Show $(ev_alpha)_{alpha∈mathbb{R}}$ is linearly independent
Show $(ev_alpha)_{alpha∈mathbb{R}}$ is linearly independent
$(ev_alpha): mathbb{R}[X]rightarrowmathbb{R}, fmapsto f(alpha)$
So I need to show, maybe with a counter-example, that $lambda=0$ in $sumlambda ev_alpha=0$
How can I do it?
linear-algebra
edited Nov 21 '18 at 20:49
John B
12.2k51840
asked Nov 21 '18 at 20:47
Dada
7010
add a comment |
Show $(ev_alpha)_{alpha∈mathbb{R}}$ is linearly independent
$(ev_alpha): mathbb{R}[X]rightarrowmathbb{R}, fmapsto f(alpha)$
So I need to show, maybe with a counter-example, that $lambda=0$ in $sumlambda ev_alpha=0$
How can I do it?
linear-algebra
edited Nov 21 '18 at 20:49
John B
12.2k51840
asked Nov 21 '18 at 20:47
Dada
7010
1
It's going to be difficult: how come you want to give a counterexample of something that you want to prove? PS: You need to replace $lambda$ by $lambda_alpha$.
– John B
Nov 21 '18 at 20:51
add a comment |
Show $(ev_alpha)_{alpha∈mathbb{R}}$ is linearly independent
$(ev_alpha): mathbb{R}[X]rightarrowmathbb{R}, fmapsto f(alpha)$
So I need to show, maybe with a counter-example, that $lambda=0$ in $sumlambda ev_alpha=0$
How can I do it?
linear-algebra
edited Nov 21 '18 at 20:49
John B
12.2k51840
asked Nov 21 '18 at 20:47
Dada
7010
Show $(ev_alpha)_{alpha∈mathbb{R}}$ is linearly independent
$(ev_alpha): mathbb{R}[X]rightarrowmathbb{R}, fmapsto f(alpha)$
So I need to show, maybe with a counter-example, that $lambda=0$ in $sumlambda ev_alpha=0$
How can I do it?
linear-algebra
linear-algebra
edited Nov 21 '18 at 20:49
John B
12.2k51840
asked Nov 21 '18 at 20:47
Dada
7010
edited Nov 21 '18 at 20:49
John B
12.2k51840
asked Nov 21 '18 at 20:47
Dada
7010
edited Nov 21 '18 at 20:49
John B
12.2k51840
edited Nov 21 '18 at 20:49
John B
12.2k51840
edited Nov 21 '18 at 20:49
John B
12.2k51840
12.2k51840
asked Nov 21 '18 at 20:47
Dada
7010
asked Nov 21 '18 at 20:47
Dada
7010
asked Nov 21 '18 at 20:47
Dada
7010
7010
1
It's going to be difficult: how come you want to give a counterexample of something that you want to prove? PS: You need to replace $lambda$ by $lambda_alpha$.
– John B
Nov 21 '18 at 20:51
add a comment |
1
It's going to be difficult: how come you want to give a counterexample of something that you want to prove? PS: You need to replace $lambda$ by $lambda_alpha$.
– John B
Nov 21 '18 at 20:51
1
1
It's going to be difficult: how come you want to give a counterexample of something that you want to prove? PS: You need to replace $lambda$ by $lambda_alpha$.
– John B
Nov 21 '18 at 20:51
It's going to be difficult: how come you want to give a counterexample of something that you want to prove? PS: You need to replace $lambda$ by $lambda_alpha$.
– John B
Nov 21 '18 at 20:51
add a comment |
1 Answer
1
active
oldest
votes
An uncountable set is linearly independent if and only if every finite subset is linearly independent.
So suppose $U subset mathbb{R}$ is finite and there exists $lambda_alpha$ for each $alpha in U$ such that $ displaystyle sum_{alpha in U} lambda_alpha ev_alpha = 0$.
Then by definition of $ev_alpha$ we have $displaystyle sum_{alpha in U} lambda_alpha f(alpha) = 0$ for any $f in mathbb{R}[X]$.
Choose some $alpha' in U$ and define $displaystyle f_{alpha'}(x) =prod_{alpha in U - { alpha' }}(x-alpha)$.
Then by construction we have $f_{alpha'}(alpha') neq 0$, but $f_{alpha'}(alpha) = 0$ for all $alpha in U - {alpha'}$.
Then $displaystyle 0= sum_{alpha in U} lambda_alpha f_{alpha'}(alpha) = lambda_{alpha'}f_{alpha'}(alpha')$ and since $f_{alpha'}(alpha')neq 0$ we have $lambda_{alpha'}=0$.
Since our choice of $alpha'$ was arbitrary, we have $lambda_alpha=0$ for all $alpha in U$.
Thus the finite subset $(ev_alpha)_{alpha in U}$ is linearly independent.
And since our choice of $U$ was arbitrary, every finite subset is linearly independent.
Therefore $(ev_alpha)_{alpha in mathbb{R}}$ is linearly independent.
answered Nov 27 '18 at 23:29
sfmiller940
3615
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
oldest
votes
An uncountable set is linearly independent if and only if every finite subset is linearly independent.
So suppose $U subset mathbb{R}$ is finite and there exists $lambda_alpha$ for each $alpha in U$ such that $ displaystyle sum_{alpha in U} lambda_alpha ev_alpha = 0$.
Then by definition of $ev_alpha$ we have $displaystyle sum_{alpha in U} lambda_alpha f(alpha) = 0$ for any $f in mathbb{R}[X]$.
Choose some $alpha' in U$ and define $displaystyle f_{alpha'}(x) =prod_{alpha in U - { alpha' }}(x-alpha)$.
