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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The Binding of Isaac: Rebirth/Afterbirth

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Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
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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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