Basis for Nullspace of matrix Null(A)












0












$begingroup$


I could only get a part of the solution and not quite it. The problem is:



A=$left[begin{matrix} 1 & -1 & 2 &1 \ 2 & 1 & 0 & 1 end{matrix}right]$



Which of the following sets is a basis for Null(A)?



The correct answer is :{$left[begin{matrix} 0\ 1\ 1\ -1 end{matrix}right]$, $left[begin{matrix} -4\ 5\ 3\ 3 end{matrix}right]$}



And what I got is : {$left[begin{matrix} -2\ 4\ 3\ 0 end{matrix}right]$,$left[begin{matrix} -2\ 1\ 0\ 3 end{matrix}right]$}



The 2nd vector of the correct answer is 2 of mine added together, I don't know why and I'm still missing the first one. I looked at this question and got the correct answer for it by trying to do it the same way I did mine. Find the basis for the null space of the given matrix and give nullity(A)?






linear-algebra







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    I could only get a part of the solution and not quite it. The problem is:



    A=$left[begin{matrix} 1 & -1 & 2 &1 \ 2 & 1 & 0 & 1 end{matrix}right]$



    Which of the following sets is a basis for Null(A)?



    The correct answer is :{$left[begin{matrix} 0\ 1\ 1\ -1 end{matrix}right]$, $left[begin{matrix} -4\ 5\ 3\ 3 end{matrix}right]$}



    And what I got is : {$left[begin{matrix} -2\ 4\ 3\ 0 end{matrix}right]$,$left[begin{matrix} -2\ 1\ 0\ 3 end{matrix}right]$}



    The 2nd vector of the correct answer is 2 of mine added together, I don't know why and I'm still missing the first one. I looked at this question and got the correct answer for it by trying to do it the same way I did mine. Find the basis for the null space of the given matrix and give nullity(A)?






    linear-algebra







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I could only get a part of the solution and not quite it. The problem is:



      A=$left[begin{matrix} 1 & -1 & 2 &1 \ 2 & 1 & 0 & 1 end{matrix}right]$



      Which of the following sets is a basis for Null(A)?



      The correct answer is :{$left[begin{matrix} 0\ 1\ 1\ -1 end{matrix}right]$, $left[begin{matrix} -4\ 5\ 3\ 3 end{matrix}right]$}



      And what I got is : {$left[begin{matrix} -2\ 4\ 3\ 0 end{matrix}right]$,$left[begin{matrix} -2\ 1\ 0\ 3 end{matrix}right]$}



      The 2nd vector of the correct answer is 2 of mine added together, I don't know why and I'm still missing the first one. I looked at this question and got the correct answer for it by trying to do it the same way I did mine. Find the basis for the null space of the given matrix and give nullity(A)?






      linear-algebra







      share|cite|improve this question









      $endgroup$




      I could only get a part of the solution and not quite it. The problem is:



      A=$left[begin{matrix} 1 & -1 & 2 &1 \ 2 & 1 & 0 & 1 end{matrix}right]$



      Which of the following sets is a basis for Null(A)?



      The correct answer is :{$left[begin{matrix} 0\ 1\ 1\ -1 end{matrix}right]$, $left[begin{matrix} -4\ 5\ 3\ 3 end{matrix}right]$}



      And what I got is : {$left[begin{matrix} -2\ 4\ 3\ 0 end{matrix}right]$,$left[begin{matrix} -2\ 1\ 0\ 3 end{matrix}right]$}



      The 2nd vector of the correct answer is 2 of mine added together, I don't know why and I'm still missing the first one. I looked at this question and got the correct answer for it by trying to do it the same way I did mine. Find the basis for the null space of the given matrix and give nullity(A)?






      linear-algebra




      linear-algebra






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      asked Jan 9 at 15:58









      Antoni MaleckiAntoni Malecki

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          3 Answers
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          $begingroup$

          Both answers are correct. That space has infinitely many bases, of course.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Now I see it, silly me , if you subtract my 2nd vector from the 1st and then divide by 3 you get the 1st vector of the answer.
            $endgroup$
            – Antoni Malecki
            Jan 9 at 16:07



















          0












          $begingroup$

          This looks like it came from a multiple-choice question. As has been pointed out, bases are not unique, so there’s no particular reason to expect that the basis that you computed is going to match any of the proposed bases in the question.



          A better strategy would’ve been to go through the possible answers and eliminate those that can’t be a null space basis. First, it’s clear from inspection that $A$’s nullity is 2, so if any of the possible answers don’t consist of two vectors, you can throw those out. Next, the vectors in a basis must be linearly independent, which is trivial to check for a pair of vectors. Discard any of the remaining answers that are linearly dependent. For the remaining viable candidates, multiply each of the vectors by $A$ and check that you get zero for each one. Whatever’s left must be a basis for $N(A)$.



