Linear transformation surjective and/ or injective?












0












$begingroup$


I am new to linear algebra, and would like some feedback regarding the following answer to a textbook question:




Let $T:M_{2times 3}(mathbb R) to M_{3times 3}(mathbb R)$ be a linear
transformation. Answer the following two questions:




  1. Is it possible for T to be surjective?

  2. Is it possible for T to be injective?




My answer is the following:




  1. Yes, because it is possible for each element in $M_{3times 3}(mathbb R)$ to be the result of linear transformation $T$ applied to $M_{2times 3}(mathbb R)$.


  2. No, because there are more rows in $M_{3times 3}(mathbf R)$ than in $M_{2times 3}(mathbf R)$, and therefore the elements in $M_{3times 3}(mathbb R)$ cannot be the exclusive result of the elements in $M_{2times3}(mathbb R)$.



I have no idea whether I am on the right track?



Thank you!






linear-algebra linear-transformations







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$endgroup$












  • $begingroup$
    @DietrichBurde criticism accepted, but can you give me a hint as to how to go about 1 and how to redefine 2?
    $endgroup$
    – dalta
    Jan 1 at 16:49






  • 1




    $begingroup$
    Vector spaces of matrices can be confusing because the vectors are the matrices. Notice that a two-by-three matrix is equivalent to a $6$-dimensional vector because it is described by $6$ real numbers. Change the problem to $T:mathbb R^6tomathbb R^9$ and then try to answer the questions.
    $endgroup$
    – John Douma
    Jan 1 at 16:59
















0












$begingroup$


I am new to linear algebra, and would like some feedback regarding the following answer to a textbook question:




Let $T:M_{2times 3}(mathbb R) to M_{3times 3}(mathbb R)$ be a linear
transformation. Answer the following two questions:




  1. Is it possible for T to be surjective?

  2. Is it possible for T to be injective?




My answer is the following:




  1. Yes, because it is possible for each element in $M_{3times 3}(mathbb R)$ to be the result of linear transformation $T$ applied to $M_{2times 3}(mathbb R)$.


  2. No, because there are more rows in $M_{3times 3}(mathbf R)$ than in $M_{2times 3}(mathbf R)$, and therefore the elements in $M_{3times 3}(mathbb R)$ cannot be the exclusive result of the elements in $M_{2times3}(mathbb R)$.



I have no idea whether I am on the right track?



Thank you!






linear-algebra linear-transformations







share|cite|improve this question











$endgroup$












  • $begingroup$
    @DietrichBurde criticism accepted, but can you give me a hint as to how to go about 1 and how to redefine 2?
    $endgroup$
    – dalta
    Jan 1 at 16:49






  • 1




    $begingroup$
    Vector spaces of matrices can be confusing because the vectors are the matrices. Notice that a two-by-three matrix is equivalent to a $6$-dimensional vector because it is described by $6$ real numbers. Change the problem to $T:mathbb R^6tomathbb R^9$ and then try to answer the questions.
    $endgroup$
    – John Douma
    Jan 1 at 16:59














0












0








0





$begingroup$


I am new to linear algebra, and would like some feedback regarding the following answer to a textbook question:




Let $T:M_{2times 3}(mathbb R) to M_{3times 3}(mathbb R)$ be a linear
transformation. Answer the following two questions:




  1. Is it possible for T to be surjective?

  2. Is it possible for T to be injective?




My answer is the following:




  1. Yes, because it is possible for each element in $M_{3times 3}(mathbb R)$ to be the result of linear transformation $T$ applied to $M_{2times 3}(mathbb R)$.


  2. No, because there are more rows in $M_{3times 3}(mathbf R)$ than in $M_{2times 3}(mathbf R)$, and therefore the elements in $M_{3times 3}(mathbb R)$ cannot be the exclusive result of the elements in $M_{2times3}(mathbb R)$.



I have no idea whether I am on the right track?



Thank you!






linear-algebra linear-transformations







share|cite|improve this question











$endgroup$




I am new to linear algebra, and would like some feedback regarding the following answer to a textbook question:




Let $T:M_{2times 3}(mathbb R) to M_{3times 3}(mathbb R)$ be a linear
transformation. Answer the following two questions:




  1. Is it possible for T to be surjective?

  2. Is it possible for T to be injective?




My answer is the following:




  1. Yes, because it is possible for each element in $M_{3times 3}(mathbb R)$ to be the result of linear transformation $T$ applied to $M_{2times 3}(mathbb R)$.


  2. No, because there are more rows in $M_{3times 3}(mathbf R)$ than in $M_{2times 3}(mathbf R)$, and therefore the elements in $M_{3times 3}(mathbb R)$ cannot be the exclusive result of the elements in $M_{2times3}(mathbb R)$.



