Find bases for N(T) and R(T) and verify the Dimension Theorem












1












$begingroup$


$Tpmatrix{a11&a12&a13\a21&a22&a23\a31&a32&a33}$= $pmatrix{a11+a12+a13&a21+a22+a23\a21+a22+a23&0}$



I know what the Dimension Theorem is and by looking at the equation I can tell that a11+a12=-a13 and so on shows that the null space contains more than just the zero vector, however I struggle in figuring out how to express the bases of N(T) and R(T).






linear-algebra







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$endgroup$








  • 3




    $begingroup$
    It should not be hard to find some elements of the range, and then to prove that those elements are linearly independent and span the range. Try it!
    $endgroup$
    – Gerry Myerson
    Dec 8 '13 at 2:54
















1












$begingroup$


$Tpmatrix{a11&a12&a13\a21&a22&a23\a31&a32&a33}$= $pmatrix{a11+a12+a13&a21+a22+a23\a21+a22+a23&0}$



I know what the Dimension Theorem is and by looking at the equation I can tell that a11+a12=-a13 and so on shows that the null space contains more than just the zero vector, however I struggle in figuring out how to express the bases of N(T) and R(T).






linear-algebra







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    It should not be hard to find some elements of the range, and then to prove that those elements are linearly independent and span the range. Try it!
    $endgroup$
    – Gerry Myerson
    Dec 8 '13 at 2:54














1












1








1





$begingroup$


$Tpmatrix{a11&a12&a13\a21&a22&a23\a31&a32&a33}$= $pmatrix{a11+a12+a13&a21+a22+a23\a21+a22+a23&0}$



I know what the Dimension Theorem is and by looking at the equation I can tell that a11+a12=-a13 and so on shows that the null space contains more than just the zero vector, however I struggle in figuring out how to express the bases of N(T) and R(T).






linear-algebra







share|cite|improve this question









$endgroup$




$Tpmatrix{a11&a12&a13\a21&a22&a23\a31&a32&a33}$= $pmatrix{a11+a12+a13&a21+a22+a23\a21+a22+a23&0}$



I know what the Dimension Theorem is and by looking at the equation I can tell that a11+a12=-a13 and so on shows that the null space contains more than just the zero vector, however I struggle in figuring out how to express the bases of N(T) and R(T).






linear-algebra




linear-algebra






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




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asked Dec 8 '13 at 2:52









user114220user114220

59117




59117








  • 3




    $begingroup$
    It should not be hard to find some elements of the range, and then to prove that those elements are linearly independent and span the range. Try it!
    $endgroup$
    – Gerry Myerson
    Dec 8 '13 at 2:54














  • 3




    $begingroup$
    It should not be hard to find some elements of the range, and then to prove that those elements are linearly independent and span the range. Try it!
    $endgroup$
    – Gerry Myerson
    Dec 8 '13 at 2:54








3




3




$begingroup$
It should not be hard to find some elements of the range, and then to prove that those elements are linearly independent and span the range. Try it!
$endgroup$
– Gerry Myerson
Dec 8 '13 at 2:54




$begingroup$
It should not be hard to find some elements of the range, and then to prove that those elements are linearly independent and span the range. Try it!
$endgroup$
– Gerry Myerson
Dec 8 '13 at 2:54










2 Answers
2






active

oldest

votes


















0












$begingroup$

You can just try some small numbers, easiest to play with $1$ and $0$.



For $N(T)$, as you started you get these $3$ equations ($2$ of them are the same) as $-a_{11}-a_{12} =a_{13}$, so that $a_{11}$, $a_{12}$ and $a_{21}$, $a_{22}$ can be anything, only $a_{13}$ and $a_{23}$ are bound to them. Also, $a_{31}, a_{32}, a_{33}$ can be anything.



So, consider these elements of $N(T)$:




  • $a_{11}=1$, rest is $0$ ($implies a_{13}=-1$)

  • $a_{12}=1$, rest is $0$ ($implies a_{13}=-1$)

  • $a_{21}=1$, rest is $0$ ($implies a_{23}=-1$)

  • and so on...






