Rank of differential at a given point is a local minimum











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I have the following exercise I`m struggling with :



Let $fin C^1(mathbb{R}^n,mathbb{R}^m)$, and $a in mathbb{R}^n$, I need to show that $rk(D_f(x))geq rk(D_f(a))$ for some neighbourhood of $a$.



I first tried looking at non-trivial linear combinations of $nabla f_i(x)$ but it didnt get me anywhere.



I also tried defining $Ker_x = {vin mathbb{R}^n | D_f(x)v=0}$, and tried showing $Ker_xsubset Ker_a$:



Let $xin B(a,delta$), and $vin Ker_x$,



$$|D_f(a)v|=|D_f(a)v-D_f(x)v+D_f(x)v|leq|D_f(a)v-D_f(x)v| +|D_f(x)v|=|D_f(a)v-D_f(x)v|leqepsilon|v|$$



Where the last inequality is from the continuity of $D_f(x)$.The problem is however, $delta$ is not fixed if I want $|D_f(a)v|=0$ so I can`t see how to find a neighbourhood of $a$ out of these inequalities.



I`ll be glad if anyone can either help me fix my solution or suggest another way of approching this problem.










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    I have the following exercise I`m struggling with :



    Let $fin C^1(mathbb{R}^n,mathbb{R}^m)$, and $a in mathbb{R}^n$, I need to show that $rk(D_f(x))geq rk(D_f(a))$ for some neighbourhood of $a$.



    I first tried looking at non-trivial linear combinations of $nabla f_i(x)$ but it didnt get me anywhere.



    I also tried defining $Ker_x = {vin mathbb{R}^n | D_f(x)v=0}$, and tried showing $Ker_xsubset Ker_a$:



    Let $xin B(a,delta$), and $vin Ker_x$,



    $$|D_f(a)v|=|D_f(a)v-D_f(x)v+D_f(x)v|leq|D_f(a)v-D_f(x)v| +|D_f(x)v|=|D_f(a)v-D_f(x)v|leqepsilon|v|$$



    Where the last inequality is from the continuity of $D_f(x)$.The problem is however, $delta$ is not fixed if I want $|D_f(a)v|=0$ so I can`t see how to find a neighbourhood of $a$ out of these inequalities.



    I`ll be glad if anyone can either help me fix my solution or suggest another way of approching this problem.










    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I have the following exercise I`m struggling with :



      Let $fin C^1(mathbb{R}^n,mathbb{R}^m)$, and $a in mathbb{R}^n$, I need to show that $rk(D_f(x))geq rk(D_f(a))$ for some neighbourhood of $a$.



      I first tried looking at non-trivial linear combinations of $nabla f_i(x)$ but it didnt get me anywhere.



      I also tried defining $Ker_x = {vin mathbb{R}^n | D_f(x)v=0}$, and tried showing $Ker_xsubset Ker_a$:



      Let $xin B(a,delta$), and $vin Ker_x$,



      $$|D_f(a)v|=|D_f(a)v-D_f(x)v+D_f(x)v|leq|D_f(a)v-D_f(x)v| +|D_f(x)v|=|D_f(a)v-D_f(x)v|leqepsilon|v|$$



      Where the last inequality is from the continuity of $D_f(x)$.The problem is however, $delta$ is not fixed if I want $|D_f(a)v|=0$ so I can`t see how to find a neighbourhood of $a$ out of these inequalities.



      I`ll be glad if anyone can either help me fix my solution or suggest another way of approching this problem.










      share|cite|improve this question













      I have the following exercise I`m struggling with :



      Let $fin C^1(mathbb{R}^n,mathbb{R}^m)$, and $a in mathbb{R}^n$, I need to show that $rk(D_f(x))geq rk(D_f(a))$ for some neighbourhood of $a$.



      I first tried looking at non-trivial linear combinations of $nabla f_i(x)$ but it didnt get me anywhere.



