Mixing
Linear independence of continuous functions in a neighborhood
Let $f_{1},f_{2}colonmathbb{R}^{n}tomathbb{R}^{m}$ be two continuous
functions and let $ainmathbb{R}^{n}$ be a point such that $f_{1}left(aright),f_{2}left(aright)$
are linearly independent. That is
$$
gamma_{1}f_{1}left(aright)+gamma_{2}f_{2}left(aright)=0qquadRightarrowqquadgamma_{1}=gamma_{2}=0
$$
I would like to know if there is also a neighborhood of $a$ such
that $f_{1},f_{2}$ are linearly independent there. That is to show
the existence of a $delta>0$ such that for all $xinmathbb{R}^{n}$
with $left|x-aright|<delta$ we get that $f_{1}left(xright),f_{2}left(xright)$
are also linearly independent. Meaning
$$
gamma_{1}f_{1}left(xright)+gamma_{2}f_{2}left(xright)=0qquadRightarrowqquadgamma_{1}=gamma_{2}=0
$$
Is this right? How can I prove it?
calculus
asked Nov 20 '18 at 12:52
Jon
557413
add a comment |
Let $f_{1},f_{2}colonmathbb{R}^{n}tomathbb{R}^{m}$ be two continuous
functions and let $ainmathbb{R}^{n}$ be a point such that $f_{1}left(aright),f_{2}left(aright)$
are linearly independent. That is
$$
gamma_{1}f_{1}left(aright)+gamma_{2}f_{2}left(aright)=0qquadRightarrowqquadgamma_{1}=gamma_{2}=0
$$
I would like to know if there is also a neighborhood of $a$ such
that $f_{1},f_{2}$ are linearly independent there. That is to show
the existence of a $delta>0$ such that for all $xinmathbb{R}^{n}$
with $left|x-aright|<delta$ we get that $f_{1}left(xright),f_{2}left(xright)$
are also linearly independent. Meaning
$$
gamma_{1}f_{1}left(xright)+gamma_{2}f_{2}left(xright)=0qquadRightarrowqquadgamma_{1}=gamma_{2}=0
$$
Is this right? How can I prove it?
calculus
asked Nov 20 '18 at 12:52
Jon
557413
add a comment |
Let $f_{1},f_{2}colonmathbb{R}^{n}tomathbb{R}^{m}$ be two continuous
functions and let $ainmathbb{R}^{n}$ be a point such that $f_{1}left(aright),f_{2}left(aright)$
are linearly independent. That is
$$
gamma_{1}f_{1}left(aright)+gamma_{2}f_{2}left(aright)=0qquadRightarrowqquadgamma_{1}=gamma_{2}=0
$$
I would like to know if there is also a neighborhood of $a$ such
that $f_{1},f_{2}$ are linearly independent there. That is to show
the existence of a $delta>0$ such that for all $xinmathbb{R}^{n}$
with $left|x-aright|<delta$ we get that $f_{1}left(xright),f_{2}left(xright)$
are also linearly independent. Meaning
$$
gamma_{1}f_{1}left(xright)+gamma_{2}f_{2}left(xright)=0qquadRightarrowqquadgamma_{1}=gamma_{2}=0
$$
Is this right? How can I prove it?
calculus
asked Nov 20 '18 at 12:52
Jon
557413
Let $f_{1},f_{2}colonmathbb{R}^{n}tomathbb{R}^{m}$ be two continuous
functions and let $ainmathbb{R}^{n}$ be a point such that $f_{1}left(aright),f_{2}left(aright)$
are linearly independent. That is
$$
gamma_{1}f_{1}left(aright)+gamma_{2}f_{2}left(aright)=0qquadRightarrowqquadgamma_{1}=gamma_{2}=0
$$
I would like to know if there is also a neighborhood of $a$ such
that $f_{1},f_{2}$ are linearly independent there. That is to show
the existence of a $delta>0$ such that for all $xinmathbb{R}^{n}$
with $left|x-aright|<delta$ we get that $f_{1}left(xright),f_{2}left(xright)$
are also linearly independent. Meaning
$$
gamma_{1}f_{1}left(xright)+gamma_{2}f_{2}left(xright)=0qquadRightarrowqquadgamma_{1}=gamma_{2}=0
$$
Is this right? How can I prove it?
