Show that $min M$ is unique












0












$begingroup$


Let $S$ be a subspace of a normed linear space, $X$ and $x_0in Xbackslash S$. Consider the subspace spanned by $M,$ i.e.
begin{align} M:=[Scup {x_0}]={m=x+alpha,x_0:,xin S,;text{for some};alphainBbb{R}} end{align}
I want to show that $m$ is unique.



MY TRIAL



Let $min M,$ then there exists $alphainBbb{R} $ such that $m=x+alpha,x_0.$ Suppose we have another representation, then there exists $betain Bbb{R},;betaneq alpha,$ such that $m=x+beta,x_0.$ Thus,
begin{align} x_0=0 ;text{which implies};x_0in S,;text{contradiction, since }; x_0notin S.end{align}
Please, I'm I right? If not, could you please, provide an alternative proof?






linear-algebra functional-analysis analysis proof-writing proof-explanation







share|cite|improve this question











$endgroup$












  • $begingroup$
    I'm not 100% sure I understand the question. As I understand it, you're trying to show that, if you have $x + alpha x_0 = y + beta x_0$, where $x, y in S$, then $x = y$ and $alpha = beta$?
    $endgroup$
    – Theo Bendit
    Jan 10 at 5:01










  • $begingroup$
    @Theo Bendit: That's true! That almost shows that I'm not correct!
    $endgroup$
    – Omojola Micheal
    Jan 10 at 5:02










  • $begingroup$
    You're almost correct; instead consider $x - y = (beta - alpha)x_0$.
    $endgroup$
    – Theo Bendit
    Jan 10 at 5:03










  • $begingroup$
    @Theo Bendit: Okay, let me try that!
    $endgroup$
    – Omojola Micheal
    Jan 10 at 5:04










  • $begingroup$
    If you get, write an answer. :-)
    $endgroup$
    – Theo Bendit
    Jan 10 at 5:05
















0












$begingroup$


Let $S$ be a subspace of a normed linear space, $X$ and $x_0in Xbackslash S$. Consider the subspace spanned by $M,$ i.e.
begin{align} M:=[Scup {x_0}]={m=x+alpha,x_0:,xin S,;text{for some};alphainBbb{R}} end{align}
I want to show that $m$ is unique.



MY TRIAL



Let $min M,$ then there exists $alphainBbb{R} $ such that $m=x+alpha,x_0.$ Suppose we have another representation, then there exists $betain Bbb{R},;betaneq alpha,$ such that $m=x+beta,x_0.$ Thus,
begin{align} x_0=0 ;text{which implies};x_0in S,;text{contradiction, since }; x_0notin S.end{align}
Please, I'm I right? If not, could you please, provide an alternative proof?






linear-algebra functional-analysis analysis proof-writing proof-explanation







share|cite|improve this question











$endgroup$












  • $begingroup$
    I'm not 100% sure I understand the question. As I understand it, you're trying to show that, if you have $x + alpha x_0 = y + beta x_0$, where $x, y in S$, then $x = y$ and $alpha = beta$?
    $endgroup$
    – Theo Bendit
    Jan 10 at 5:01










  • $begingroup$
    @Theo Bendit: That's true! That almost shows that I'm not correct!
    $endgroup$
    – Omojola Micheal
    Jan 10 at 5:02










  • $begingroup$
    You're almost correct; instead consider $x - y = (beta - alpha)x_0$.
    $endgroup$
    – Theo Bendit
    Jan 10 at 5:03










  • $begingroup$
    @Theo Bendit: Okay, let me try that!
    $endgroup$
    – Omojola Micheal
    Jan 10 at 5:04










  • $begingroup$
    If you get, write an answer. :-)
    $endgroup$
    – Theo Bendit
    Jan 10 at 5:05














0












0








0


1



$begingroup$


Let $S$ be a subspace of a normed linear space, $X$ and $x_0in Xbackslash S$. Consider the subspace spanned by $M,$ i.e.
begin{align} M:=[Scup {x_0}]={m=x+alpha,x_0:,xin S,;text{for some};alphainBbb{R}} end{align}
I want to show that $m$ is unique.



MY TRIAL



Let $min M,$ then there exists $alphainBbb{R} $ such that $m=x+alpha,x_0.$ Suppose we have another representation, then there exists $betain Bbb{R},;betaneq alpha,$ such that $m=x+beta,x_0.$ Thus,
begin{align} x_0=0 ;text{which implies};x_0in S,;text{contradiction, since }; x_0notin S.end{align}
Please, I'm I right? If not, could you please, provide an alternative proof?






linear-algebra functional-analysis analysis proof-writing proof-explanation







share|cite|improve this question











$endgroup$




Let $S$ be a subspace of a normed linear space, $X$ and $x_0in Xbackslash S$. Consider the subspace spanned by $M,$ i.e.
begin{align} M:=[Scup {x_0}]={m=x+alpha,x_0:,xin S,;text{for some};alphainBbb{R}} end{align}
I want to show that $m$ is unique.



