if $A in C^{2015,2015}$ and $rank(A) 15$












4












$begingroup$


I want to solve that thesis:

if $A in C^{2015,2015}$ and $rank(A) < 1000$ proof that $dim(ker(A+A^T)) > 15 $



from the fact that $$dim(im(A)) = dim(im(A^T))$$ and
$$ dim(ker(A))+dim(im(A))=n $$
it follows that
$$dim(ker(A)) = dim(ker(A^T))$$
We know that $rank(A) < 1000$. Hence $dim(ker(A))>1015$
But there I have stuck...
thanks for your time






linear-algebra matrices matrix-rank







share|cite|improve this question











$endgroup$

















    4












    $begingroup$


    I want to solve that thesis:

    if $A in C^{2015,2015}$ and $rank(A) < 1000$ proof that $dim(ker(A+A^T)) > 15 $



    from the fact that $$dim(im(A)) = dim(im(A^T))$$ and
    $$ dim(ker(A))+dim(im(A))=n $$
    it follows that
    $$dim(ker(A)) = dim(ker(A^T))$$
    We know that $rank(A) < 1000$. Hence $dim(ker(A))>1015$
    But there I have stuck...
    thanks for your time






    linear-algebra matrices matrix-rank







    share|cite|improve this question











    $endgroup$















      4












      4








      4





      $begingroup$


      I want to solve that thesis:

      if $A in C^{2015,2015}$ and $rank(A) < 1000$ proof that $dim(ker(A+A^T)) > 15 $



      from the fact that $$dim(im(A)) = dim(im(A^T))$$ and
      $$ dim(ker(A))+dim(im(A))=n $$
      it follows that
      $$dim(ker(A)) = dim(ker(A^T))$$
      We know that $rank(A) < 1000$. Hence $dim(ker(A))>1015$
      But there I have stuck...
      thanks for your time






      linear-algebra matrices matrix-rank







      share|cite|improve this question











      $endgroup$




      I want to solve that thesis:

      if $A in C^{2015,2015}$ and $rank(A) < 1000$ proof that $dim(ker(A+A^T)) > 15 $



      from the fact that $$dim(im(A)) = dim(im(A^T))$$ and
      $$ dim(ker(A))+dim(im(A))=n $$
      it follows that
      $$dim(ker(A)) = dim(ker(A^T))$$
      We know that $rank(A) < 1000$. Hence $dim(ker(A))>1015$
      But there I have stuck...
      thanks for your time






      linear-algebra matrices matrix-rank




      linear-algebra matrices matrix-rank






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 30 '18 at 15:20









      Yanko

      8,4692830




      8,4692830










      asked Dec 30 '18 at 15:18









      VirtualUserVirtualUser

      1,319317




      1,319317






















          2 Answers
          2






          active

          oldest

          votes


















          6












          $begingroup$

          You're one line from the solution:



          You showed that $dim (ker A) > 1015$ (so $A$ sends a vector space of dimension $1015$ to zero).



          Moreover since $dim (Im A ) = dim (Im A^T)$ you also have that the dimensions of the kernels equal and so $dim(ker A^T)>1015$.



          Therefore by the dimensions theorem we have that the dimension of $ker A cap ker A^T$ is at least $15$ (because $dim(ker A + ker A^T)leq 2015$). This completes the proof, because $ker (A+A^T)$ contains $ker Acap ker A^T$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Sorry if it is stupid question but I am sitting and I have no idea why $ dim(ker A + ker A^T)leq 2015$. Could I ask you for an explanation?
            $endgroup$
            – VirtualUser
            Dec 30 '18 at 15:45










          • $begingroup$
            @VirtualUser Both $ker A$ and $ker A^T$ are subspaces of $mathbb{C}^{2015}$. Therefore so is their sum.
            $endgroup$
            – Yanko
            Dec 30 '18 at 15:47












          • $begingroup$
            "Therefore by the dimensions theorem we have that the dimension of $ker A cap ker A^T$ is at least $15$ (because $dim(ker A + ker A^T)leq 2015$)." Ok but how from dimensions theorem I get this proposal? Dimension theorem says that $dim(im(T)) + dim(kernel(T)) = n $, but there, If we assume that T is $ ker A + ker A^T$ I haven't got neither dim(im(T)) nor dim(kernel(T))
            $endgroup$
            – VirtualUser
            Dec 30 '18 at 16:04












          • $begingroup$
            @VirtualUser This is the other dimension theorem. I'm talking about this $dim (U+W) = dim U + dim W - dim Ucap W$. Insert $U=ker A$ and $W=ker A^T$.
            $endgroup$
            – Yanko
            Dec 30 '18 at 16:05