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    quite sure:



    basis for $N(T) = $$pmatrix{1&0&-1\0&0&0\0&0&0}$
    $pmatrix{0&1&-1\0&0&0\0&0&0}$$pmatrix{0&0&0\1&0&-1\0&0&0}$$pmatrix{0&0&0\0&1&-1\0&0&0}$$pmatrix{0&0&0\0&0&0\1&0&0}$$pmatrix{0&0&0\0&0&0\0&1&0}$$pmatrix{0&0&0\0&0&0\0&0&1}$



    basis for $R(T) = pmatrix{1&0\0&0},pmatrix{0&1\1&0}$



    edit: I think this is what Berci said






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Yes, almost. I just haven't found my 'matrix' switch on the keyboard :) The last two ones of $N(T)$ are not correct, $a_{31}, a_{32}, a_{33}$ are free.
      $endgroup$
      – Berci
      Dec 8 '13 at 3:24












    • $begingroup$
      ah, you're right!
      $endgroup$
      – Sam Creamer
      Dec 8 '13 at 3:36






    • 1




      $begingroup$
      Did you notice that every matrix in the range is symmetric?
      $endgroup$
      – Gerry Myerson
      Dec 8 '13 at 3:39










    • $begingroup$
      I corrected according to the comments.
      $endgroup$
      – Berci
      Dec 8 '13 at 12:44











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    2 Answers
    2






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    2 Answers
    2






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    active

    oldest

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    0












    $begingroup$

    You can just try some small numbers, easiest to play with $1$ and $0$.



    For $N(T)$, as you started you get these $3$ equations ($2$ of them are the same) as $-a_{11}-a_{12} =a_{13}$, so that $a_{11}$, $a_{12}$ and $a_{21}$, $a_{22}$ can be anything, only $a_{13}$ and $a_{23}$ are bound to them. Also, $a_{31}, a_{32}, a_{33}$ can be anything.



    So, consider these elements of $N(T)$:




    • $a_{11}=1$, rest is $0$ ($implies a_{13}=-1$)

    • $a_{12}=1$, rest is $0$ ($implies a_{13}=-1$)

    • $a_{21}=1$, rest is $0$ ($implies a_{23}=-1$)

    • and so on...






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      You can just try some small numbers, easiest to play with $1$ and $0$.



      For $N(T)$, as you started you get these $3$ equations ($2$ of them are the same) as $-a_{11}-a_{12} =a_{13}$, so that $a_{11}$, $a_{12}$ and $a_{21}$, $a_{22}$ can be anything, only $a_{13}$ and $a_{23}$ are bound to them. Also, $a_{31}, a_{32}, a_{33}$ can be anything.



      So, consider these elements of $N(T)$:




      • $a_{11}=1$, rest is $0$ ($implies a_{13}=-1$)

      • $a_{12}=1$, rest is $0$ ($implies a_{13}=-1$)

      • $a_{21}=1$, rest is $0$ ($implies a_{23}=-1$)

      • and so on...






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        You can just try some small numbers, easiest to play with $1$ and $0$.



        For $N(T)$, as you started you get these $3$ equations ($2$ of them are the same) as $-a_{11}-a_{12} =a_{13}$, so that $a_{11}$, $a_{12}$ and $a_{21}$, $a_{22}$ can be anything, only $a_{13}$ and $a_{23}$ are bound to them. Also, $a_{31}, a_{32}, a_{33}$ can be anything.



        So, consider these elements of $N(T)$:




        • $a_{11}=1$, rest is $0$ ($implies a_{13}=-1$)

        • $a_{12}=1$, rest is $0$ ($implies a_{13}=-1$)

        • $a_{21}=1$, rest is $0$ ($implies a_{23}=-1$)

        • and so on...






        share|cite|improve this answer











        $endgroup$



        You can just try some small numbers, easiest to play with $1$ and $0$.