calculus
calculus
asked Nov 20 '18 at 12:52
Jon
557413
asked Nov 20 '18 at 12:52
Jon
557413
asked Nov 20 '18 at 12:52
Jon
557413
asked Nov 20 '18 at 12:52
Jon
557413
asked Nov 20 '18 at 12:52
Jon
557413
557413
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
If $m = 2$:
consider $$g(x)=det(f_1(x),f_2(x))$$
Since $g(a) ne 0$, $exists epsilon > 0$ s.t. $0 notin I=]g(a)-epsilon, g(a)+epsilon[$
Since $I$ is an open set of $mathbb{R}$ and $g$ is continuous, $g^{-1}(I)$ is an open set of $mathbb{R}^n$
$$forall x in g^{-1}(I), g(x)ne 0 iff f_1(x), f_2(x) text{ independent}$$
If $m = 1$:
$f_1(a)$ and $f_2(a)$ cannot be independent
If $m > 2$:
There exists $(e_2, .., e_m)$ such that $(f_1(a), f_2(a), e_2, .., e_m)$ is a base of $mathbb{R^m}$
Consider $$g(x)=det(f_1(x),f_2(x), e_2, .., e_m)$$ and apply the same processus.
answered Nov 20 '18 at 13:45
stity
3,150513
add a comment |
Just take the Gram-Determinant of $f_1(a)$ and $f_2(a)$, that's here
$$|f_1(a)|^2|f_2(a)|^2-langle f_1(a),f_2(a)rangle^2$$
see https://en.wikipedia.org/wiki/Gramian_matrix#Gram_determinant. If it's non-zero at $a$, due to continuity it will be non-zero in a neighbourhood of $a$.
answered Nov 20 '18 at 17:29
Michael Hoppe
10.8k31834
add a comment |
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2 Answers
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
If $m = 2$:
consider $$g(x)=det(f_1(x),f_2(x))$$
Since $g(a) ne 0$, $exists epsilon > 0$ s.t. $0 notin I=]g(a)-epsilon, g(a)+epsilon[$
Since $I$ is an open set of $mathbb{R}$ and $g$ is continuous, $g^{-1}(I)$ is an open set of $mathbb{R}^n$
$$forall x in g^{-1}(I), g(x)ne 0 iff f_1(x), f_2(x) text{ independent}$$
If $m = 1$:
$f_1(a)$ and $f_2(a)$ cannot be independent
If $m > 2$:
There exists $(e_2, .., e_m)$ such that $(f_1(a), f_2(a), e_2, .., e_m)$ is a base of $mathbb{R^m}$
Consider $$g(x)=det(f_1(x),f_2(x), e_2, .., e_m)$$ and apply the same processus.
answered Nov 20 '18 at 13:45
stity
3,150513
add a comment |
If $m = 2$:
consider $$g(x)=det(f_1(x),f_2(x))$$
Since $g(a) ne 0$, $exists epsilon > 0$ s.t. $0 notin I=]g(a)-epsilon, g(a)+epsilon[$
Since $I$ is an open set of $mathbb{R}$ and $g$ is continuous, $g^{-1}(I)$ is an open set of $mathbb{R}^n$
$$forall x in g^{-1}(I), g(x)ne 0 iff f_1(x), f_2(x) text{ independent}$$
If $m = 1$:
$f_1(a)$ and $f_2(a)$ cannot be independent
If $m > 2$:
There exists $(e_2, .., e_m)$ such that $(f_1(a), f_2(a), e_2, .., e_m)$ is a base of $mathbb{R^m}$
Consider $$g(x)=det(f_1(x),f_2(x), e_2, .., e_m)$$ and apply the same processus.