          • $begingroup$
            Aaah, now I understand, you deserve a lot of thanks!
            $endgroup$
            – VirtualUser
            Dec 30 '18 at 16:07



















          2












          $begingroup$

          Hint: $operatorname{dim}(operatorname {im}(A+A^t)=operatorname{dim}(operatorname{im}A)+operatorname{dim}(operatorname{im}A^t)-operatorname{dim}(operatorname{im}Acapoperatorname{im}A^t)lt2000$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            No, using the hypothesis, less than $1000+1000=2000$.
            $endgroup$
            – Chris Custer
            Dec 30 '18 at 15:43












          Your Answer








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          2 Answers
          2






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          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          6












          $begingroup$

          You're one line from the solution:



          You showed that $dim (ker A) > 1015$ (so $A$ sends a vector space of dimension $1015$ to zero).



          Moreover since $dim (Im A ) = dim (Im A^T)$ you also have that the dimensions of the kernels equal and so $dim(ker A^T)>1015$.



          Therefore by the dimensions theorem we have that the dimension of $ker A cap ker A^T$ is at least $15$ (because $dim(ker A + ker A^T)leq 2015$). This completes the proof, because $ker (A+A^T)$ contains $ker Acap ker A^T$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Sorry if it is stupid question but I am sitting and I have no idea why $ dim(ker A + ker A^T)leq 2015$. Could I ask you for an explanation?
            $endgroup$
            – VirtualUser
            Dec 30 '18 at 15:45










          • $begingroup$
            @VirtualUser Both $ker A$ and $ker A^T$ are subspaces of $mathbb{C}^{2015}$. Therefore so is their sum.
            $endgroup$
            – Yanko
            Dec 30 '18 at 15:47












          • $begingroup$
            "Therefore by the dimensions theorem we have that the dimension of $ker A cap ker A^T$ is at least $15$ (because $dim(ker A + ker A^T)leq 2015$)." Ok but how from dimensions theorem I get this proposal? Dimension theorem says that $dim(im(T)) + dim(kernel(T)) = n $, but there, If we assume that T is $ ker A + ker A^T$ I haven't got neither dim(im(T)) nor dim(kernel(T))
            $endgroup$
            – VirtualUser
            Dec 30 '18 at 16:04












          • $begingroup$
            @VirtualUser This is the other dimension theorem. I'm talking about this $dim (U+W) = dim U + dim W - dim Ucap W$. Insert $U=ker A$ and $W=ker A^T$.
            $endgroup$
            – Yanko
            Dec 30 '18 at 16:05










          • $begingroup$
            Aaah, now I understand, you deserve a lot of thanks!
            $endgroup$
            – VirtualUser
            Dec 30 '18 at 16:07
















          6












          $begingroup$

          You're one line from the solution:



          You showed that $dim (ker A) > 1015$ (so $A$ sends a vector space of dimension $1015$ to zero).



          Moreover since $dim (Im A ) = dim (Im A^T)$ you also have that the dimensions of the kernels equal and so $dim(ker A^T)>1015$.



          Therefore by the dimensions theorem we have that the dimension of $ker A cap ker A^T$ is at least $15$ (because $dim(ker A + ker A^T)leq 2015$). This completes the proof, because $ker (A+A^T)$ contains $ker Acap ker A^T$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Sorry if it is stupid question but I am sitting and I have no idea why $ dim(ker A + ker A^T)leq 2015$. Could I ask you for an explanation?
            $endgroup$
            – VirtualUser
            Dec 30 '18 at 15:45










          • $begingroup$
            @VirtualUser Both $ker A$ and $ker A^T$ are subspaces of $mathbb{C}^{2015}$. Therefore so is their sum.
            $endgroup$
            – Yanko
            Dec 30 '18 at 15:47












          • $begingroup$
            "Therefore by the dimensions theorem we have that the dimension of $ker A cap ker A^T$ is at least $15$ (because $dim(ker A + ker A^T)leq 2015$)." Ok but how from dimensions theorem I get this proposal? Dimension theorem says that $dim(im(T)) + dim(kernel(T)) = n $, but there, If we assume that T is $ ker A + ker A^T$ I haven't got neither dim(im(T)) nor dim(kernel(T))
            $endgroup$
            – VirtualUser
            Dec 30 '18 at 16:04












          • $begingroup$
            @VirtualUser This is the other dimension theorem. I'm talking about this $dim (U+W) = dim U + dim W - dim Ucap W$. Insert $U=ker A$ and $W=ker A^T$.
            $endgroup$
            – Yanko
            Dec 30 '18 at 16:05










          • $begingroup$
            Aaah, now I understand, you deserve a lot of thanks!
            $endgroup$
            – VirtualUser
            Dec 30 '18 at 16:07














          6












          6








          6





          $begingroup$

          You're one line from the solution:



          You showed that $dim (ker A) > 1015$ (so $A$ sends a vector space of dimension $1015$ to zero).