        For $N(T)$, as you started you get these $3$ equations ($2$ of them are the same) as $-a_{11}-a_{12} =a_{13}$, so that $a_{11}$, $a_{12}$ and $a_{21}$, $a_{22}$ can be anything, only $a_{13}$ and $a_{23}$ are bound to them. Also, $a_{31}, a_{32}, a_{33}$ can be anything.



        So, consider these elements of $N(T)$:




        • $a_{11}=1$, rest is $0$ ($implies a_{13}=-1$)

        • $a_{12}=1$, rest is $0$ ($implies a_{13}=-1$)

        • $a_{21}=1$, rest is $0$ ($implies a_{23}=-1$)

        • and so on...







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 8 '13 at 3:26

























        answered Dec 8 '13 at 3:11









        BerciBerci

        61.7k23674




        61.7k23674























            0












            $begingroup$

            quite sure:



            basis for $N(T) = $$pmatrix{1&0&-1\0&0&0\0&0&0}$
            $pmatrix{0&1&-1\0&0&0\0&0&0}$$pmatrix{0&0&0\1&0&-1\0&0&0}$$pmatrix{0&0&0\0&1&-1\0&0&0}$$pmatrix{0&0&0\0&0&0\1&0&0}$$pmatrix{0&0&0\0&0&0\0&1&0}$$pmatrix{0&0&0\0&0&0\0&0&1}$



            basis for $R(T) = pmatrix{1&0\0&0},pmatrix{0&1\1&0}$



            edit: I think this is what Berci said






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Yes, almost. I just haven't found my 'matrix' switch on the keyboard :) The last two ones of $N(T)$ are not correct, $a_{31}, a_{32}, a_{33}$ are free.
              $endgroup$
              – Berci
              Dec 8 '13 at 3:24












            • $begingroup$
              ah, you're right!
              $endgroup$
              – Sam Creamer
              Dec 8 '13 at 3:36






            • 1




              $begingroup$
              Did you notice that every matrix in the range is symmetric?
              $endgroup$
              – Gerry Myerson
              Dec 8 '13 at 3:39










            • $begingroup$
              I corrected according to the comments.
              $endgroup$
              – Berci
              Dec 8 '13 at 12:44
















            0












            $begingroup$

            quite sure:



            basis for $N(T) = $$pmatrix{1&0&-1\0&0&0\0&0&0}$
            $pmatrix{0&1&-1\0&0&0\0&0&0}$$pmatrix{0&0&0\1&0&-1\0&0&0}$$pmatrix{0&0&0\0&1&-1\0&0&0}$$pmatrix{0&0&0\0&0&0\1&0&0}$$pmatrix{0&0&0\0&0&0\0&1&0}$$pmatrix{0&0&0\0&0&0\0&0&1}$



            basis for $R(T) = pmatrix{1&0\0&0},pmatrix{0&1\1&0}$



            edit: I think this is what Berci said






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Yes, almost. I just haven't found my 'matrix' switch on the keyboard :) The last two ones of $N(T)$ are not correct, $a_{31}, a_{32}, a_{33}$ are free.
              $endgroup$
              – Berci
              Dec 8 '13 at 3:24












            • $begingroup$
              ah, you're right!
              $endgroup$
              – Sam Creamer
              Dec 8 '13 at 3:36






            • 1




              $begingroup$
              Did you notice that every matrix in the range is symmetric?
              $endgroup$
              – Gerry Myerson
              Dec 8 '13 at 3:39










            • $begingroup$
              I corrected according to the comments.
              $endgroup$
              – Berci
              Dec 8 '13 at 12:44














