W1 + W3 = W2 + W3. Then W1 = W2
Claim. Let $V$ be a vector space over $F$, and suppose that $W_1, W_2,$ and $W_3$ are subspaces
of $V$ such that $W_1 + W_3 = W_2 + W_3.$ Then $W_1 = W_2.$
Proof
$W_1 + W_3 - W_3 = W_2 + W_3 - W_3$
$therefore$ $W_1 = W_2$
My proof feels like cheating, is it even valid? Also is there anyway to 'say', prove or demonstrate the same thing in a more rigorous way i.e. set builder notation?
In regards to the rules of the site is my question to simple, if so what can I do to make it better?
proof-verification elementary-set-theory vector-spaces alternative-proof
add a comment |
Claim. Let $V$ be a vector space over $F$, and suppose that $W_1, W_2,$ and $W_3$ are subspaces
of $V$ such that $W_1 + W_3 = W_2 + W_3.$ Then $W_1 = W_2.$
Proof
$W_1 + W_3 - W_3 = W_2 + W_3 - W_3$
$therefore$ $W_1 = W_2$
My proof feels like cheating, is it even valid? Also is there anyway to 'say', prove or demonstrate the same thing in a more rigorous way i.e. set builder notation?
In regards to the rules of the site is my question to simple, if so what can I do to make it better?
proof-verification elementary-set-theory vector-spaces alternative-proof
Subtraction between subspaces of a vector space is not defined.
– Batominovski
Nov 25 at 13:52
2
Your claim is clearly false: ${0}+V=V+V$, for every vector space $V$. If your claim is true, it implies that every vector space just consists of the zero vector. Your attempt at a proof is essentially the same as concluding that $1=2$ from the fact that $1cdot0=2cdot0$ and multiplying both sides by $0^{-1}$.
– egreg
Nov 25 at 13:54
add a comment |
Claim. Let $V$ be a vector space over $F$, and suppose that $W_1, W_2,$ and $W_3$ are subspaces
of $V$ such that $W_1 + W_3 = W_2 + W_3.$ Then $W_1 = W_2.$
Proof
$W_1 + W_3 - W_3 = W_2 + W_3 - W_3$
$therefore$ $W_1 = W_2$
My proof feels like cheating, is it even valid? Also is there anyway to 'say', prove or demonstrate the same thing in a more rigorous way i.e. set builder notation?
In regards to the rules of the site is my question to simple, if so what can I do to make it better?
proof-verification elementary-set-theory vector-spaces alternative-proof
Claim. Let $V$ be a vector space over $F$, and suppose that $W_1, W_2,$ and $W_3$ are subspaces
of $V$ such that $W_1 + W_3 = W_2 + W_3.$ Then $W_1 = W_2.$
Proof
$W_1 + W_3 - W_3 = W_2 + W_3 - W_3$
$therefore$ $W_1 = W_2$
My proof feels like cheating, is it even valid? Also is there anyway to 'say', prove or demonstrate the same thing in a more rigorous way i.e. set builder notation?
In regards to the rules of the site is my question to simple, if so what can I do to make it better?
proof-verification elementary-set-theory vector-spaces alternative-proof
proof-verification elementary-set-theory vector-spaces alternative-proof
asked Nov 25 at 13:36
oypus
518
518
Subtraction between subspaces of a vector space is not defined.
– Batominovski
Nov 25 at 13:52
2
Your claim is clearly false: ${0}+V=V+V$, for every vector space $V$. If your claim is true, it implies that every vector space just consists of the zero vector. Your attempt at a proof is essentially the same as concluding that $1=2$ from the fact that $1cdot0=2cdot0$ and multiplying both sides by $0^{-1}$.
– egreg
Nov 25 at 13:54
add a comment |
Subtraction between subspaces of a vector space is not defined.
– Batominovski
Nov 25 at 13:52
2
Your claim is clearly false: ${0}+V=V+V$, for every vector space $V$. If your claim is true, it implies that every vector space just consists of the zero vector. Your attempt at a proof is essentially the same as concluding that $1=2$ from the fact that $1cdot0=2cdot0$ and multiplying both sides by $0^{-1}$.
– egreg
Nov 25 at 13:54
Subtraction between subspaces of a vector space is not defined.
– Batominovski
Nov 25 at 13:52
Subtraction between subspaces of a vector space is not defined.
– Batominovski
Nov 25 at 13:52
2
2
Your claim is clearly false: ${0}+V=V+V$, for every vector space $V$. If your claim is true, it implies that every vector space just consists of the zero vector. Your attempt at a proof is essentially the same as concluding that $1=2$ from the fact that $1cdot0=2cdot0$ and multiplying both sides by $0^{-1}$.
