Proving a matrix is surjective











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Is a $n times n$ matrix always surjective? If it is, how can I prove it using the rank-nullity theorem?






linear-algebra matrices







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    Small language point for the future. Matrices are not surjective The linear functions they determine may be.
    – Ethan Bolker
    3 hours ago















up vote
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down vote

favorite












Is a $n times n$ matrix always surjective? If it is, how can I prove it using the rank-nullity theorem?






linear-algebra matrices







share|cite|improve this question









New contributor




Elliot Silver is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 4




    Small language point for the future. Matrices are not surjective The linear functions they determine may be.
    – Ethan Bolker
    3 hours ago













up vote
1
down vote

favorite









up vote
1
down vote

favorite











Is a $n times n$ matrix always surjective? If it is, how can I prove it using the rank-nullity theorem?






linear-algebra matrices







share|cite|improve this question









New contributor




Elliot Silver is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











Is a $n times n$ matrix always surjective? If it is, how can I prove it using the rank-nullity theorem?






linear-algebra matrices




linear-algebra matrices






share|cite|improve this question









New contributor




Elliot Silver is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









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Elliot Silver is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









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edited 9 hours ago









gimusi

86.9k74393




86.9k74393






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Elliot Silver is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Elliot Silver is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 4




    Small language point for the future. Matrices are not surjective The linear functions they determine may be.
    – Ethan Bolker
    3 hours ago














  • 4




    Small language point for the future. Matrices are not surjective The linear functions they determine may be.
    – Ethan Bolker
    3 hours ago








4




4




Small language point for the future. Matrices are not surjective The linear functions they determine may be.
– Ethan Bolker
3 hours ago




Small language point for the future. Matrices are not surjective The linear functions they determine may be.
– Ethan Bolker
3 hours ago










3 Answers
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4
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No it's not. To see this consider the zero matrix as a map from $mathbb{R}^nrightarrow mathbb{R}^n$ which is clearly not surjective.






share|cite|improve this answer




























    up vote
    3
    down vote













    Recall that surjective means that for any $bin mathbb{R^n}$ the system



    $$Ax=b$$



    has solution, that is true if and only if $dim(Im(A))=n$ that if and only if $A$ is full rank.






    share|cite|improve this answer






























      up vote
      1
      down vote













      Maybe your getting confused because there is a fact that says that a $ntimes n$ matrix which has null kernel is always surjective. This is a classical applications of rank-nullity theorem because $$n=dim Ker (A) + dim Im (A)$$ and its clear that $dim Ker(A)=0 implies dim Im(A)=n$ which in this case means that $A$ is surjective.






      share|cite|improve this answer























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        3 Answers
        3






        active

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        3 Answers
        3






        active

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        active

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        up vote
        4
        down vote













        No it's not. To see this consider the zero matrix as a map from $mathbb{R}^nrightarrow mathbb{R}^n$ which is clearly not surjective.






        share|cite|improve this answer

























          up vote
          4
          down vote













          No it's not. To see this consider the zero matrix as a map from $mathbb{R}^nrightarrow mathbb{R}^n$ which is clearly not surjective.






          share|cite|improve this answer























            up vote
            4
            down vote










            up vote
            4
            down vote









            No it's not. To see this consider the zero matrix as a map from $mathbb{R}^nrightarrow mathbb{R}^n$ which is clearly not surjective.






            share|cite|improve this answer












            No it's not. To see this consider the zero matrix as a map from $mathbb{R}^nrightarrow mathbb{R}^n$ which is clearly not surjective.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 9 hours ago









            CyclotomicField

            2,0691312




            2,0691312






















                up vote
                3
                down vote













                Recall that surjective means that for any $bin mathbb{R^n}$ the system



                $$Ax=b$$



                has solution, that is true if and only if $dim(Im(A))=n$ that if and only if $A$ is full rank.






                share|cite|improve this answer



























                  up vote
                  3
                  down vote













                  Recall that surjective means that for any $bin mathbb{R^n}$ the system



                  $$Ax=b$$



                  has solution, that is true if and only if $dim(Im(A))=n$ that if and only if $A$ is full rank.






                  share|cite|improve this answer

























                    up vote
                    3
                    down vote










                    up vote
                    3
                    down vote









                    Recall that surjective means that for any $bin mathbb{R^n}$ the system



                    $$Ax=b$$



                    has solution, that is true if and only if $dim(Im(A))=n$ that if and only if $A$ is full rank.






                    share|cite|improve this answer














                    Recall that surjective means that for any $bin mathbb{R^n}$ the system



                    $$Ax=b$$



                    has solution, that is true if and only if $dim(Im(A))=n$ that if and only if $A$ is full rank.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 9 hours ago

























                    answered 9 hours ago









                    gimusi

                    86.9k74393




                    86.9k74393






















                        up vote
                        1
                        down vote













                        Maybe your getting confused because there is a fact that says that a $ntimes n$ matrix which has null kernel is always surjective. This is a classical applications of rank-nullity theorem because $$n=dim Ker (A) + dim Im (A)$$ and its clear that $dim Ker(A)=0 implies dim Im(A)=n$ which in this case means that $A$ is surjective.






                        share|cite|improve this answer



























                          up vote
                          1
                          down vote













                          Maybe your getting confused because there is a fact that says that a $ntimes n$ matrix which has null kernel is always surjective. This is a classical applications of rank-nullity theorem because $$n=dim Ker (A) + dim Im (A)$$ and its clear that $dim Ker(A)=0 implies dim Im(A)=n$ which in this case means that $A$ is surjective.






                          share|cite|improve this answer

























                            up vote
                            1
                            down vote










                            up vote
                            1
                            down vote









                            Maybe your getting confused because there is a fact that says that a $ntimes n$ matrix which has null kernel is always surjective. This is a classical applications of rank-nullity theorem because $$n=dim Ker (A) + dim Im (A)$$ and its clear that $dim Ker(A)=0 implies dim Im(A)=n$ which in this case means that $A$ is surjective.






                            share|cite|improve this answer














                            Maybe your getting confused because there is a fact that says that a $ntimes n$ matrix which has null kernel is always surjective. This is a classical applications of rank-nullity theorem because $$n=dim Ker (A) + dim Im (A)$$ and its clear that $dim Ker(A)=0 implies dim Im(A)=n$ which in this case means that $A$ is surjective.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited 3 hours ago

























                            answered 3 hours ago









                            Robson

                            688221




                            688221






















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