improve this question











$endgroup$




In the following textbook example I understand how they get Null(L) but not how they get Range(L) which they say is clear to see. Can anyone elucidate the method of finding the Range of a linear mapping? Thank you!



enter image description here






linear-algebra linear-transformations




linear-algebra linear-transformations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Jan 1 at 18:10







P.ython

















asked Jan 1 at 18:02









P.ythonP.ython

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205












  • $begingroup$
    Exactly where does the expression “clear to see” appear?
    $endgroup$
    – José Carlos Santos
    Jan 1 at 18:04


















  • $begingroup$
    Exactly where does the expression “clear to see” appear?
    $endgroup$
    – José Carlos Santos
    Jan 1 at 18:04
















$begingroup$
Exactly where does the expression “clear to see” appear?
$endgroup$
– José Carlos Santos
Jan 1 at 18:04




$begingroup$
Exactly where does the expression “clear to see” appear?
$endgroup$
– José Carlos Santos
Jan 1 at 18:04










3 Answers
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The range of $L, R(L)subseteqBbb P_3$. Every element in $R(L)$ is of the form $k_1x+k_2x^3$, which is a linear combination of $x$ and $x^3$. So ${x,x^3}$ is a basis of $R(L)$, giving its dimension as $2$.






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    0












    $begingroup$

    You know $dim M(2,2)=4$ and $dim null(L)=2$, so by the Bank nullity theorem, $dim rank(L)=2$. But all the polynomials which are image of L are of the form $alpha x + beta x^3$, so you can get all of them with a linear combination of the polynomials $x$ and $x^3$. Since $dim rank(L)=2$ and $x,x^3$ are two linearly independent polynomials, it follows that $rank(L)$ is the subspace of $P_3$ generated by $x,x^3$.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      ${x,x^3}$ is a linearly independent set and spans the range, as can be seen by looking at $cx+(a+b)x^3$. That's why it says "clearly".



      In general the range of a linear map is the span of the columns of the matrix rel the standard basis, but we are able to take a short cut in this case.






      share|cite|improve this answer











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        3 Answers
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        3 Answers
        3






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        0












        $begingroup$

        The range of $L, R(L)subseteqBbb P_3$. Every element in $R(L)$ is of the form $k_1x+k_2x^3$, which is a linear combination of $x$ and $x^3$. So ${x,x^3}$ is a basis of $R(L)$, giving its dimension as $2$.






        share|cite|improve this answer









        $endgroup$


















          0












          $begingroup$

          The range of $L, R(L)subseteqBbb P_3$. Every element in $R(L)$ is of the form $k_1x+k_2x^3$, which is a linear combination of $x$ and $x^3$. So ${x,x^3}$ is a basis of $R(L)$, giving its dimension as $2$.






          share|cite|improve this answer









          $endgroup$
















            0












            0








            0





            $begingroup$

            The range of $L, R(L)subseteqBbb P_3$. Every element in $R(L)$ is of the form $k_1x+k_2x^3$, which is a linear combination of $x$ and $x^3$. So ${x,x^3}$ is a basis of $R(L)$, giving its dimension as $2$.






            share|cite|improve this answer









            $endgroup$



            The range of $L, R(L)subseteqBbb P_3$. Every element in $R(L)$ is of the form $k_1x+k_2x^3$, which is a linear combination of $x$ and $x^3$. So ${x,x^3}$ is a basis of $R(L)$, giving its dimension as $2$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 1 at 18:38









            Shubham JohriShubham Johri

            5,204718




            5,204718























                0












                $begingroup$

                You know $dim M(2,2)=4$ and $dim null(L)=2$, so by the Bank nullity theorem, $dim rank(L)=2$. But all the polynomials which are image of L are of the form $alpha x + beta x^3$, so you can get all of them with a linear combination of the polynomials $x$ and $x^3$. Since $dim rank(L)=2$ and $x,x^3$ are two linearly independent polynomials, it follows that $rank(L)$ is the subspace of $P_3$ generated by $x,x^3$.






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  You know $dim M(2,2)=4$ and $dim null(L)=2$, so by the Bank nullity theorem, $dim rank(L)=2$. But all the polynomials which are image of L are of the form $alpha x + beta x^3$, so you can get all of them with a linear combination of the polynomials $x$ and $x^3$. Since $dim rank(L)=2$ and $x,x^3$ are two linearly independent polynomials, it follows that $rank(L)$ is the subspace of $P_3$ generated by $x,x^3$.






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    You know $dim M(2,2)=4$ and $dim null(L)=2$, so by the Bank nullity theorem, $dim rank(L)=2$. But all the polynomials which are image of L are of the form $alpha x + beta x^3$, so you can get all of them with a linear combination of the polynomials $x$ and $x^3$. Since $dim rank(L)=2$ and $x,x^3$ are two linearly independent polynomials, it follows that $rank(L)$ is the subspace of $P_3$ generated by $x,x^3$.






                    share|cite|improve this answer









                    $endgroup$



                    You know $dim M(2,2)=4$ and $dim null(L)=2$, so by the Bank nullity theorem, $dim rank(L)=2$. But all the polynomials which are image of L are of the form $alpha x + beta x^3$, so you can get all of them with a linear combination of the polynomials $x$ and $x^3$. Since $dim rank(L)=2$ and $x,x^3$ are two linearly independent polynomials, it follows that $rank(L)$ is the subspace of $P_3$ generated by $x,x^3$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 1 at 18:48









                    R. N. MarleyR. N. Marley

                    788




                    788























                        0












                        $begingroup$

                        ${x,x^3}$ is a linearly independent set and spans the range, as can be seen by looking at $cx+(a+b)x^3$. That's why it says "clearly".



                        In general the range of a linear map is the span of the columns of the matrix rel the standard basis, but we are able to take a short cut in this case.






                        share|cite|improve this answer











                        $endgroup$


















                          0












                          $begingroup$

                          ${x,x^3}$ is a linearly independent set and spans the range, as can be seen by looking at $cx+(a+b)x^3$. That's why it says "clearly".



                          In general the range of a linear map is the span of the columns of the matrix rel the standard basis, but we are able to take a short cut in this case.






                          share|cite|improve this answer











                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            ${x,x^3}$ is a linearly independent set and spans the range, as can be seen by looking at $cx+(a+b)x^3$. That's why it says "clearly".



                            In general the range of a linear map is the span of the columns of the matrix rel the standard basis, but we are able to take a short cut in this case.






                            share|cite|improve this answer











                            $endgroup$



                            ${x,x^3}$ is a linearly independent set and spans the range, as can be seen by looking at $cx+(a+b)x^3$. That's why it says "clearly".



                            In general the range of a linear map is the span of the columns of the matrix rel the standard basis, but we are able to take a short cut in this case.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Jan 1 at 19:10

























                            answered Jan 1 at 19:01









                            Chris CusterChris Custer

                            14k3827




                            14k3827






























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