MY TRIAL



Let $min M,$ then there exists $alphainBbb{R} $ such that $m=x+alpha,x_0.$ Suppose we have another representation, then there exists $betain Bbb{R},;betaneq alpha,$ such that $m=x+beta,x_0.$ Thus,
begin{align} x_0=0 ;text{which implies};x_0in S,;text{contradiction, since }; x_0notin S.end{align}
Please, I'm I right? If not, could you please, provide an alternative proof?






linear-algebra functional-analysis analysis proof-writing proof-explanation




linear-algebra functional-analysis analysis proof-writing proof-explanation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 10 at 7:06







Omojola Micheal

















asked Jan 10 at 4:54









Omojola MichealOmojola Micheal

1,824324




1,824324












  • $begingroup$
    I'm not 100% sure I understand the question. As I understand it, you're trying to show that, if you have $x + alpha x_0 = y + beta x_0$, where $x, y in S$, then $x = y$ and $alpha = beta$?
    $endgroup$
    – Theo Bendit
    Jan 10 at 5:01










  • $begingroup$
    @Theo Bendit: That's true! That almost shows that I'm not correct!
    $endgroup$
    – Omojola Micheal
    Jan 10 at 5:02










  • $begingroup$
    You're almost correct; instead consider $x - y = (beta - alpha)x_0$.
    $endgroup$
    – Theo Bendit
    Jan 10 at 5:03










  • $begingroup$
    @Theo Bendit: Okay, let me try that!
    $endgroup$
    – Omojola Micheal
    Jan 10 at 5:04










  • $begingroup$
    If you get, write an answer. :-)
    $endgroup$
    – Theo Bendit
    Jan 10 at 5:05


















  • $begingroup$
    I'm not 100% sure I understand the question. As I understand it, you're trying to show that, if you have $x + alpha x_0 = y + beta x_0$, where $x, y in S$, then $x = y$ and $alpha = beta$?
    $endgroup$
    – Theo Bendit
    Jan 10 at 5:01










  • $begingroup$
    @Theo Bendit: That's true! That almost shows that I'm not correct!
    $endgroup$
    – Omojola Micheal
    Jan 10 at 5:02










  • $begingroup$
    You're almost correct; instead consider $x - y = (beta - alpha)x_0$.
    $endgroup$
    – Theo Bendit
    Jan 10 at 5:03










  • $begingroup$
    @Theo Bendit: Okay, let me try that!
    $endgroup$
    – Omojola Micheal
    Jan 10 at 5:04










  • $begingroup$
    If you get, write an answer. :-)
    $endgroup$
    – Theo Bendit
    Jan 10 at 5:05
















$begingroup$
I'm not 100% sure I understand the question. As I understand it, you're trying to show that, if you have $x + alpha x_0 = y + beta x_0$, where $x, y in S$, then $x = y$ and $alpha = beta$?
$endgroup$
– Theo Bendit
Jan 10 at 5:01




$begingroup$
I'm not 100% sure I understand the question. As I understand it, you're trying to show that, if you have $x + alpha x_0 = y + beta x_0$, where $x, y in S$, then $x = y$ and $alpha = beta$?
$endgroup$
– Theo Bendit
Jan 10 at 5:01












$begingroup$
@Theo Bendit: That's true! That almost shows that I'm not correct!
$endgroup$
– Omojola Micheal
Jan 10 at 5:02




$begingroup$
@Theo Bendit: That's true! That almost shows that I'm not correct!
$endgroup$
– Omojola Micheal
Jan 10 at 5:02












$begingroup$
You're almost correct; instead consider $x - y = (beta - alpha)x_0$.
$endgroup$
– Theo Bendit
Jan 10 at 5:03




$begingroup$
You're almost correct; instead consider $x - y = (beta - alpha)x_0$.
$endgroup$
– Theo Bendit
Jan 10 at 5:03












$begingroup$
@Theo Bendit: Okay, let me try that!
$endgroup$
– Omojola Micheal
Jan 10 at 5:04




$begingroup$
@Theo Bendit: Okay, let me try that!
$endgroup$
– Omojola Micheal
Jan 10 at 5:04












$begingroup$
If you get, write an answer. :-)
$endgroup$
– Theo Bendit
Jan 10 at 5:05




$begingroup$
If you get, write an answer. :-)
$endgroup$
– Theo Bendit
Jan 10 at 5:05










1 Answer
1






active

oldest

votes


















1












$begingroup$

Credits to @Theo Bendit for the hints.