I have no idea whether I am on the right track?



Thank you!






linear-algebra linear-transformations




linear-algebra linear-transformations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 1 at 17:04









amWhy

1




1










asked Jan 1 at 16:43









daltadalta

1168




1168












  • $begingroup$
    @DietrichBurde criticism accepted, but can you give me a hint as to how to go about 1 and how to redefine 2?
    $endgroup$
    – dalta
    Jan 1 at 16:49






  • 1




    $begingroup$
    Vector spaces of matrices can be confusing because the vectors are the matrices. Notice that a two-by-three matrix is equivalent to a $6$-dimensional vector because it is described by $6$ real numbers. Change the problem to $T:mathbb R^6tomathbb R^9$ and then try to answer the questions.
    $endgroup$
    – John Douma
    Jan 1 at 16:59


















  • $begingroup$
    @DietrichBurde criticism accepted, but can you give me a hint as to how to go about 1 and how to redefine 2?
    $endgroup$
    – dalta
    Jan 1 at 16:49






  • 1




    $begingroup$
    Vector spaces of matrices can be confusing because the vectors are the matrices. Notice that a two-by-three matrix is equivalent to a $6$-dimensional vector because it is described by $6$ real numbers. Change the problem to $T:mathbb R^6tomathbb R^9$ and then try to answer the questions.
    $endgroup$
    – John Douma
    Jan 1 at 16:59
















$begingroup$
@DietrichBurde criticism accepted, but can you give me a hint as to how to go about 1 and how to redefine 2?
$endgroup$
– dalta
Jan 1 at 16:49




$begingroup$
@DietrichBurde criticism accepted, but can you give me a hint as to how to go about 1 and how to redefine 2?
$endgroup$
– dalta
Jan 1 at 16:49




1




1




$begingroup$
Vector spaces of matrices can be confusing because the vectors are the matrices. Notice that a two-by-three matrix is equivalent to a $6$-dimensional vector because it is described by $6$ real numbers. Change the problem to $T:mathbb R^6tomathbb R^9$ and then try to answer the questions.
$endgroup$
– John Douma
Jan 1 at 16:59




$begingroup$
Vector spaces of matrices can be confusing because the vectors are the matrices. Notice that a two-by-three matrix is equivalent to a $6$-dimensional vector because it is described by $6$ real numbers. Change the problem to $T:mathbb R^6tomathbb R^9$ and then try to answer the questions.
$endgroup$
– John Douma
Jan 1 at 16:59










2 Answers
2






active

oldest

votes


















0












$begingroup$

Hint:
For 1. use the rank-nullity theorem for $T$.



For 2. consider the embedding of $M_{2,3}(K)$ into $M_{3,3}(K)$.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Let $A:Uto V$ be a linear operator between finite-dimensional vector spaces $U$ and $V$.



    (1) $text{rank}(A)leqtext{dim}(V)$.



    (2) $text{rank}(A)+text{nullity}(A)=text{dim}(U)$.



    (3) $A$ is injective iff $text{nullity}(A)=0$.



    (4) $A$ is surjective iff $text{rank}(A)=text{dim}(V)$



    (5) The dimension of $M_{mtimes n}(mathbb{R})$ is $mn$.



    That is more than enough to answer both questions.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Right, which is why I used a different symbol to denote an arbitrary linear operator between $U$ and $V$
      $endgroup$
      – Ben W
      Jan 1 at 16:58










    • $begingroup$
      Ordinarily I would agree, but $T$ was already taken. It would be like fixing $n=4$ and then trying to use the notation $sum_{n=1}^infty frac{1}{n^2}$ or something.
      $endgroup$
      – Ben W
      Jan 1 at 17:00










    • $begingroup$
      Thank you for your answer! Just want to check that I got this right, because the answer it leads to is the opposite of what I had originally thought: according to the above, if A is surjective, then rank(A)=9, but rank(A)+nullity(A)=dim(U), giving 9+nullity(A)=6, which is impossible: so A cannot be surjective. However, if A is injective, Rank A = 6 (because dimKerT = 0), which is indeed less than 9, so there is no contradiction, and A can be injective… Did I understand this correctly?
      $endgroup$
      – dalta
      Jan 1 at 18:07










    • $begingroup$
      You have the first part right. But for the second part, merely showing that it doesn't contradict the rank-nullity theorem isn't enough. You have to actually exhibit some kind of example, even if it's not explicit. For example, let ${u_1,cdots,u_6}$ be a basis for $M_{2times 3}$ and ${v_1,cdots,v_9}$ a basis for $M_{3times 3}$. Then define $T$ by the rule $T(u_i)=v_i$ for $i=1,cdots 6$.
      $endgroup$
      – Ben W
      Jan 1 at 18:10










    • $begingroup$
      and the remaining $v_7-v_9$ are simply unrelated to $M_{2x3}$?
      $endgroup$
      – dalta
      Jan 1 at 18:16











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    2 Answers
    2






    active

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

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    active

    oldest

    votes









    0












    $begingroup$

    Hint:
    For 1. use the rank-nullity theorem for $T$.



