If we have an $n times m$ matrix $A$, is Col(A) a subspace of $mathbbR^m$? [closed]

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I am reading through Axler's Linear Algebra book.



I believe that the answer to my question is no.



This link shows that if we have an $n times m$ matrix $A$, then Col(A) is a subspace of $mathbbR^n$, but I cannot come up with a similar proof to show that the same holds for $mathbbR^m$. For example, consder $m > n$.



Thanks.




linear-algebra




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closed as off-topic by John Ma, Xander Henderson, Claude Leibovici, Taroccoesbrocco, Mostafa Ayaz Jul 29 at 19:54


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Xander Henderson, Claude Leibovici, Taroccoesbrocco, Mostafa Ayaz
If this question can be reworded to fit the rules in the help center, please edit the question.
















    up vote
    0
    down vote

    favorite
    1












    I am reading through Axler's Linear Algebra book.



    I believe that the answer to my question is no.



    This link shows that if we have an $n times m$ matrix $A$, then Col(A) is a subspace of $mathbbR^n$, but I cannot come up with a similar proof to show that the same holds for $mathbbR^m$. For example, consder $m > n$.



    Thanks.




    linear-algebra




    share|cite|improve this question











    closed as off-topic by John Ma, Xander Henderson, Claude Leibovici, Taroccoesbrocco, Mostafa Ayaz Jul 29 at 19:54


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Xander Henderson, Claude Leibovici, Taroccoesbrocco, Mostafa Ayaz
    If this question can be reworded to fit the rules in the help center, please edit the question.














      up vote
      0
      down vote

      favorite
      1









      up vote
      0
      down vote

      favorite
      1






      1





      I am reading through Axler's Linear Algebra book.



      I believe that the answer to my question is no.



      This link shows that if we have an $n times m$ matrix $A$, then Col(A) is a subspace of $mathbbR^n$, but I cannot come up with a similar proof to show that the same holds for $mathbbR^m$. For example, consder $m > n$.



      Thanks.




      linear-algebra




      share|cite|improve this question











      I am reading through Axler's Linear Algebra book.



      I believe that the answer to my question is no.



      This link shows that if we have an $n times m$ matrix $A$, then Col(A) is a subspace of $mathbbR^n$, but I cannot come up with a similar proof to show that the same holds for $mathbbR^m$. For example, consder $m > n$.



      Thanks.





      linear-algebra





      share|cite|improve this question










      share|cite|improve this question




      share|cite|improve this question









      asked Jul 23 at 21:32









      Hat

      789115




      789115




      closed as off-topic by John Ma, Xander Henderson, Claude Leibovici, Taroccoesbrocco, Mostafa Ayaz Jul 29 at 19:54


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Xander Henderson, Claude Leibovici, Taroccoesbrocco, Mostafa Ayaz
      If this question can be reworded to fit the rules in the help center, please edit the question.




      closed as off-topic by John Ma, Xander Henderson, Claude Leibovici, Taroccoesbrocco, Mostafa Ayaz Jul 29 at 19:54


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Xander Henderson, Claude Leibovici, Taroccoesbrocco, Mostafa Ayaz
      If this question can be reworded to fit the rules in the help center, please edit the question.




















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          Note that for a n-by m matrix any vectors in Col(A) is in $mathbbR^n$ therefore Col(A) is a subspace of $mathbbR^n$ and Row(A) is a subspace of $mathbbR^m$.






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          • That's what I thought, thanks.
            – Hat
            Jul 23 at 21:35










          • @Hat That was a nice thought! You are welcome, Bye.
            – gimusi
            Jul 23 at 21:36

















          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          1
          down vote



          accepted










          Note that for a n-by m matrix any vectors in Col(A) is in $mathbbR^n$ therefore Col(A) is a subspace of $mathbbR^n$ and Row(A) is a subspace of $mathbbR^m$.






          share|cite|improve this answer





















          • That's what I thought, thanks.
            – Hat
            Jul 23 at 21:35










          • @Hat That was a nice thought! You are welcome, Bye.
            – gimusi
            Jul 23 at 21:36














          up vote
          1
          down vote



          accepted










          Note that for a n-by m matrix any vectors in Col(A) is in $mathbbR^n$ therefore Col(A) is a subspace of $mathbbR^n$ and Row(A) is a subspace of $mathbbR^m$.






          share|cite|improve this answer





















          • That's what I thought, thanks.
            – Hat
            Jul 23 at 21:35










          • @Hat That was a nice thought! You are welcome, Bye.
            – gimusi
            Jul 23 at 21:36












          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          Note that for a n-by m matrix any vectors in Col(A) is in $mathbbR^n$ therefore Col(A) is a subspace of $mathbbR^n$ and Row(A) is a subspace of $mathbbR^m$.






          share|cite|improve this answer













          Note that for a n-by m matrix any vectors in Col(A) is in $mathbbR^n$ therefore Col(A) is a subspace of $mathbbR^n$ and Row(A) is a subspace of $mathbbR^m$.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Jul 23 at 21:35









          gimusi

          65.2k73583




          65.2k73583











          • That's what I thought, thanks.
            – Hat
            Jul 23 at 21:35










          • @Hat That was a nice thought! You are welcome, Bye.
            – gimusi
            Jul 23 at 21:36
















          • That's what I thought, thanks.
            – Hat
            Jul 23 at 21:35










          • @Hat That was a nice thought! You are welcome, Bye.
            – gimusi
            Jul 23 at 21:36















          That's what I thought, thanks.
          – Hat
          Jul 23 at 21:35




          That's what I thought, thanks.
          – Hat
          Jul 23 at 21:35












          @Hat That was a nice thought! You are welcome, Bye.
          – gimusi
          Jul 23 at 21:36




          @Hat That was a nice thought! You are welcome, Bye.
          – gimusi
          Jul 23 at 21:36


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