Corrected:



Let $x_0in Xbackslash S$ be arbitrary. Suppose that $min M, $ then there exists $xin S;text{and};alphainBbb{R} $ such that $m=x+alpha,x_0.$ Assume there is another representation, then there exists $yin S;text{and};betain Bbb{R}$ such that $m=y+beta,x_0,$ where $xneq y$ or $alphaneqbeta$. If $xneq y$, then $beta neq alpha.$ Otherwise, if $beta neq alpha$ then $x=y$ or $xneq y.$ In both cases, $beta neq alpha$. Thus,
begin{align} left(x+alpha,x_0=y+beta,x_0right)&implies (x-y)=(beta-alpha),x_0\&implies x_0=dfrac{1}{beta-alpha}(x-y)in S,;text{where};beta-alphaneq 0;text{and};x-yin S,\&implies x_0in S,;text{since };S;text{is a subspace of a normed linear space }\&impliestext{a contradiction }end{align}
Hence, the representation $m=x+alpha x_0,;$ for $;min M$ is unique.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    You should make the final implication more clear. What's wrong with $alpha - beta neq 0$? You should be able to use this assumption to conclude $x_0 in S$ (though you won't be able to conclude $x_0 = 0$).
    $endgroup$
    – Theo Bendit
    Jan 10 at 5:17










  • $begingroup$
    @Theo Bendit: Okay, thanks for the guidiance.
    $endgroup$
    – Omojola Micheal
    Jan 10 at 5:18










  • $begingroup$
    Hint: subspaces are closed under subtraction and scalar multipliaction.
    $endgroup$
    – Theo Bendit
    Jan 10 at 5:38










  • $begingroup$
    @Theo Bendit: Okay, thanks, once again, for the guidiance. You have my respect!
    $endgroup$
    – Omojola Micheal
    Jan 10 at 5:51








  • 1




    $begingroup$
    @Theo Bendit: Thanks a lot for your time and efforts.
    $endgroup$
    – Omojola Micheal
    Jan 10 at 6:11











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Credits to @Theo Bendit for the hints.



Corrected:



Let $x_0in Xbackslash S$ be arbitrary. Suppose that $min M, $ then there exists $xin S;text{and};alphainBbb{R} $ such that $m=x+alpha,x_0.$ Assume there is another representation, then there exists $yin S;text{and};betain Bbb{R}$ such that $m=y+beta,x_0,$ where $xneq y$ or $alphaneqbeta$. If $xneq y$, then $beta neq alpha.$ Otherwise, if $beta neq alpha$ then $x=y$ or $xneq y.$ In both cases, $beta neq alpha$. Thus,
begin{align} left(x+alpha,x_0=y+beta,x_0right)&implies (x-y)=(beta-alpha),x_0\&implies x_0=dfrac{1}{beta-alpha}(x-y)in S,;text{where};beta-alphaneq 0;text{and};x-yin S,\&implies x_0in S,;text{since };S;text{is a subspace of a normed linear space }\&impliestext{a contradiction }end{align}
Hence, the representation $m=x+alpha x_0,;$ for $;min M$ is unique.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    You should make the final implication more clear. What's wrong with $alpha - beta neq 0$? You should be able to use this assumption to conclude $x_0 in S$ (though you won't be able to conclude $x_0 = 0$).
    $endgroup$
    – Theo Bendit
    Jan 10 at 5:17










  • $begingroup$
    @Theo Bendit: Okay, thanks for the guidiance.
    $endgroup$
    – Omojola Micheal
    Jan 10 at 5:18










  • $begingroup$
    Hint: subspaces are closed under subtraction and scalar multipliaction.
    $endgroup$
    – Theo Bendit
    Jan 10 at 5:38










  • $begingroup$
    @Theo Bendit: Okay, thanks, once again, for the guidiance. You have my respect!
    $endgroup$
    – Omojola Micheal
    Jan 10 at 5:51








  • 1




    $begingroup$
    @Theo Bendit: Thanks a lot for your time and efforts.
    $endgroup$
    – Omojola Micheal
    Jan 10 at 6:11
















1












$begingroup$

Credits to @Theo Bendit for the hints.



