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.
calculus real-analysis
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up vote
0
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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.
calculus real-analysis
add a comment |
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.
calculus real-analysis
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
calculus real-analysis
asked Nov 16 at 11:57
Sar
44012
44012
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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 $a$.
That's just a great idea. Thank you !
– Sar
Nov 16 at 14:35
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 $a$.
That's just a great idea. Thank you !
– Sar
Nov 16 at 14:35
add a comment |
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 $a$.
That's just a great idea. Thank you !
– Sar
Nov 16 at 14:35
add a comment |
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 $a$.
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 $a$.
answered Nov 16 at 13:09
Tsemo Aristide
54.7k11444
54.7k11444
That's just a great idea. Thank you !
– Sar
Nov 16 at 14:35
add a comment |
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
add a comment |
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