          Moreover since $dim (Im A ) = dim (Im A^T)$ you also have that the dimensions of the kernels equal and so $dim(ker A^T)>1015$.



          Therefore by the dimensions theorem we have that the dimension of $ker A cap ker A^T$ is at least $15$ (because $dim(ker A + ker A^T)leq 2015$). This completes the proof, because $ker (A+A^T)$ contains $ker Acap ker A^T$.






          share|cite|improve this answer









          $endgroup$



          You're one line from the solution:



          You showed that $dim (ker A) > 1015$ (so $A$ sends a vector space of dimension $1015$ to zero).



          Moreover since $dim (Im A ) = dim (Im A^T)$ you also have that the dimensions of the kernels equal and so $dim(ker A^T)>1015$.



          Therefore by the dimensions theorem we have that the dimension of $ker A cap ker A^T$ is at least $15$ (because $dim(ker A + ker A^T)leq 2015$). This completes the proof, because $ker (A+A^T)$ contains $ker Acap ker A^T$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 30 '18 at 15:23









          YankoYanko

          8,4692830




          8,4692830












          • $begingroup$
            Sorry if it is stupid question but I am sitting and I have no idea why $ dim(ker A + ker A^T)leq 2015$. Could I ask you for an explanation?
            $endgroup$
            – VirtualUser
            Dec 30 '18 at 15:45










          • $begingroup$
            @VirtualUser Both $ker A$ and $ker A^T$ are subspaces of $mathbb{C}^{2015}$. Therefore so is their sum.
            $endgroup$
            – Yanko
            Dec 30 '18 at 15:47












          • $begingroup$
            "Therefore by the dimensions theorem we have that the dimension of $ker A cap ker A^T$ is at least $15$ (because $dim(ker A + ker A^T)leq 2015$)." Ok but how from dimensions theorem I get this proposal? Dimension theorem says that $dim(im(T)) + dim(kernel(T)) = n $, but there, If we assume that T is $ ker A + ker A^T$ I haven't got neither dim(im(T)) nor dim(kernel(T))
            $endgroup$
            – VirtualUser
            Dec 30 '18 at 16:04












          • $begingroup$
            @VirtualUser This is the other dimension theorem. I'm talking about this $dim (U+W) = dim U + dim W - dim Ucap W$. Insert $U=ker A$ and $W=ker A^T$.
            $endgroup$
            – Yanko
            Dec 30 '18 at 16:05










          • $begingroup$
            Aaah, now I understand, you deserve a lot of thanks!
            $endgroup$
            – VirtualUser
            Dec 30 '18 at 16:07


















          • $begingroup$
            Sorry if it is stupid question but I am sitting and I have no idea why $ dim(ker A + ker A^T)leq 2015$. Could I ask you for an explanation?
            $endgroup$
            – VirtualUser
            Dec 30 '18 at 15:45










          • $begingroup$
            @VirtualUser Both $ker A$ and $ker A^T$ are subspaces of $mathbb{C}^{2015}$. Therefore so is their sum.
            $endgroup$
            – Yanko
            Dec 30 '18 at 15:47












          • $begingroup$
            "Therefore by the dimensions theorem we have that the dimension of $ker A cap ker A^T$ is at least $15$ (because $dim(ker A + ker A^T)leq 2015$)." Ok but how from dimensions theorem I get this proposal? Dimension theorem says that $dim(im(T)) + dim(kernel(T)) = n $, but there, If we assume that T is $ ker A + ker A^T$ I haven't got neither dim(im(T)) nor dim(kernel(T))
            $endgroup$
            – VirtualUser
            Dec 30 '18 at 16:04












          • $begingroup$
            @VirtualUser This is the other dimension theorem. I'm talking about this $dim (U+W) = dim U + dim W - dim Ucap W$. Insert $U=ker A$ and $W=ker A^T$.
            $endgroup$
            – Yanko
            Dec 30 '18 at 16:05










          • $begingroup$
            Aaah, now I understand, you deserve a lot of thanks!
            $endgroup$
            – VirtualUser
            Dec 30 '18 at 16:07
















          $begingroup$
          Sorry if it is stupid question but I am sitting and I have no idea why $ dim(ker A + ker A^T)leq 2015$. Could I ask you for an explanation?
          $endgroup$
          – VirtualUser
          Dec 30 '18 at 15:45




          $begingroup$
          Sorry if it is stupid question but I am sitting and I have no idea why $ dim(ker A + ker A^T)leq 2015$. Could I ask you for an explanation?
          $endgroup$
          – VirtualUser
          Dec 30 '18 at 15:45