    For 2. consider the embedding of $M_{2,3}(K)$ into $M_{3,3}(K)$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Hint:
      For 1. use the rank-nullity theorem for $T$.



      For 2. consider the embedding of $M_{2,3}(K)$ into $M_{3,3}(K)$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Hint:
        For 1. use the rank-nullity theorem for $T$.



        For 2. consider the embedding of $M_{2,3}(K)$ into $M_{3,3}(K)$.






        share|cite|improve this answer









        $endgroup$



        Hint:
        For 1. use the rank-nullity theorem for $T$.



        For 2. consider the embedding of $M_{2,3}(K)$ into $M_{3,3}(K)$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 1 at 16:54









        Dietrich BurdeDietrich Burde

        80.1k647104




        80.1k647104























            1












            $begingroup$

            Let $A:Uto V$ be a linear operator between finite-dimensional vector spaces $U$ and $V$.



            (1) $text{rank}(A)leqtext{dim}(V)$.



            (2) $text{rank}(A)+text{nullity}(A)=text{dim}(U)$.



            (3) $A$ is injective iff $text{nullity}(A)=0$.



            (4) $A$ is surjective iff $text{rank}(A)=text{dim}(V)$



            (5) The dimension of $M_{mtimes n}(mathbb{R})$ is $mn$.



            That is more than enough to answer both questions.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Right, which is why I used a different symbol to denote an arbitrary linear operator between $U$ and $V$
              $endgroup$
              – Ben W
              Jan 1 at 16:58










            • $begingroup$
              Ordinarily I would agree, but $T$ was already taken. It would be like fixing $n=4$ and then trying to use the notation $sum_{n=1}^infty frac{1}{n^2}$ or something.
              $endgroup$
              – Ben W
              Jan 1 at 17:00










            • $begingroup$
              Thank you for your answer! Just want to check that I got this right, because the answer it leads to is the opposite of what I had originally thought: according to the above, if A is surjective, then rank(A)=9, but rank(A)+nullity(A)=dim(U), giving 9+nullity(A)=6, which is impossible: so A cannot be surjective. However, if A is injective, Rank A = 6 (because dimKerT = 0), which is indeed less than 9, so there is no contradiction, and A can be injective… Did I understand this correctly?
              $endgroup$
              – dalta
              Jan 1 at 18:07










            • $begingroup$
              You have the first part right. But for the second part, merely showing that it doesn't contradict the rank-nullity theorem isn't enough. You have to actually exhibit some kind of example, even if it's not explicit. For example, let ${u_1,cdots,u_6}$ be a basis for $M_{2times 3}$ and ${v_1,cdots,v_9}$ a basis for $M_{3times 3}$. Then define $T$ by the rule $T(u_i)=v_i$ for $i=1,cdots 6$.
              $endgroup$
              – Ben W
              Jan 1 at 18:10










            • $begingroup$
              and the remaining $v_7-v_9$ are simply unrelated to $M_{2x3}$?
              $endgroup$
              – dalta
              Jan 1 at 18:16
















            1












            $begingroup$

            Let $A:Uto V$ be a linear operator between finite-dimensional vector spaces $U$ and $V$.



            (1) $text{rank}(A)leqtext{dim}(V)$.



            (2) $text{rank}(A)+text{nullity}(A)=text{dim}(U)$.



            (3) $A$ is injective iff $text{nullity}(A)=0$.



            (4) $A$ is surjective iff $text{rank}(A)=text{dim}(V)$



            (5) The dimension of $M_{mtimes n}(mathbb{R})$ is $mn$.