answered Nov 20 '18 at 13:45
stity
3,150513
add a comment |
If $m = 2$:
consider $$g(x)=det(f_1(x),f_2(x))$$
Since $g(a) ne 0$, $exists epsilon > 0$ s.t. $0 notin I=]g(a)-epsilon, g(a)+epsilon[$
Since $I$ is an open set of $mathbb{R}$ and $g$ is continuous, $g^{-1}(I)$ is an open set of $mathbb{R}^n$
$$forall x in g^{-1}(I), g(x)ne 0 iff f_1(x), f_2(x) text{ independent}$$
If $m = 1$:
$f_1(a)$ and $f_2(a)$ cannot be independent
If $m > 2$:
There exists $(e_2, .., e_m)$ such that $(f_1(a), f_2(a), e_2, .., e_m)$ is a base of $mathbb{R^m}$
Consider $$g(x)=det(f_1(x),f_2(x), e_2, .., e_m)$$ and apply the same processus.
answered Nov 20 '18 at 13:45
stity
3,150513
If $m = 2$:
consider $$g(x)=det(f_1(x),f_2(x))$$
Since $g(a) ne 0$, $exists epsilon > 0$ s.t. $0 notin I=]g(a)-epsilon, g(a)+epsilon[$
Since $I$ is an open set of $mathbb{R}$ and $g$ is continuous, $g^{-1}(I)$ is an open set of $mathbb{R}^n$
$$forall x in g^{-1}(I), g(x)ne 0 iff f_1(x), f_2(x) text{ independent}$$
If $m = 1$:
$f_1(a)$ and $f_2(a)$ cannot be independent
If $m > 2$:
There exists $(e_2, .., e_m)$ such that $(f_1(a), f_2(a), e_2, .., e_m)$ is a base of $mathbb{R^m}$
Consider $$g(x)=det(f_1(x),f_2(x), e_2, .., e_m)$$ and apply the same processus.
answered Nov 20 '18 at 13:45
stity
3,150513
answered Nov 20 '18 at 13:45
stity
3,150513
answered Nov 20 '18 at 13:45
stity
3,150513
answered Nov 20 '18 at 13:45
stity
3,150513
3,150513
add a comment |
add a comment |
Just take the Gram-Determinant of $f_1(a)$ and $f_2(a)$, that's here
$$|f_1(a)|^2|f_2(a)|^2-langle f_1(a),f_2(a)rangle^2$$
see https://en.wikipedia.org/wiki/Gramian_matrix#Gram_determinant. If it's non-zero at $a$, due to continuity it will be non-zero in a neighbourhood of $a$.
answered Nov 20 '18 at 17:29
Michael Hoppe
10.8k31834
add a comment |
Just take the Gram-Determinant of $f_1(a)$ and $f_2(a)$, that's here
$$|f_1(a)|^2|f_2(a)|^2-langle f_1(a),f_2(a)rangle^2$$
see https://en.wikipedia.org/wiki/Gramian_matrix#Gram_determinant. If it's non-zero at $a$, due to continuity it will be non-zero in a neighbourhood of $a$.
answered Nov 20 '18 at 17:29
Michael Hoppe
10.8k31834
add a comment |
Just take the Gram-Determinant of $f_1(a)$ and $f_2(a)$, that's here
$$|f_1(a)|^2|f_2(a)|^2-langle f_1(a),f_2(a)rangle^2$$
see https://en.wikipedia.org/wiki/Gramian_matrix#Gram_determinant. If it's non-zero at $a$, due to continuity it will be non-zero in a neighbourhood of $a$.
answered Nov 20 '18 at 17:29
Michael Hoppe
10.8k31834
Just take the Gram-Determinant of $f_1(a)$ and $f_2(a)$, that's here
$$|f_1(a)|^2|f_2(a)|^2-langle f_1(a),f_2(a)rangle^2$$
see https://en.wikipedia.org/wiki/Gramian_matrix#Gram_determinant. If it's non-zero at $a$, due to continuity it will be non-zero in a neighbourhood of $a$.
answered Nov 20 '18 at 17:29
Michael Hoppe
10.8k31834
answered Nov 20 '18 at 17:29
Michael Hoppe
10.8k31834
answered Nov 20 '18 at 17:29
Michael Hoppe
10.8k31834
answered Nov 20 '18 at 17:29
Michael Hoppe
10.8k31834
10.8k31834
add a comment |
add a comment |
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