          In this case, the two vectors in the given answer are obviously linearly independent, and their products with $A$ are both zero, so they are indeed a basis for $N(A)$ (as is the set you found).






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            You probably went for the RREF of $A$, which is
            $$
            begin{bmatrix}
            1 & 0 & -2/3 & -2/3 \
            0 & 1 & 4/3 & 1/3
            end{bmatrix}
            $$
            With $x_3=3$ and $x_4=0$ you get the first vector, with $x_3=0$ and $x_3=3$ you get the second one. Your solution is correct, good work.



            If you try $x_3=1$ and $x_4=-1$, you find the first vector in the solution; with $x_3=3$ and $x_3=3$, you find the second one. I'm not sure how they got their vectors, to be honest.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              It was from a multiple choice so probably was done to cause confusion.
              $endgroup$
              – Antoni Malecki
              Jan 10 at 10:20











            Your Answer





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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            Both answers are correct. That space has infinitely many bases, of course.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Now I see it, silly me , if you subtract my 2nd vector from the 1st and then divide by 3 you get the 1st vector of the answer.
              $endgroup$
              – Antoni Malecki
              Jan 9 at 16:07
















            1












            $begingroup$

            Both answers are correct. That space has infinitely many bases, of course.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Now I see it, silly me , if you subtract my 2nd vector from the 1st and then divide by 3 you get the 1st vector of the answer.
              $endgroup$
              – Antoni Malecki
              Jan 9 at 16:07














            1












            1








            1





            $begingroup$

            Both answers are correct. That space has infinitely many bases, of course.






            share|cite|improve this answer









            $endgroup$



            Both answers are correct. That space has infinitely many bases, of course.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 9 at 16:00









            José Carlos SantosJosé Carlos Santos

            164k22131234




            164k22131234












            • $begingroup$
              Now I see it, silly me , if you subtract my 2nd vector from the 1st and then divide by 3 you get the 1st vector of the answer.
              $endgroup$
              – Antoni Malecki
              Jan 9 at 16:07


















            • $begingroup$
              Now I see it, silly me , if you subtract my 2nd vector from the 1st and then divide by 3 you get the 1st vector of the answer.
              $endgroup$
              – Antoni Malecki
              Jan 9 at 16:07
















            $begingroup$
            Now I see it, silly me , if you subtract my 2nd vector from the 1st and then divide by 3 you get the 1st vector of the answer.
            $endgroup$
            – Antoni Malecki
            Jan 9 at 16:07




            $begingroup$
            Now I see it, silly me , if you subtract my 2nd vector from the 1st and then divide by 3 you get the 1st vector of the answer.
            $endgroup$
            – Antoni Malecki
            Jan 9 at 16:07











            0












            $begingroup$

            This looks like it came from a multiple-choice question. As has been pointed out, bases are not unique, so there’s no particular reason to expect that the basis that you computed is going to match any of the proposed bases in the question.



            A better strategy would’ve been to go through the possible answers and eliminate those that can’t be a null space basis. First, it’s clear from inspection that $A$’s nullity is 2, so if any of the possible answers don’t consist of two vectors, you can throw those out. Next, the vectors in a basis must be linearly independent, which is trivial to check for a pair of vectors. Discard any of the remaining answers that are linearly dependent. For the remaining viable candidates, multiply each of the vectors by $A$ and check that you get zero for each one. Whatever’s left must be a basis for $N(A)$.



            In this case, the two vectors in the given answer are obviously linearly independent, and their products with $A$ are both zero, so they are indeed a basis for $N(A)$ (as is the set you found).






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              This looks like it came from a multiple-choice question. As has been pointed out, bases are not unique, so there’s no particular reason to expect that the basis that you computed is going to match any of the proposed bases in the question.



              A better strategy would’ve been to go through the possible answers and eliminate those that can’t be a null space basis. First, it’s clear from inspection that $A$’s nullity is 2, so if any of the possible answers don’t consist of two vectors, you can throw those out. Next, the vectors in a basis must be linearly independent, which is trivial to check for a pair of vectors. Discard any of the remaining answers that are linearly dependent. For the remaining viable candidates, multiply each of the vectors by $A$ and check that you get zero for each one. Whatever’s left must be a basis for $N(A)$.



              In this case, the two vectors in the given answer are obviously linearly independent, and their products with $A$ are both zero, so they are indeed a basis for $N(A)$ (as is the set you found).






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                This looks like it came from a multiple-choice question. As has been pointed out, bases are not unique, so there’s no particular reason to expect that the basis that you computed is going to match any of the proposed bases in the question.



                A better strategy would’ve been to go through the possible answers and eliminate those that can’t be a null space basis. First, it’s clear from inspection that $A$’s nullity is 2, so if any of the possible answers don’t consist of two vectors, you can throw those out. Next, the vectors in a basis must be linearly independent, which is trivial to check for a pair of vectors. Discard any of the remaining answers that are linearly dependent. For the remaining viable candidates, multiply each of the vectors by $A$ and check that you get zero for each one. Whatever’s left must be a basis for $N(A)$.



