Mixing
Let$~${$ f_1 , f_2 , .. , f_n$} be functions lineary indepenedante in $ I$ Prove that every Subset …
$begingroup$
Let$~${$ f_1 , f_2 , .. , f_n$} be functions lineary indepenedante in $ I$ Prove that every Subset of the group is also lineary independante in $ I $ ( interval).
how do i prove such claim with differential equations ? say with the wronsekian .
here is my prof ( might not be right ) :
lets suppose that there exist subset such that
without the loss of generality {$ f_1 , f_2 , .. , f_k$} , $k<n$ dependant.
so $ sum_i^k alpha_i^*f_i = 0$ such that $alpha_1^* ,alpha_2^*,alpha_3^*,...,alpha_k^* $ not all $0$ ,suppose at least $alpha_1^* not= 0$.
so there exist $alpha_1^* ,alpha_2^*,alpha_3^*,...,alpha_k^*,0,0,0,0,...,0 $ not all $0$ such that :
$ sum_1^n alpha_if_i = sum_1^k alpha_i^*f_i + 0f_{k+1}+0f_{k+2}+...+0f_{n} = 0$
so its lineary dependant but we know that its not. QED
functional-analysis ordinary-differential-equations
asked Jan 2 at 18:08
Mather Mather
1877
$endgroup$
add a comment |
$begingroup$
Let$~${$ f_1 , f_2 , .. , f_n$} be functions lineary indepenedante in $ I$ Prove that every Subset of the group is also lineary independante in $ I $ ( interval).
how do i prove such claim with differential equations ? say with the wronsekian .
here is my prof ( might not be right ) :
lets suppose that there exist subset such that
without the loss of generality {$ f_1 , f_2 , .. , f_k$} , $k<n$ dependant.
so $ sum_i^k alpha_i^*f_i = 0$ such that $alpha_1^* ,alpha_2^*,alpha_3^*,...,alpha_k^* $ not all $0$ ,suppose at least $alpha_1^* not= 0$.
so there exist $alpha_1^* ,alpha_2^*,alpha_3^*,...,alpha_k^*,0,0,0,0,...,0 $ not all $0$ such that :
$ sum_1^n alpha_if_i = sum_1^k alpha_i^*f_i + 0f_{k+1}+0f_{k+2}+...+0f_{n} = 0$
so its lineary dependant but we know that its not. QED
functional-analysis ordinary-differential-equations
asked Jan 2 at 18:08
Mather Mather
1877
$endgroup$
3
$begingroup$
Your proof is correct.
$endgroup$
– John Douma
Jan 2 at 18:40
add a comment |
$begingroup$
Let$~${$ f_1 , f_2 , .. , f_n$} be functions lineary indepenedante in $ I$ Prove that every Subset of the group is also lineary independante in $ I $ ( interval).
how do i prove such claim with differential equations ? say with the wronsekian .
here is my prof ( might not be right ) :
lets suppose that there exist subset such that
without the loss of generality {$ f_1 , f_2 , .. , f_k$} , $k<n$ dependant.
so $ sum_i^k alpha_i^*f_i = 0$ such that $alpha_1^* ,alpha_2^*,alpha_3^*,...,alpha_k^* $ not all $0$ ,suppose at least $alpha_1^* not= 0$.
so there exist $alpha_1^* ,alpha_2^*,alpha_3^*,...,alpha_k^*,0,0,0,0,...,0 $ not all $0$ such that :
$ sum_1^n alpha_if_i = sum_1^k alpha_i^*f_i + 0f_{k+1}+0f_{k+2}+...+0f_{n} = 0$
so its lineary dependant but we know that its not. QED
functional-analysis ordinary-differential-equations
asked Jan 2 at 18:08
Mather Mather
1877
$endgroup$
Let$~${$ f_1 , f_2 , .. , f_n$} be functions lineary indepenedante in $ I$ Prove that every Subset of the group is also lineary independante in $ I $ ( interval).
how do i prove such claim with differential equations ? say with the wronsekian .
here is my prof ( might not be right ) :
lets suppose that there exist subset such that
without the loss of generality {$ f_1 , f_2 , .. , f_k$} , $k<n$ dependant.
so $ sum_i^k alpha_i^*f_i = 0$ such that $alpha_1^* ,alpha_2^*,alpha_3^*,...,alpha_k^* $ not all $0$ ,suppose at least $alpha_1^* not= 0$.
so there exist $alpha_1^* ,alpha_2^*,alpha_3^*,...,alpha_k^*,0,0,0,0,...,0 $ not all $0$ such that :
$ sum_1^n alpha_if_i = sum_1^k alpha_i^*f_i + 0f_{k+1}+0f_{k+2}+...+0f_{n} = 0$
so its lineary dependant but we know that its not. QED
functional-analysis ordinary-differential-equations
functional-analysis ordinary-differential-equations
asked Jan 2 at 18:08
Mather Mather
1877
asked Jan 2 at 18:08
Mather Mather
1877
edited Jan 2 at 19:02
Mather
asked Jan 2 at 18:08
Mather Mather
1877
asked Jan 2 at 18:08
Mather Mather
1877
asked Jan 2 at 18:08
Mather Mather
1877
1877
3
$begingroup$
Your proof is correct.
$endgroup$
– John Douma
Jan 2 at 18:40
add a comment |
3
$begingroup$
Your proof is correct.
$endgroup$
– John Douma
Jan 2 at 18:40
3
3
$begingroup$
Your proof is correct.
$endgroup$
– John Douma
Jan 2 at 18:40
$begingroup$
Your proof is correct.
$endgroup$
– John Douma
Jan 2 at 18:40
add a comment |
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Your proof is correct.
$endgroup$
– John Douma
Jan 2 at 18:40