Why can a 5x4 matrix never be surjective?












0












$begingroup$


My thought is that since a matrix can only be surjective if null(T)= n-m, no matrix can have a nullity of -1 therefore a 5x4 matrix cannot be surjective. Is this a valid proof?






linear-algebra matrices







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$endgroup$












  • $begingroup$
    Recall that rank(T)= dim( Im(T)).
    $endgroup$
    – gimusi
    Dec 4 '18 at 15:07
















0












$begingroup$


My thought is that since a matrix can only be surjective if null(T)= n-m, no matrix can have a nullity of -1 therefore a 5x4 matrix cannot be surjective. Is this a valid proof?






linear-algebra matrices







share|cite|improve this question









$endgroup$












  • $begingroup$
    Recall that rank(T)= dim( Im(T)).
    $endgroup$
    – gimusi
    Dec 4 '18 at 15:07














0












0








0





$begingroup$


My thought is that since a matrix can only be surjective if null(T)= n-m, no matrix can have a nullity of -1 therefore a 5x4 matrix cannot be surjective. Is this a valid proof?






linear-algebra matrices







share|cite|improve this question









$endgroup$




My thought is that since a matrix can only be surjective if null(T)= n-m, no matrix can have a nullity of -1 therefore a 5x4 matrix cannot be surjective. Is this a valid proof?






linear-algebra matrices




linear-algebra matrices






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 4 '18 at 15:03









Elliot SilverElliot Silver

304




304












  • $begingroup$
    Recall that rank(T)= dim( Im(T)).
    $endgroup$
    – gimusi
    Dec 4 '18 at 15:07


















  • $begingroup$
    Recall that rank(T)= dim( Im(T)).
    $endgroup$
    – gimusi
    Dec 4 '18 at 15:07
















$begingroup$
Recall that rank(T)= dim( Im(T)).
$endgroup$
– gimusi
Dec 4 '18 at 15:07




$begingroup$
Recall that rank(T)= dim( Im(T)).
$endgroup$
– gimusi
Dec 4 '18 at 15:07










2 Answers
2






active

oldest

votes


















1












$begingroup$

A $5×4$ matrix represents a linear transformation $T$ from a $4$ dimensional vector space to a $5$ dimensional vector space. If $T$ is surjective, then the rank of $T=5$. But rank nullity theorem implies that $operatorname{rank}(T)+operatorname{nullity}(T)=4$, i.e.
$operatorname{nullity}(T)=-1,$a contradiction.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    HINT



    Recall that a basis for $mathbb{R^5}$ requires exactly $5$ linearly independent vectors.



    What is the maximum rank for $Ain M_{5times 4}$?






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I'm supposed to justify it using the rank nullity theorem
      $endgroup$
      – Elliot Silver
      Dec 4 '18 at 15:05










    • $begingroup$
      @ElliotSilver Yes that's a direct consequance, I've added an additional tip for that.
      $endgroup$
      – gimusi
      Dec 4 '18 at 15:05










    • $begingroup$
      The max rank would be 5?
      $endgroup$
      – Elliot Silver
      Dec 4 '18 at 15:08










    • $begingroup$
      @ElliotSilver Recall that for a $ntimes m$ matrix $rank(A) le min (m,n)$. That's an important concept you should clarify. Refer to Rank.
      $endgroup$
      – gimusi
      Dec 4 '18 at 15:09












    • $begingroup$
      Thank you for trying to help, but I don't think I understand what you mean, the rank is 5 which is less or equal to 5?
      $endgroup$
      – Elliot Silver
      Dec 4 '18 at 15:12