            0












            0








            0





            $begingroup$

            quite sure:



            basis for $N(T) = $$pmatrix{1&0&-1\0&0&0\0&0&0}$
            $pmatrix{0&1&-1\0&0&0\0&0&0}$$pmatrix{0&0&0\1&0&-1\0&0&0}$$pmatrix{0&0&0\0&1&-1\0&0&0}$$pmatrix{0&0&0\0&0&0\1&0&0}$$pmatrix{0&0&0\0&0&0\0&1&0}$$pmatrix{0&0&0\0&0&0\0&0&1}$



            basis for $R(T) = pmatrix{1&0\0&0},pmatrix{0&1\1&0}$



            edit: I think this is what Berci said






            share|cite|improve this answer











            $endgroup$



            quite sure:



            basis for $N(T) = $$pmatrix{1&0&-1\0&0&0\0&0&0}$
            $pmatrix{0&1&-1\0&0&0\0&0&0}$$pmatrix{0&0&0\1&0&-1\0&0&0}$$pmatrix{0&0&0\0&1&-1\0&0&0}$$pmatrix{0&0&0\0&0&0\1&0&0}$$pmatrix{0&0&0\0&0&0\0&1&0}$$pmatrix{0&0&0\0&0&0\0&0&1}$



            basis for $R(T) = pmatrix{1&0\0&0},pmatrix{0&1\1&0}$



            edit: I think this is what Berci said







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 8 '13 at 12:44









            Berci

            61.7k23674




            61.7k23674










            answered Dec 8 '13 at 3:20









            Sam CreamerSam Creamer

            209414




            209414












            • $begingroup$
              Yes, almost. I just haven't found my 'matrix' switch on the keyboard :) The last two ones of $N(T)$ are not correct, $a_{31}, a_{32}, a_{33}$ are free.
              $endgroup$
              – Berci
              Dec 8 '13 at 3:24












            • $begingroup$
              ah, you're right!
              $endgroup$
              – Sam Creamer
              Dec 8 '13 at 3:36






            • 1




              $begingroup$
              Did you notice that every matrix in the range is symmetric?
              $endgroup$
              – Gerry Myerson
              Dec 8 '13 at 3:39










            • $begingroup$
              I corrected according to the comments.
              $endgroup$
              – Berci
              Dec 8 '13 at 12:44


















            • $begingroup$
              Yes, almost. I just haven't found my 'matrix' switch on the keyboard :) The last two ones of $N(T)$ are not correct, $a_{31}, a_{32}, a_{33}$ are free.
              $endgroup$
              – Berci
              Dec 8 '13 at 3:24












            • $begingroup$
              ah, you're right!
              $endgroup$
              – Sam Creamer
              Dec 8 '13 at 3:36






            • 1




              $begingroup$
              Did you notice that every matrix in the range is symmetric?
              $endgroup$
              – Gerry Myerson
              Dec 8 '13 at 3:39










            • $begingroup$
              I corrected according to the comments.
              $endgroup$
              – Berci
              Dec 8 '13 at 12:44
















            $begingroup$
            Yes, almost. I just haven't found my 'matrix' switch on the keyboard :) The last two ones of $N(T)$ are not correct, $a_{31}, a_{32}, a_{33}$ are free.
            $endgroup$
            – Berci
            Dec 8 '13 at 3:24






            $begingroup$
            Yes, almost. I just haven't found my 'matrix' switch on the keyboard :) The last two ones of $N(T)$ are not correct, $a_{31}, a_{32}, a_{33}$ are free.
            $endgroup$
            – Berci
            Dec 8 '13 at 3:24














            $begingroup$
            ah, you're right!
            $endgroup$
            – Sam Creamer
            Dec 8 '13 at 3:36




            $begingroup$
            ah, you're right!
            $endgroup$
            – Sam Creamer
            Dec 8 '13 at 3:36




            1




            1




            $begingroup$
            Did you notice that every matrix in the range is symmetric?
            $endgroup$
            – Gerry Myerson
            Dec 8 '13 at 3:39




            $begingroup$
            Did you notice that every matrix in the range is symmetric?
            $endgroup$
            – Gerry Myerson
            Dec 8 '13 at 3:39












            $begingroup$
            I corrected according to the comments.
            $endgroup$
            – Berci
            Dec 8 '13 at 12:44




            $begingroup$
            I corrected according to the comments.
            $endgroup$
            – Berci
            Dec 8 '13 at 12:44


















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