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If A is an m by n matrix, prove that the set of vectors b that are not in C(A) forms a subspace.



The Next CEO of Stack OverflowProve that S forms a subspace of R^3Prove that is a subspaceWhy do these vectors not belong to the same vector space?Am I correctly determining whether the vectors are in the subspace?Prove the following set of vectors is a subspaceCan a subspace S containing vectors with a finite number of nonzero components contain the zero vector?Vector subspace of two linear transformationsProve that if the following two vectors are not parallel that the following holds.Are the polynomials of form $a_0 + a_1x + a_2x^2 +a_3x^3$, with $a_i$ rational, a subspace of $P_3$?Why are all vectors with exactly one nonzero component not a subspace of $mathbbR^3$?










2












$begingroup$


If A is an m by n matrix, prove that the set of vectors b that are not in C(A) forms a subspace.



I would like to first understand if I am interpreting the question correctly. My understanding of this question is that I need to first prove that the set of vectors b are equal to the 0 vector, and if b1 and b2 are members of the subspace, then the sum of set b should be a member, and so should some scalar C multiplied by the set of vectors b. I just don't understand how to actually prove this






linear-algebra matrices vector-spaces







share|cite|improve this question











$endgroup$
















    2












    $begingroup$


    If A is an m by n matrix, prove that the set of vectors b that are not in C(A) forms a subspace.



    I would like to first understand if I am interpreting the question correctly. My understanding of this question is that I need to first prove that the set of vectors b are equal to the 0 vector, and if b1 and b2 are members of the subspace, then the sum of set b should be a member, and so should some scalar C multiplied by the set of vectors b. I just don't understand how to actually prove this






    linear-algebra matrices vector-spaces







    share|cite|improve this question











    $endgroup$














      2












      2








      2





      $begingroup$


      If A is an m by n matrix, prove that the set of vectors b that are not in C(A) forms a subspace.



      I would like to first understand if I am interpreting the question correctly. My understanding of this question is that I need to first prove that the set of vectors b are equal to the 0 vector, and if b1 and b2 are members of the subspace, then the sum of set b should be a member, and so should some scalar C multiplied by the set of vectors b. I just don't understand how to actually prove this






      linear-algebra matrices vector-spaces







      share|cite|improve this question











      $endgroup$




      If A is an m by n matrix, prove that the set of vectors b that are not in C(A) forms a subspace.



      I would like to first understand if I am interpreting the question correctly. My understanding of this question is that I need to first prove that the set of vectors b are equal to the 0 vector, and if b1 and b2 are members of the subspace, then the sum of set b should be a member, and so should some scalar C multiplied by the set of vectors b. I just don't understand how to actually prove this






      linear-algebra matrices vector-spaces




      linear-algebra matrices vector-spaces






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 2 days ago









      YuiTo Cheng

      2,1863937




      2,1863937










      asked 2 days ago









      AdamAdam

      345




      345




















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          Your result is wrong. Maybe this picture can help you figure out.



          enter image description here



          Since every subspace must contain the zero vector, noted by $underline0$.



          We know that $C(A)$ is a subspace for $mathbbR^m$, so $underline0in C(A) subseteq mathbbR^m$, that means the zero vector are inside the column space and also $mathbbR^m$.



          But if we consider $mathbbR^msetminus C(A)$, by the picture, it means we are now cancelling the red circle out, so the zero vector is no longer inside this "space".



          Therefore, it can not form a subspace.






          share|cite|improve this answer









          $endgroup$




















            5












            $begingroup$

            Your first condition should be $0$ is in a subspace.



            Also, the result is not true.



            Let $0_m$ be the zero vector in $mathbbR^m$. We know that $0_m in C(A)$ since $sum_i=1^n A_i cdot 0=0_m$.



            $0_m$ is in $C(A)$, $0_m$ can't be in the set of vector that are not in $C(A)$. Hence the set of vectors that are not in $C(A)$ can't form a subspace.






            share|cite|improve this answer











            $endgroup$












            • $begingroup$
              How do I know that 0 is in C(A)?
              $endgroup$
              – Adam
              2 days ago











            • $begingroup$
              $C(A)$ is the column space right?
              $endgroup$
              – Siong Thye Goh
              2 days ago










            • $begingroup$
              Correct. I am new to linear algebra and am trying to understand the definitions Column space, null space, subspace, etc...Perhaps I am not understanding the proper definition of column space
              $endgroup$
              – Adam
              2 days ago






            • 1




              $begingroup$
              I have included an explaination of why $0_m in C(A)$.
              $endgroup$
              – Siong Thye Goh
              2 days ago











            Your Answer





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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            Your result is wrong. Maybe this picture can help you figure out.



            enter image description here



            Since every subspace must contain the zero vector, noted by $underline0$.