            That is more than enough to answer both questions.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Right, which is why I used a different symbol to denote an arbitrary linear operator between $U$ and $V$
              $endgroup$
              – Ben W
              Jan 1 at 16:58










            • $begingroup$
              Ordinarily I would agree, but $T$ was already taken. It would be like fixing $n=4$ and then trying to use the notation $sum_{n=1}^infty frac{1}{n^2}$ or something.
              $endgroup$
              – Ben W
              Jan 1 at 17:00










            • $begingroup$
              Thank you for your answer! Just want to check that I got this right, because the answer it leads to is the opposite of what I had originally thought: according to the above, if A is surjective, then rank(A)=9, but rank(A)+nullity(A)=dim(U), giving 9+nullity(A)=6, which is impossible: so A cannot be surjective. However, if A is injective, Rank A = 6 (because dimKerT = 0), which is indeed less than 9, so there is no contradiction, and A can be injective… Did I understand this correctly?
              $endgroup$
              – dalta
              Jan 1 at 18:07










            • $begingroup$
              You have the first part right. But for the second part, merely showing that it doesn't contradict the rank-nullity theorem isn't enough. You have to actually exhibit some kind of example, even if it's not explicit. For example, let ${u_1,cdots,u_6}$ be a basis for $M_{2times 3}$ and ${v_1,cdots,v_9}$ a basis for $M_{3times 3}$. Then define $T$ by the rule $T(u_i)=v_i$ for $i=1,cdots 6$.
              $endgroup$
              – Ben W
              Jan 1 at 18:10










            • $begingroup$
              and the remaining $v_7-v_9$ are simply unrelated to $M_{2x3}$?
              $endgroup$
              – dalta
              Jan 1 at 18:16














            1












            1








            1





            $begingroup$

            Let $A:Uto V$ be a linear operator between finite-dimensional vector spaces $U$ and $V$.



            (1) $text{rank}(A)leqtext{dim}(V)$.



            (2) $text{rank}(A)+text{nullity}(A)=text{dim}(U)$.



            (3) $A$ is injective iff $text{nullity}(A)=0$.



            (4) $A$ is surjective iff $text{rank}(A)=text{dim}(V)$



            (5) The dimension of $M_{mtimes n}(mathbb{R})$ is $mn$.



            That is more than enough to answer both questions.






            share|cite|improve this answer









            $endgroup$



            Let $A:Uto V$ be a linear operator between finite-dimensional vector spaces $U$ and $V$.



            (1) $text{rank}(A)leqtext{dim}(V)$.



            (2) $text{rank}(A)+text{nullity}(A)=text{dim}(U)$.



            (3) $A$ is injective iff $text{nullity}(A)=0$.



            (4) $A$ is surjective iff $text{rank}(A)=text{dim}(V)$



            (5) The dimension of $M_{mtimes n}(mathbb{R})$ is $mn$.



            That is more than enough to answer both questions.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 1 at 16:53









            Ben WBen W

            2,322615




            2,322615












            • $begingroup$
              Right, which is why I used a different symbol to denote an arbitrary linear operator between $U$ and $V$
              $endgroup$
              – Ben W
              Jan 1 at 16:58










            • $begingroup$
              Ordinarily I would agree, but $T$ was already taken. It would be like fixing $n=4$ and then trying to use the notation $sum_{n=1}^infty frac{1}{n^2}$ or something.
              $endgroup$
              – Ben W
              Jan 1 at 17:00










            • $begingroup$
              Thank you for your answer! Just want to check that I got this right, because the answer it leads to is the opposite of what I had originally thought: according to the above, if A is surjective, then rank(A)=9, but rank(A)+nullity(A)=dim(U), giving 9+nullity(A)=6, which is impossible: so A cannot be surjective. However, if A is injective, Rank A = 6 (because dimKerT = 0), which is indeed less than 9, so there is no contradiction, and A can be injective… Did I understand this correctly?
              $endgroup$
              – dalta
              Jan 1 at 18:07










            • $begingroup$
              You have the first part right. But for the second part, merely showing that it doesn't contradict the rank-nullity theorem isn't enough. You have to actually exhibit some kind of example, even if it's not explicit. For example, let ${u_1,cdots,u_6}$ be a basis for $M_{2times 3}$ and ${v_1,cdots,v_9}$ a basis for $M_{3times 3}$. Then define $T$ by the rule $T(u_i)=v_i$ for $i=1,cdots 6$.
              $endgroup$
              – Ben W
              Jan 1 at 18:10










            • $begingroup$
              and the remaining $v_7-v_9$ are simply unrelated to $M_{2x3}$?
              $endgroup$
              – dalta
              Jan 1 at 18:16


















            • $begingroup$
              Right, which is why I used a different symbol to denote an arbitrary linear operator between $U$ and $V$
              $endgroup$
              – Ben W
              Jan 1 at 16:58










            • $begingroup$
              Ordinarily I would agree, but $T$ was already taken. It would be like fixing $n=4$ and then trying to use the notation $sum_{n=1}^infty frac{1}{n^2}$ or something.
              $endgroup$
              – Ben W
              Jan 1 at 17:00