                In this case, the two vectors in the given answer are obviously linearly independent, and their products with $A$ are both zero, so they are indeed a basis for $N(A)$ (as is the set you found).






                share|cite|improve this answer









                $endgroup$



                This looks like it came from a multiple-choice question. As has been pointed out, bases are not unique, so there’s no particular reason to expect that the basis that you computed is going to match any of the proposed bases in the question.



                A better strategy would’ve been to go through the possible answers and eliminate those that can’t be a null space basis. First, it’s clear from inspection that $A$’s nullity is 2, so if any of the possible answers don’t consist of two vectors, you can throw those out. Next, the vectors in a basis must be linearly independent, which is trivial to check for a pair of vectors. Discard any of the remaining answers that are linearly dependent. For the remaining viable candidates, multiply each of the vectors by $A$ and check that you get zero for each one. Whatever’s left must be a basis for $N(A)$.



                In this case, the two vectors in the given answer are obviously linearly independent, and their products with $A$ are both zero, so they are indeed a basis for $N(A)$ (as is the set you found).







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 9 at 21:12









                amdamd

                30.6k21050




                30.6k21050























                    0












                    $begingroup$

                    You probably went for the RREF of $A$, which is
                    $$
                    begin{bmatrix}
                    1 & 0 & -2/3 & -2/3 \
                    0 & 1 & 4/3 & 1/3
                    end{bmatrix}
                    $$
                    With $x_3=3$ and $x_4=0$ you get the first vector, with $x_3=0$ and $x_3=3$ you get the second one. Your solution is correct, good work.



                    If you try $x_3=1$ and $x_4=-1$, you find the first vector in the solution; with $x_3=3$ and $x_3=3$, you find the second one. I'm not sure how they got their vectors, to be honest.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      It was from a multiple choice so probably was done to cause confusion.
                      $endgroup$
                      – Antoni Malecki
                      Jan 10 at 10:20
















                    0












                    $begingroup$

                    You probably went for the RREF of $A$, which is
                    $$
                    begin{bmatrix}
                    1 & 0 & -2/3 & -2/3 \
                    0 & 1 & 4/3 & 1/3
                    end{bmatrix}
                    $$
                    With $x_3=3$ and $x_4=0$ you get the first vector, with $x_3=0$ and $x_3=3$ you get the second one. Your solution is correct, good work.



                    If you try $x_3=1$ and $x_4=-1$, you find the first vector in the solution; with $x_3=3$ and $x_3=3$, you find the second one. I'm not sure how they got their vectors, to be honest.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      It was from a multiple choice so probably was done to cause confusion.
                      $endgroup$
                      – Antoni Malecki
                      Jan 10 at 10:20














                    0












                    0








                    0





                    $begingroup$

                    You probably went for the RREF of $A$, which is
                    $$
                    begin{bmatrix}
                    1 & 0 & -2/3 & -2/3 \
                    0 & 1 & 4/3 & 1/3
                    end{bmatrix}
                    $$
                    With $x_3=3$ and $x_4=0$ you get the first vector, with $x_3=0$ and $x_3=3$ you get the second one. Your solution is correct, good work.



                    If you try $x_3=1$ and $x_4=-1$, you find the first vector in the solution; with $x_3=3$ and $x_3=3$, you find the second one. I'm not sure how they got their vectors, to be honest.






                    share|cite|improve this answer









                    $endgroup$



                    You probably went for the RREF of $A$, which is
                    $$
                    begin{bmatrix}
                    1 & 0 & -2/3 & -2/3 \
                    0 & 1 & 4/3 & 1/3
                    end{bmatrix}
                    $$
                    With $x_3=3$ and $x_4=0$ you get the first vector, with $x_3=0$ and $x_3=3$ you get the second one. Your solution is correct, good work.



                    If you try $x_3=1$ and $x_4=-1$, you find the first vector in the solution; with $x_3=3$ and $x_3=3$, you find the second one. I'm not sure how they got their vectors, to be honest.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 9 at 23:23









                    egregegreg

                    183k1486205




                    183k1486205












                    • $begingroup$
                      It was from a multiple choice so probably was done to cause confusion.
                      $endgroup$
                      – Antoni Malecki
                      Jan 10 at 10:20


















                    • $begingroup$
                      It was from a multiple choice so probably was done to cause confusion.
                      $endgroup$
                      – Antoni Malecki
                      Jan 10 at 10:20
















                    $begingroup$
                    It was from a multiple choice so probably was done to cause confusion.
                    $endgroup$
                    – Antoni Malecki
                    Jan 10 at 10:20




                    $begingroup$
                    It was from a multiple choice so probably was done to cause confusion.
                    $endgroup$
                    – Antoni Malecki
                    Jan 10 at 10:20


















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