Corrected:



Let $x_0in Xbackslash S$ be arbitrary. Suppose that $min M, $ then there exists $xin S;text{and};alphainBbb{R} $ such that $m=x+alpha,x_0.$ Assume there is another representation, then there exists $yin S;text{and};betain Bbb{R}$ such that $m=y+beta,x_0,$ where $xneq y$ or $alphaneqbeta$. If $xneq y$, then $beta neq alpha.$ Otherwise, if $beta neq alpha$ then $x=y$ or $xneq y.$ In both cases, $beta neq alpha$. Thus,
begin{align} left(x+alpha,x_0=y+beta,x_0right)&implies (x-y)=(beta-alpha),x_0\&implies x_0=dfrac{1}{beta-alpha}(x-y)in S,;text{where};beta-alphaneq 0;text{and};x-yin S,\&implies x_0in S,;text{since };S;text{is a subspace of a normed linear space }\&impliestext{a contradiction }end{align}
Hence, the representation $m=x+alpha x_0,;$ for $;min M$ is unique.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    You should make the final implication more clear. What's wrong with $alpha - beta neq 0$? You should be able to use this assumption to conclude $x_0 in S$ (though you won't be able to conclude $x_0 = 0$).
    $endgroup$
    – Theo Bendit
    Jan 10 at 5:17










  • $begingroup$
    @Theo Bendit: Okay, thanks for the guidiance.
    $endgroup$
    – Omojola Micheal
    Jan 10 at 5:18










  • $begingroup$
    Hint: subspaces are closed under subtraction and scalar multipliaction.
    $endgroup$
    – Theo Bendit
    Jan 10 at 5:38










  • $begingroup$
    @Theo Bendit: Okay, thanks, once again, for the guidiance. You have my respect!
    $endgroup$
    – Omojola Micheal
    Jan 10 at 5:51








  • 1




    $begingroup$
    @Theo Bendit: Thanks a lot for your time and efforts.
    $endgroup$
    – Omojola Micheal
    Jan 10 at 6:11














1












1








1





$begingroup$

Credits to @Theo Bendit for the hints.



Corrected:



Let $x_0in Xbackslash S$ be arbitrary. Suppose that $min M, $ then there exists $xin S;text{and};alphainBbb{R} $ such that $m=x+alpha,x_0.$ Assume there is another representation, then there exists $yin S;text{and};betain Bbb{R}$ such that $m=y+beta,x_0,$ where $xneq y$ or $alphaneqbeta$. If $xneq y$, then $beta neq alpha.$ Otherwise, if $beta neq alpha$ then $x=y$ or $xneq y.$ In both cases, $beta neq alpha$. Thus,
begin{align} left(x+alpha,x_0=y+beta,x_0right)&implies (x-y)=(beta-alpha),x_0\&implies x_0=dfrac{1}{beta-alpha}(x-y)in S,;text{where};beta-alphaneq 0;text{and};x-yin S,\&implies x_0in S,;text{since };S;text{is a subspace of a normed linear space }\&impliestext{a contradiction }end{align}
Hence, the representation $m=x+alpha x_0,;$ for $;min M$ is unique.






share|cite|improve this answer











$endgroup$



Credits to @Theo Bendit for the hints.



Corrected:



Let $x_0in Xbackslash S$ be arbitrary. Suppose that $min M, $ then there exists $xin S;text{and};alphainBbb{R} $ such that $m=x+alpha,x_0.$ Assume there is another representation, then there exists $yin S;text{and};betain Bbb{R}$ such that $m=y+beta,x_0,$ where $xneq y$ or $alphaneqbeta$. If $xneq y$, then $beta neq alpha.$ Otherwise, if $beta neq alpha$ then $x=y$ or $xneq y.$ In both cases, $beta neq alpha$. Thus,
begin{align} left(x+alpha,x_0=y+beta,x_0right)&implies (x-y)=(beta-alpha),x_0\&implies x_0=dfrac{1}{beta-alpha}(x-y)in S,;text{where};beta-alphaneq 0;text{and};x-yin S,\&implies x_0in S,;text{since };S;text{is a subspace of a normed linear space }\&impliestext{a contradiction }end{align}
Hence, the representation $m=x+alpha x_0,;$ for $;min M$ is unique.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 10 at 6:44

























answered Jan 10 at 5:14









Omojola MichealOmojola Micheal

1,824324




1,824324








  • 1




    $begingroup$
    You should make the final implication more clear. What's wrong with $alpha - beta neq 0$? You should be able to use this assumption to conclude $x_0 in S$ (though you won't be able to conclude $x_0 = 0$).
    $endgroup$
    – Theo Bendit
    Jan 10 at 5:17










  • $begingroup$
    @Theo Bendit: Okay, thanks for the guidiance.
    $endgroup$
    – Omojola Micheal
    Jan 10 at 5:18