      I also tried defining $Ker_x = {vin mathbb{R}^n | D_f(x)v=0}$, and tried showing $Ker_xsubset Ker_a$:



      Let $xin B(a,delta$), and $vin Ker_x$,



      $$|D_f(a)v|=|D_f(a)v-D_f(x)v+D_f(x)v|leq|D_f(a)v-D_f(x)v| +|D_f(x)v|=|D_f(a)v-D_f(x)v|leqepsilon|v|$$



      Where the last inequality is from the continuity of $D_f(x)$.The problem is however, $delta$ is not fixed if I want $|D_f(a)v|=0$ so I can`t see how to find a neighbourhood of $a$ out of these inequalities.



      I`ll be glad if anyone can either help me fix my solution or suggest another way of approching this problem.







      calculus real-analysis






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      asked Nov 16 at 11:57









      Sar

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      48811






















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          If $rk(D_f(a))=p$, this is equivalent to saying that there exists a $ptimes p$-minor $A_p(x)$ (submatrix) of $D_f(x)$ such that $g(a)=det(A_p(a))neq 0$, since $f$ is $C^1$, $D_f$ and henceforth $g$ are continuous,.Let $I$ be an open interval containing $det(A_p(a)$, but not $0$, $g^{-1}(I)$ is an open subset containing $a$.






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          • That's just a great idea. Thank you !
            – Sar
            Nov 16 at 14:35











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          accepted










          If $rk(D_f(a))=p$, this is equivalent to saying that there exists a $ptimes p$-minor $A_p(x)$ (submatrix) of $D_f(x)$ such that $g(a)=det(A_p(a))neq 0$, since $f$ is $C^1$, $D_f$ and henceforth $g$ are continuous,.Let $I$ be an open interval containing $det(A_p(a)$, but not $0$, $g^{-1}(I)$ is an open subset containing $a$.






          share|cite|improve this answer





















          • That's just a great idea. Thank you !
            – Sar
            Nov 16 at 14:35















          up vote
          1
          down vote



          accepted










          If $rk(D_f(a))=p$, this is equivalent to saying that there exists a $ptimes p$-minor $A_p(x)$ (submatrix) of $D_f(x)$ such that $g(a)=det(A_p(a))neq 0$, since $f$ is $C^1$, $D_f$ and henceforth $g$ are continuous,.Let $I$ be an open interval containing $det(A_p(a)$, but not $0$, $g^{-1}(I)$ is an open subset containing $a$.






          share|cite|improve this answer





















          • That's just a great idea. Thank you !
            – Sar
            Nov 16 at 14:35













          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          If $rk(D_f(a))=p$, this is equivalent to saying that there exists a $ptimes p$-minor $A_p(x)$ (submatrix) of $D_f(x)$ such that $g(a)=det(A_p(a))neq 0$, since $f$ is $C^1$, $D_f$ and henceforth $g$ are continuous,.Let $I$ be an open interval containing $det(A_p(a)$, but not $0$, $g^{-1}(I)$ is an open subset containing $a$.






          share|cite|improve this answer












          If $rk(D_f(a))=p$, this is equivalent to saying that there exists a $ptimes p$-minor $A_p(x)$ (submatrix) of $D_f(x)$ such that $g(a)=det(A_p(a))neq 0$, since $f$ is $C^1$, $D_f$ and henceforth $g$ are continuous,.Let $I$ be an open interval containing $det(A_p(a)$, but not $0$, $g^{-1}(I)$ is an open subset containing $a$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 16 at 13:09









          Tsemo Aristide

          54.4k11344




          54.4k11344












          • That's just a great idea. Thank you !
            – Sar
            Nov 16 at 14:35


















          • That's just a great idea. Thank you !
            – Sar
            Nov 16 at 14:35
















          That's just a great idea. Thank you !
          – Sar
          Nov 16 at 14:35




          That's just a great idea. Thank you !
          – Sar
          Nov 16 at 14:35


















           

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