Then by construction we have $f_{alpha'}(alpha') neq 0$, but $f_{alpha'}(alpha) = 0$ for all $alpha in U - {alpha'}$.
Then $displaystyle 0= sum_{alpha in U} lambda_alpha f_{alpha'}(alpha) = lambda_{alpha'}f_{alpha'}(alpha')$ and since $f_{alpha'}(alpha')neq 0$ we have $lambda_{alpha'}=0$.
Since our choice of $alpha'$ was arbitrary, we have $lambda_alpha=0$ for all $alpha in U$.
Thus the finite subset $(ev_alpha)_{alpha in U}$ is linearly independent.
And since our choice of $U$ was arbitrary, every finite subset is linearly independent.
Therefore $(ev_alpha)_{alpha in mathbb{R}}$ is linearly independent.
answered Nov 27 '18 at 23:29
sfmiller940
3615
add a comment |
An uncountable set is linearly independent if and only if every finite subset is linearly independent.
So suppose $U subset mathbb{R}$ is finite and there exists $lambda_alpha$ for each $alpha in U$ such that $ displaystyle sum_{alpha in U} lambda_alpha ev_alpha = 0$.
Then by definition of $ev_alpha$ we have $displaystyle sum_{alpha in U} lambda_alpha f(alpha) = 0$ for any $f in mathbb{R}[X]$.
Choose some $alpha' in U$ and define $displaystyle f_{alpha'}(x) =prod_{alpha in U - { alpha' }}(x-alpha)$.
Then by construction we have $f_{alpha'}(alpha') neq 0$, but $f_{alpha'}(alpha) = 0$ for all $alpha in U - {alpha'}$.
Then $displaystyle 0= sum_{alpha in U} lambda_alpha f_{alpha'}(alpha) = lambda_{alpha'}f_{alpha'}(alpha')$ and since $f_{alpha'}(alpha')neq 0$ we have $lambda_{alpha'}=0$.
Since our choice of $alpha'$ was arbitrary, we have $lambda_alpha=0$ for all $alpha in U$.
Thus the finite subset $(ev_alpha)_{alpha in U}$ is linearly independent.
And since our choice of $U$ was arbitrary, every finite subset is linearly independent.
Therefore $(ev_alpha)_{alpha in mathbb{R}}$ is linearly independent.
answered Nov 27 '18 at 23:29
sfmiller940
3615
add a comment |
An uncountable set is linearly independent if and only if every finite subset is linearly independent.
So suppose $U subset mathbb{R}$ is finite and there exists $lambda_alpha$ for each $alpha in U$ such that $ displaystyle sum_{alpha in U} lambda_alpha ev_alpha = 0$.
Then by definition of $ev_alpha$ we have $displaystyle sum_{alpha in U} lambda_alpha f(alpha) = 0$ for any $f in mathbb{R}[X]$.
Choose some $alpha' in U$ and define $displaystyle f_{alpha'}(x) =prod_{alpha in U - { alpha' }}(x-alpha)$.
Then by construction we have $f_{alpha'}(alpha') neq 0$, but $f_{alpha'}(alpha) = 0$ for all $alpha in U - {alpha'}$.
Then $displaystyle 0= sum_{alpha in U} lambda_alpha f_{alpha'}(alpha) = lambda_{alpha'}f_{alpha'}(alpha')$ and since $f_{alpha'}(alpha')neq 0$ we have $lambda_{alpha'}=0$.
Since our choice of $alpha'$ was arbitrary, we have $lambda_alpha=0$ for all $alpha in U$.
Thus the finite subset $(ev_alpha)_{alpha in U}$ is linearly independent.
And since our choice of $U$ was arbitrary, every finite subset is linearly independent.
Therefore $(ev_alpha)_{alpha in mathbb{R}}$ is linearly independent.
answered Nov 27 '18 at 23:29
sfmiller940
3615
An uncountable set is linearly independent if and only if every finite subset is linearly independent.
So suppose $U subset mathbb{R}$ is finite and there exists $lambda_alpha$ for each $alpha in U$ such that $ displaystyle sum_{alpha in U} lambda_alpha ev_alpha = 0$.
Then by definition of $ev_alpha$ we have $displaystyle sum_{alpha in U} lambda_alpha f(alpha) = 0$ for any $f in mathbb{R}[X]$.
Choose some $alpha' in U$ and define $displaystyle f_{alpha'}(x) =prod_{alpha in U - { alpha' }}(x-alpha)$.
Then by construction we have $f_{alpha'}(alpha') neq 0$, but $f_{alpha'}(alpha) = 0$ for all $alpha in U - {alpha'}$.
Then $displaystyle 0= sum_{alpha in U} lambda_alpha f_{alpha'}(alpha) = lambda_{alpha'}f_{alpha'}(alpha')$ and since $f_{alpha'}(alpha')neq 0$ we have $lambda_{alpha'}=0$.
Since our choice of $alpha'$ was arbitrary, we have $lambda_alpha=0$ for all $alpha in U$.
Thus the finite subset $(ev_alpha)_{alpha in U}$ is linearly independent.
And since our choice of $U$ was arbitrary, every finite subset is linearly independent.
Therefore $(ev_alpha)_{alpha in mathbb{R}}$ is linearly independent.
answered Nov 27 '18 at 23:29
sfmiller940
3615
answered Nov 27 '18 at 23:29
sfmiller940
3615
answered Nov 27 '18 at 23:29
sfmiller940
3615
answered Nov 27 '18 at 23:29
sfmiller940
3615
3615
add a comment |
add a comment |
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It's going to be difficult: how come you want to give a counterexample of something that you want to prove? PS: You need to replace $lambda$ by $lambda_alpha$.
– John B
Nov 21 '18 at 20:51