  • $begingroup$
    Hint: subspaces are closed under subtraction and scalar multipliaction.
    $endgroup$
    – Theo Bendit
    Jan 10 at 5:38










  • $begingroup$
    @Theo Bendit: Okay, thanks, once again, for the guidiance. You have my respect!
    $endgroup$
    – Omojola Micheal
    Jan 10 at 5:51








  • 1




    $begingroup$
    @Theo Bendit: Thanks a lot for your time and efforts.
    $endgroup$
    – Omojola Micheal
    Jan 10 at 6:11














  • 1




    $begingroup$
    You should make the final implication more clear. What's wrong with $alpha - beta neq 0$? You should be able to use this assumption to conclude $x_0 in S$ (though you won't be able to conclude $x_0 = 0$).
    $endgroup$
    – Theo Bendit
    Jan 10 at 5:17










  • $begingroup$
    @Theo Bendit: Okay, thanks for the guidiance.
    $endgroup$
    – Omojola Micheal
    Jan 10 at 5:18










  • $begingroup$
    Hint: subspaces are closed under subtraction and scalar multipliaction.
    $endgroup$
    – Theo Bendit
    Jan 10 at 5:38










  • $begingroup$
    @Theo Bendit: Okay, thanks, once again, for the guidiance. You have my respect!
    $endgroup$
    – Omojola Micheal
    Jan 10 at 5:51








  • 1




    $begingroup$
    @Theo Bendit: Thanks a lot for your time and efforts.
    $endgroup$
    – Omojola Micheal
    Jan 10 at 6:11








1




1




$begingroup$
You should make the final implication more clear. What's wrong with $alpha - beta neq 0$? You should be able to use this assumption to conclude $x_0 in S$ (though you won't be able to conclude $x_0 = 0$).
$endgroup$
– Theo Bendit
Jan 10 at 5:17




$begingroup$
You should make the final implication more clear. What's wrong with $alpha - beta neq 0$? You should be able to use this assumption to conclude $x_0 in S$ (though you won't be able to conclude $x_0 = 0$).
$endgroup$
– Theo Bendit
Jan 10 at 5:17












$begingroup$
@Theo Bendit: Okay, thanks for the guidiance.
$endgroup$
– Omojola Micheal
Jan 10 at 5:18




$begingroup$
@Theo Bendit: Okay, thanks for the guidiance.
$endgroup$
– Omojola Micheal
Jan 10 at 5:18












$begingroup$
Hint: subspaces are closed under subtraction and scalar multipliaction.
$endgroup$
– Theo Bendit
Jan 10 at 5:38




$begingroup$
Hint: subspaces are closed under subtraction and scalar multipliaction.
$endgroup$
– Theo Bendit
Jan 10 at 5:38












$begingroup$
@Theo Bendit: Okay, thanks, once again, for the guidiance. You have my respect!
$endgroup$
– Omojola Micheal
Jan 10 at 5:51






$begingroup$
@Theo Bendit: Okay, thanks, once again, for the guidiance. You have my respect!
$endgroup$
– Omojola Micheal
Jan 10 at 5:51






1




1




$begingroup$
@Theo Bendit: Thanks a lot for your time and efforts.
$endgroup$
– Omojola Micheal
Jan 10 at 6:11




$begingroup$
@Theo Bendit: Thanks a lot for your time and efforts.
$endgroup$
– Omojola Micheal
Jan 10 at 6:11


















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Mario Kart Wii

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Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro
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What does “Dominus providebit” mean?

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5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1
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Antonio Litta Visconti Arese

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Antonio Litta Visconti Arese I duca Litta, VII marchese di Gambolò In carica 1797 – 1820 Predecessore Pompeo Litta Visconti Arese Successore Alfonso Litta Visconti Arese Trattamento Sua grazia Altri titoli Duca Litta Conte del Regno napoleonico d'Italia Patrizio milanese Don Nascita Milano, 2 maggio 1748 Morte Milano, 14 agosto 1820 Dinastia Litta Visconti Arese Padre Pompeo Litta Visconti Arese Madre Maria Elisabetta Visconti Borromeo Consorte Barbara Barbiano di Belgioioso Religione cattolicesimo Antonio Litta Visconti Arese , I duca Litta, VII marchese di Gambolò (Milano, 2 maggio 1748 – Milano, 14 agosto 1820), è stato un nobile e politico italiano. Composizione "Il pastore svizzero" di Pietro Morlacchi, con dedica Indice 1 Biografia 2 Onorificenze 3 Araldica 4 Albero genealogico 5 Note 6 Altri progetti Biografia | Antonio Litta Visco
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