            • $begingroup$
              Thank you for your answer! Just want to check that I got this right, because the answer it leads to is the opposite of what I had originally thought: according to the above, if A is surjective, then rank(A)=9, but rank(A)+nullity(A)=dim(U), giving 9+nullity(A)=6, which is impossible: so A cannot be surjective. However, if A is injective, Rank A = 6 (because dimKerT = 0), which is indeed less than 9, so there is no contradiction, and A can be injective… Did I understand this correctly?
              $endgroup$
              – dalta
              Jan 1 at 18:07










            • $begingroup$
              You have the first part right. But for the second part, merely showing that it doesn't contradict the rank-nullity theorem isn't enough. You have to actually exhibit some kind of example, even if it's not explicit. For example, let ${u_1,cdots,u_6}$ be a basis for $M_{2times 3}$ and ${v_1,cdots,v_9}$ a basis for $M_{3times 3}$. Then define $T$ by the rule $T(u_i)=v_i$ for $i=1,cdots 6$.
              $endgroup$
              – Ben W
              Jan 1 at 18:10










            • $begingroup$
              and the remaining $v_7-v_9$ are simply unrelated to $M_{2x3}$?
              $endgroup$
              – dalta
              Jan 1 at 18:16
















            $begingroup$
            Right, which is why I used a different symbol to denote an arbitrary linear operator between $U$ and $V$
            $endgroup$
            – Ben W
            Jan 1 at 16:58




            $begingroup$
            Right, which is why I used a different symbol to denote an arbitrary linear operator between $U$ and $V$
            $endgroup$
            – Ben W
            Jan 1 at 16:58












            $begingroup$
            Ordinarily I would agree, but $T$ was already taken. It would be like fixing $n=4$ and then trying to use the notation $sum_{n=1}^infty frac{1}{n^2}$ or something.
            $endgroup$
            – Ben W
            Jan 1 at 17:00




            $begingroup$
            Ordinarily I would agree, but $T$ was already taken. It would be like fixing $n=4$ and then trying to use the notation $sum_{n=1}^infty frac{1}{n^2}$ or something.
            $endgroup$
            – Ben W
            Jan 1 at 17:00












            $begingroup$
            Thank you for your answer! Just want to check that I got this right, because the answer it leads to is the opposite of what I had originally thought: according to the above, if A is surjective, then rank(A)=9, but rank(A)+nullity(A)=dim(U), giving 9+nullity(A)=6, which is impossible: so A cannot be surjective. However, if A is injective, Rank A = 6 (because dimKerT = 0), which is indeed less than 9, so there is no contradiction, and A can be injective… Did I understand this correctly?
            $endgroup$
            – dalta
            Jan 1 at 18:07




            $begingroup$
            Thank you for your answer! Just want to check that I got this right, because the answer it leads to is the opposite of what I had originally thought: according to the above, if A is surjective, then rank(A)=9, but rank(A)+nullity(A)=dim(U), giving 9+nullity(A)=6, which is impossible: so A cannot be surjective. However, if A is injective, Rank A = 6 (because dimKerT = 0), which is indeed less than 9, so there is no contradiction, and A can be injective… Did I understand this correctly?
            $endgroup$
            – dalta
            Jan 1 at 18:07












            $begingroup$
            You have the first part right. But for the second part, merely showing that it doesn't contradict the rank-nullity theorem isn't enough. You have to actually exhibit some kind of example, even if it's not explicit. For example, let ${u_1,cdots,u_6}$ be a basis for $M_{2times 3}$ and ${v_1,cdots,v_9}$ a basis for $M_{3times 3}$. Then define $T$ by the rule $T(u_i)=v_i$ for $i=1,cdots 6$.
            $endgroup$
            – Ben W
            Jan 1 at 18:10




            $begingroup$
            You have the first part right. But for the second part, merely showing that it doesn't contradict the rank-nullity theorem isn't enough. You have to actually exhibit some kind of example, even if it's not explicit. For example, let ${u_1,cdots,u_6}$ be a basis for $M_{2times 3}$ and ${v_1,cdots,v_9}$ a basis for $M_{3times 3}$. Then define $T$ by the rule $T(u_i)=v_i$ for $i=1,cdots 6$.
            $endgroup$
            – Ben W
            Jan 1 at 18:10












            $begingroup$
            and the remaining $v_7-v_9$ are simply unrelated to $M_{2x3}$?
            $endgroup$
            – dalta
            Jan 1 at 18:16




            $begingroup$
            and the remaining $v_7-v_9$ are simply unrelated to $M_{2x3}$?
            $endgroup$
            – dalta
            Jan 1 at 18:16


















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