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

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    1












    $begingroup$

    A $5×4$ matrix represents a linear transformation $T$ from a $4$ dimensional vector space to a $5$ dimensional vector space. If $T$ is surjective, then the rank of $T=5$. But rank nullity theorem implies that $operatorname{rank}(T)+operatorname{nullity}(T)=4$, i.e.
    $operatorname{nullity}(T)=-1,$a contradiction.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      A $5×4$ matrix represents a linear transformation $T$ from a $4$ dimensional vector space to a $5$ dimensional vector space. If $T$ is surjective, then the rank of $T=5$. But rank nullity theorem implies that $operatorname{rank}(T)+operatorname{nullity}(T)=4$, i.e.
      $operatorname{nullity}(T)=-1,$a contradiction.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        A $5×4$ matrix represents a linear transformation $T$ from a $4$ dimensional vector space to a $5$ dimensional vector space. If $T$ is surjective, then the rank of $T=5$. But rank nullity theorem implies that $operatorname{rank}(T)+operatorname{nullity}(T)=4$, i.e.
        $operatorname{nullity}(T)=-1,$a contradiction.






        share|cite|improve this answer











        $endgroup$



        A $5×4$ matrix represents a linear transformation $T$ from a $4$ dimensional vector space to a $5$ dimensional vector space. If $T$ is surjective, then the rank of $T=5$. But rank nullity theorem implies that $operatorname{rank}(T)+operatorname{nullity}(T)=4$, i.e.
        $operatorname{nullity}(T)=-1,$a contradiction.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 4 '18 at 15:22

























        answered Dec 4 '18 at 15:13









        Thomas ShelbyThomas Shelby

        2,408221




        2,408221























            0












            $begingroup$

            HINT



            Recall that a basis for $mathbb{R^5}$ requires exactly $5$ linearly independent vectors.



            What is the maximum rank for $Ain M_{5times 4}$?






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I'm supposed to justify it using the rank nullity theorem
              $endgroup$
              – Elliot Silver
              Dec 4 '18 at 15:05










            • $begingroup$
              @ElliotSilver Yes that's a direct consequance, I've added an additional tip for that.
              $endgroup$
              – gimusi
              Dec 4 '18 at 15:05










            • $begingroup$
              The max rank would be 5?
              $endgroup$
              – Elliot Silver
              Dec 4 '18 at 15:08










            • $begingroup$
              @ElliotSilver Recall that for a $ntimes m$ matrix $rank(A) le min (m,n)$. That's an important concept you should clarify. Refer to Rank.
              $endgroup$
              – gimusi
              Dec 4 '18 at 15:09












            • $begingroup$
              Thank you for trying to help, but I don't think I understand what you mean, the rank is 5 which is less or equal to 5?
              $endgroup$
              – Elliot Silver
              Dec 4 '18 at 15:12
















            0












            $begingroup$

            HINT



            Recall that a basis for $mathbb{R^5}$ requires exactly $5$ linearly independent vectors.



            What is the maximum rank for $Ain M_{5times 4}$?






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I'm supposed to justify it using the rank nullity theorem
              $endgroup$
              – Elliot Silver
              Dec 4 '18 at 15:05










            • $begingroup$
              @ElliotSilver Yes that's a direct consequance, I've added an additional tip for that.
              $endgroup$
              – gimusi
              Dec 4 '18 at 15:05










            • $begingroup$
              The max rank would be 5?
              $endgroup$
              – Elliot Silver
              Dec 4 '18 at 15:08










            • $begingroup$
              @ElliotSilver Recall that for a $ntimes m$ matrix $rank(A) le min (m,n)$. That's an important concept you should clarify. Refer to Rank.
              $endgroup$
              – gimusi
              Dec 4 '18 at 15:09












            • $begingroup$
              Thank you for trying to help, but I don't think I understand what you mean, the rank is 5 which is less or equal to 5?
              $endgroup$
              – Elliot Silver
              Dec 4 '18 at 15:12














            0












            0








            0





            $begingroup$

            HINT



            Recall that a basis for $mathbb{R^5}$ requires exactly $5$ linearly independent vectors.



            What is the maximum rank for $Ain M_{5times 4}$?






            share|cite|improve this answer











            $endgroup$



            HINT



            Recall that a basis for $mathbb{R^5}$ requires exactly $5$ linearly independent vectors.



            What is the maximum rank for $Ain M_{5times 4}$?