          $begingroup$
          @VirtualUser Both $ker A$ and $ker A^T$ are subspaces of $mathbb{C}^{2015}$. Therefore so is their sum.
          $endgroup$
          – Yanko
          Dec 30 '18 at 15:47






          $begingroup$
          @VirtualUser Both $ker A$ and $ker A^T$ are subspaces of $mathbb{C}^{2015}$. Therefore so is their sum.
          $endgroup$
          – Yanko
          Dec 30 '18 at 15:47














          $begingroup$
          "Therefore by the dimensions theorem we have that the dimension of $ker A cap ker A^T$ is at least $15$ (because $dim(ker A + ker A^T)leq 2015$)." Ok but how from dimensions theorem I get this proposal? Dimension theorem says that $dim(im(T)) + dim(kernel(T)) = n $, but there, If we assume that T is $ ker A + ker A^T$ I haven't got neither dim(im(T)) nor dim(kernel(T))
          $endgroup$
          – VirtualUser
          Dec 30 '18 at 16:04






          $begingroup$
          "Therefore by the dimensions theorem we have that the dimension of $ker A cap ker A^T$ is at least $15$ (because $dim(ker A + ker A^T)leq 2015$)." Ok but how from dimensions theorem I get this proposal? Dimension theorem says that $dim(im(T)) + dim(kernel(T)) = n $, but there, If we assume that T is $ ker A + ker A^T$ I haven't got neither dim(im(T)) nor dim(kernel(T))
          $endgroup$
          – VirtualUser
          Dec 30 '18 at 16:04














          $begingroup$
          @VirtualUser This is the other dimension theorem. I'm talking about this $dim (U+W) = dim U + dim W - dim Ucap W$. Insert $U=ker A$ and $W=ker A^T$.
          $endgroup$
          – Yanko
          Dec 30 '18 at 16:05




          $begingroup$
          @VirtualUser This is the other dimension theorem. I'm talking about this $dim (U+W) = dim U + dim W - dim Ucap W$. Insert $U=ker A$ and $W=ker A^T$.
          $endgroup$
          – Yanko
          Dec 30 '18 at 16:05












          $begingroup$
          Aaah, now I understand, you deserve a lot of thanks!
          $endgroup$
          – VirtualUser
          Dec 30 '18 at 16:07




          $begingroup$
          Aaah, now I understand, you deserve a lot of thanks!
          $endgroup$
          – VirtualUser
          Dec 30 '18 at 16:07











          2












          $begingroup$

          Hint: $operatorname{dim}(operatorname {im}(A+A^t)=operatorname{dim}(operatorname{im}A)+operatorname{dim}(operatorname{im}A^t)-operatorname{dim}(operatorname{im}Acapoperatorname{im}A^t)lt2000$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            No, using the hypothesis, less than $1000+1000=2000$.
            $endgroup$
            – Chris Custer
            Dec 30 '18 at 15:43
















          2












          $begingroup$

          Hint: $operatorname{dim}(operatorname {im}(A+A^t)=operatorname{dim}(operatorname{im}A)+operatorname{dim}(operatorname{im}A^t)-operatorname{dim}(operatorname{im}Acapoperatorname{im}A^t)lt2000$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            No, using the hypothesis, less than $1000+1000=2000$.
            $endgroup$
            – Chris Custer
            Dec 30 '18 at 15:43














          2












          2








          2





          $begingroup$

          Hint: $operatorname{dim}(operatorname {im}(A+A^t)=operatorname{dim}(operatorname{im}A)+operatorname{dim}(operatorname{im}A^t)-operatorname{dim}(operatorname{im}Acapoperatorname{im}A^t)lt2000$.






          share|cite|improve this answer









          $endgroup$



          Hint: $operatorname{dim}(operatorname {im}(A+A^t)=operatorname{dim}(operatorname{im}A)+operatorname{dim}(operatorname{im}A^t)-operatorname{dim}(operatorname{im}Acapoperatorname{im}A^t)lt2000$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 30 '18 at 15:34









          Chris CusterChris Custer

          14.4k3827




          14.4k3827












          • $begingroup$
            No, using the hypothesis, less than $1000+1000=2000$.
            $endgroup$
            – Chris Custer
            Dec 30 '18 at 15:43


















          • $begingroup$
            No, using the hypothesis, less than $1000+1000=2000$.
            $endgroup$
            – Chris Custer
            Dec 30 '18 at 15:43
















          $begingroup$
          No, using the hypothesis, less than $1000+1000=2000$.
          $endgroup$
          – Chris Custer
          Dec 30 '18 at 15:43




          $begingroup$
          No, using the hypothesis, less than $1000+1000=2000$.
          $endgroup$
          – Chris Custer
          Dec 30 '18 at 15:43


















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