            We know that $C(A)$ is a subspace for $mathbbR^m$, so $underline0in C(A) subseteq mathbbR^m$, that means the zero vector are inside the column space and also $mathbbR^m$.



            But if we consider $mathbbR^msetminus C(A)$, by the picture, it means we are now cancelling the red circle out, so the zero vector is no longer inside this "space".



            Therefore, it can not form a subspace.






            share|cite|improve this answer









            $endgroup$

















              2












              $begingroup$

              Your result is wrong. Maybe this picture can help you figure out.



              enter image description here



              Since every subspace must contain the zero vector, noted by $underline0$.



              We know that $C(A)$ is a subspace for $mathbbR^m$, so $underline0in C(A) subseteq mathbbR^m$, that means the zero vector are inside the column space and also $mathbbR^m$.



              But if we consider $mathbbR^msetminus C(A)$, by the picture, it means we are now cancelling the red circle out, so the zero vector is no longer inside this "space".



              Therefore, it can not form a subspace.






              share|cite|improve this answer









              $endgroup$















                2












                2








                2





                $begingroup$

                Your result is wrong. Maybe this picture can help you figure out.



                enter image description here



                Since every subspace must contain the zero vector, noted by $underline0$.



                We know that $C(A)$ is a subspace for $mathbbR^m$, so $underline0in C(A) subseteq mathbbR^m$, that means the zero vector are inside the column space and also $mathbbR^m$.



                But if we consider $mathbbR^msetminus C(A)$, by the picture, it means we are now cancelling the red circle out, so the zero vector is no longer inside this "space".



                Therefore, it can not form a subspace.






                share|cite|improve this answer









                $endgroup$



                Your result is wrong. Maybe this picture can help you figure out.



                enter image description here



                Since every subspace must contain the zero vector, noted by $underline0$.



                We know that $C(A)$ is a subspace for $mathbbR^m$, so $underline0in C(A) subseteq mathbbR^m$, that means the zero vector are inside the column space and also $mathbbR^m$.



                But if we consider $mathbbR^msetminus C(A)$, by the picture, it means we are now cancelling the red circle out, so the zero vector is no longer inside this "space".



                Therefore, it can not form a subspace.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 days ago









                Jade PangJade Pang

                1114




                1114





















                    5












                    $begingroup$

                    Your first condition should be $0$ is in a subspace.



                    Also, the result is not true.



                    Let $0_m$ be the zero vector in $mathbbR^m$. We know that $0_m in C(A)$ since $sum_i=1^n A_i cdot 0=0_m$.



                    $0_m$ is in $C(A)$, $0_m$ can't be in the set of vector that are not in $C(A)$. Hence the set of vectors that are not in $C(A)$ can't form a subspace.






                    share|cite|improve this answer











                    $endgroup$












                    • $begingroup$
                      How do I know that 0 is in C(A)?
                      $endgroup$
                      – Adam
                      2 days ago











                    • $begingroup$
                      $C(A)$ is the column space right?
                      $endgroup$
                      – Siong Thye Goh
                      2 days ago










                    • $begingroup$
                      Correct. I am new to linear algebra and am trying to understand the definitions Column space, null space, subspace, etc...Perhaps I am not understanding the proper definition of column space
                      $endgroup$
                      – Adam
                      2 days ago






                    • 1




                      $begingroup$
                      I have included an explaination of why $0_m in C(A)$.
                      $endgroup$
                      – Siong Thye Goh
                      2 days ago















                    5












                    $begingroup$

                    Your first condition should be $0$ is in a subspace.



                    Also, the result is not true.



                    Let $0_m$ be the zero vector in $mathbbR^m$. We know that $0_m in C(A)$ since $sum_i=1^n A_i cdot 0=0_m$.