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 4 '18 at 15:05

























            answered Dec 4 '18 at 15:04









            gimusigimusi

            92.9k94494




            92.9k94494












            • $begingroup$
              I'm supposed to justify it using the rank nullity theorem
              $endgroup$
              – Elliot Silver
              Dec 4 '18 at 15:05










            • $begingroup$
              @ElliotSilver Yes that's a direct consequance, I've added an additional tip for that.
              $endgroup$
              – gimusi
              Dec 4 '18 at 15:05










            • $begingroup$
              The max rank would be 5?
              $endgroup$
              – Elliot Silver
              Dec 4 '18 at 15:08










            • $begingroup$
              @ElliotSilver Recall that for a $ntimes m$ matrix $rank(A) le min (m,n)$. That's an important concept you should clarify. Refer to Rank.
              $endgroup$
              – gimusi
              Dec 4 '18 at 15:09












            • $begingroup$
              Thank you for trying to help, but I don't think I understand what you mean, the rank is 5 which is less or equal to 5?
              $endgroup$
              – Elliot Silver
              Dec 4 '18 at 15:12


















            • $begingroup$
              I'm supposed to justify it using the rank nullity theorem
              $endgroup$
              – Elliot Silver
              Dec 4 '18 at 15:05










            • $begingroup$
              @ElliotSilver Yes that's a direct consequance, I've added an additional tip for that.
              $endgroup$
              – gimusi
              Dec 4 '18 at 15:05










            • $begingroup$
              The max rank would be 5?
              $endgroup$
              – Elliot Silver
              Dec 4 '18 at 15:08










            • $begingroup$
              @ElliotSilver Recall that for a $ntimes m$ matrix $rank(A) le min (m,n)$. That's an important concept you should clarify. Refer to Rank.
              $endgroup$
              – gimusi
              Dec 4 '18 at 15:09












            • $begingroup$
              Thank you for trying to help, but I don't think I understand what you mean, the rank is 5 which is less or equal to 5?
              $endgroup$
              – Elliot Silver
              Dec 4 '18 at 15:12
















            $begingroup$
            I'm supposed to justify it using the rank nullity theorem
            $endgroup$
            – Elliot Silver
            Dec 4 '18 at 15:05




            $begingroup$
            I'm supposed to justify it using the rank nullity theorem
            $endgroup$
            – Elliot Silver
            Dec 4 '18 at 15:05












            $begingroup$
            @ElliotSilver Yes that's a direct consequance, I've added an additional tip for that.
            $endgroup$
            – gimusi
            Dec 4 '18 at 15:05




            $begingroup$
            @ElliotSilver Yes that's a direct consequance, I've added an additional tip for that.
            $endgroup$
            – gimusi
            Dec 4 '18 at 15:05












            $begingroup$
            The max rank would be 5?
            $endgroup$
            – Elliot Silver
            Dec 4 '18 at 15:08




            $begingroup$
            The max rank would be 5?
            $endgroup$
            – Elliot Silver
            Dec 4 '18 at 15:08












            $begingroup$
            @ElliotSilver Recall that for a $ntimes m$ matrix $rank(A) le min (m,n)$. That's an important concept you should clarify. Refer to Rank.
            $endgroup$
            – gimusi
            Dec 4 '18 at 15:09






            $begingroup$
            @ElliotSilver Recall that for a $ntimes m$ matrix $rank(A) le min (m,n)$. That's an important concept you should clarify. Refer to Rank.
            $endgroup$
            – gimusi
            Dec 4 '18 at 15:09














            $begingroup$
            Thank you for trying to help, but I don't think I understand what you mean, the rank is 5 which is less or equal to 5?
            $endgroup$
            – Elliot Silver
            Dec 4 '18 at 15:12




            $begingroup$
            Thank you for trying to help, but I don't think I understand what you mean, the rank is 5 which is less or equal to 5?
            $endgroup$
            – Elliot Silver
            Dec 4 '18 at 15:12


















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