                    $0_m$ is in $C(A)$, $0_m$ can't be in the set of vector that are not in $C(A)$. Hence the set of vectors that are not in $C(A)$ can't form a subspace.






                    share|cite|improve this answer











                    $endgroup$












                    • $begingroup$
                      How do I know that 0 is in C(A)?
                      $endgroup$
                      – Adam
                      2 days ago











                    • $begingroup$
                      $C(A)$ is the column space right?
                      $endgroup$
                      – Siong Thye Goh
                      2 days ago










                    • $begingroup$
                      Correct. I am new to linear algebra and am trying to understand the definitions Column space, null space, subspace, etc...Perhaps I am not understanding the proper definition of column space
                      $endgroup$
                      – Adam
                      2 days ago






                    • 1




                      $begingroup$
                      I have included an explaination of why $0_m in C(A)$.
                      $endgroup$
                      – Siong Thye Goh
                      2 days ago













                    5












                    5








                    5





                    $begingroup$

                    Your first condition should be $0$ is in a subspace.



                    Also, the result is not true.



                    Let $0_m$ be the zero vector in $mathbbR^m$. We know that $0_m in C(A)$ since $sum_i=1^n A_i cdot 0=0_m$.



                    $0_m$ is in $C(A)$, $0_m$ can't be in the set of vector that are not in $C(A)$. Hence the set of vectors that are not in $C(A)$ can't form a subspace.






                    share|cite|improve this answer











                    $endgroup$



                    Your first condition should be $0$ is in a subspace.



                    Also, the result is not true.



                    Let $0_m$ be the zero vector in $mathbbR^m$. We know that $0_m in C(A)$ since $sum_i=1^n A_i cdot 0=0_m$.



                    $0_m$ is in $C(A)$, $0_m$ can't be in the set of vector that are not in $C(A)$. Hence the set of vectors that are not in $C(A)$ can't form a subspace.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 2 days ago

























                    answered 2 days ago









                    Siong Thye GohSiong Thye Goh

                    103k1468120




                    103k1468120











                    • $begingroup$
                      How do I know that 0 is in C(A)?
                      $endgroup$
                      – Adam
                      2 days ago











                    • $begingroup$
                      $C(A)$ is the column space right?
                      $endgroup$
                      – Siong Thye Goh
                      2 days ago










                    • $begingroup$
                      Correct. I am new to linear algebra and am trying to understand the definitions Column space, null space, subspace, etc...Perhaps I am not understanding the proper definition of column space
                      $endgroup$
                      – Adam
                      2 days ago






                    • 1




                      $begingroup$
                      I have included an explaination of why $0_m in C(A)$.
                      $endgroup$
                      – Siong Thye Goh
                      2 days ago
















                    • $begingroup$
                      How do I know that 0 is in C(A)?
                      $endgroup$
                      – Adam
                      2 days ago











                    • $begingroup$
                      $C(A)$ is the column space right?
                      $endgroup$
                      – Siong Thye Goh
                      2 days ago










                    • $begingroup$
                      Correct. I am new to linear algebra and am trying to understand the definitions Column space, null space, subspace, etc...Perhaps I am not understanding the proper definition of column space
                      $endgroup$
                      – Adam
                      2 days ago






                    • 1




                      $begingroup$
                      I have included an explaination of why $0_m in C(A)$.
                      $endgroup$
                      – Siong Thye Goh
                      2 days ago















                    $begingroup$
                    How do I know that 0 is in C(A)?
                    $endgroup$
                    – Adam
                    2 days ago





                    $begingroup$
                    How do I know that 0 is in C(A)?
                    $endgroup$
                    – Adam
                    2 days ago













                    $begingroup$
                    $C(A)$ is the column space right?
                    $endgroup$
                    – Siong Thye Goh
                    2 days ago




                    $begingroup$
                    $C(A)$ is the column space right?
                    $endgroup$
                    – Siong Thye Goh
                    2 days ago












                    $begingroup$
                    Correct. I am new to linear algebra and am trying to understand the definitions Column space, null space, subspace, etc...Perhaps I am not understanding the proper definition of column space
                    $endgroup$
                    – Adam
                    2 days ago




                    $begingroup$
                    Correct. I am new to linear algebra and am trying to understand the definitions Column space, null space, subspace, etc...Perhaps I am not understanding the proper definition of column space
                    $endgroup$
                    – Adam
                    2 days ago




                    1




                    1




                    $begingroup$
                    I have included an explaination of why $0_m in C(A)$.
                    $endgroup$
                    – Siong Thye Goh
                    2 days ago




                    $begingroup$
                    I have included an explaination of why $0_m in C(A)$.
                    $endgroup$
                    – Siong Thye Goh
                    2 days ago

















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