– egreg
Nov 25 at 13:54
Your claim is clearly false: ${0}+V=V+V$, for every vector space $V$. If your claim is true, it implies that every vector space just consists of the zero vector. Your attempt at a proof is essentially the same as concluding that $1=2$ from the fact that $1cdot0=2cdot0$ and multiplying both sides by $0^{-1}$.
– egreg
Nov 25 at 13:54
add a comment |
3 Answers
3
active
oldest
votes
Consider $V=mathbb{R}^2 $ and $W_3= {(x,0) : xin mathbb{R}}$, $W_1= {(0,y) : yin mathbb{R}}$ and $W_2= {(t,t) : tin mathbb{R}}$
You should be able to verify that $W_1+W_3 = W_2 + W_3$
What does that tell you about your claim?
Does it say that the claim is false?
– oypus
Nov 25 at 13:53
add a comment |
What does your claim say if $W_3$ is the whole space $V$?
You can't subtract subspaces.
Thank you, but that only answers one component of my question, may you elaborate on you answer.
– oypus
Nov 25 at 13:43
1
@oypus This answer answers your question completely.
– Thomas
Nov 25 at 13:50
3
SInce your claim is false, there is no way to "demonstrate the same thing in a more rigorous way".Your notation is fine. Better notation would not make the claim true.
– Ethan Bolker
Nov 25 at 13:51
add a comment |
There is a very simple counterexample. Since the statement is about arbitrary subspaces $W_1$, $W_2$ and $W_3$ only subject to the condition that $W_1+W_3=W_2+W_3$, it should in particular hold for $W_1={0}$, $W_2=W_3=V$. Then from
$$
{0}+V=V+V
$$
(which is true), you'd conclude that
$$
{0}=V
$$
Now any non trivial vector space is a counterexample.
From a slightly higher point of view, the set $mathscr{L}(V)$ of subspaces of $V$ is a commutative monoid under the $+$ operation, because it is associative and has the neutral element ${0}$, because ${0}+W=W$, for every $Winmathscr{L}(V)$.
However, this monoid cannot be cancellative (that is $W_1+W_3=W_2+W_3$ implies $W_1=W_2$) for several reasons, the most important one being that it has an absorbing element, namely $V$:
$$
W+V=V
$$
for every $Winmathscr{L}(V)$. An absorbing element cannot have a symmetric element, unless the monoid is trivial.
Also, if $U$ is a subspace of $V$ and $mathscr{L}(V;subseteq U)$ denotes the set of subspaces of $V$ contained in $U$, we have $mathscr{L}(V;subseteq U)=mathscr{L}(U)$, which is thus a submonoid of $mathscr{L}(V)$. Since every $Uinmathscr{L}(V)$ is the absorbing element in a submonoid, no element can have a symmetric element, except for ${0}$.
Note. By “symmetric element” of an element $x$ in a monoid $(M,*,e)$ I mean an element $y$ such that $x*y=y*x=e$.
Correct me if I'm wrong but essentially what you are saying is that since addition isn't always 'symmetric' i.e. W+V=V we cannot use inverse operations like subtraction without a prior knowledge of the sets we are dealing with, because inverse operations imply or assume symmetry.
– oypus
Nov 25 at 14:13
Is "symmetric element" the same as "inverse element", or is there more to what you're saying?
– Henning Makholm
Nov 25 at 14:20
@HenningMakholm Yes, I used “symmetric” to be agnostic about the operation being denoted additively or multiplicatively. I added a note.
– egreg
Nov 25 at 14:24
1
@egreg: Okay. (However, in my experience "inverse" too is used for both additively and multiplicatively notated monoids).
– Henning Makholm
Nov 25 at 14:29
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Consider $V=mathbb{R}^2 $ and $W_3= {(x,0) : xin mathbb{R}}$, $W_1= {(0,y) : yin mathbb{R}}$ and $W_2= {(t,t) : tin mathbb{R}}$
You should be able to verify that $W_1+W_3 = W_2 + W_3$
What does that tell you about your claim?
Does it say that the claim is false?
– oypus
Nov 25 at 13:53
add a comment |
Consider $V=mathbb{R}^2 $ and $W_3= {(x,0) : xin mathbb{R}}$, $W_1= {(0,y) : yin mathbb{R}}$ and $W_2= {(t,t) : tin mathbb{R}}$
You should be able to verify that $W_1+W_3 = W_2 + W_3$
What does that tell you about your claim?
Does it say that the claim is false?
– oypus
Nov 25 at 13:53
add a comment |
Consider $V=mathbb{R}^2 $ and $W_3= {(x,0) : xin mathbb{R}}$, $W_1= {(0,y) : yin mathbb{R}}$ and $W_2= {(t,t) : tin mathbb{R}}$
You should be able to verify that $W_1+W_3 = W_2 + W_3$
What does that tell you about your claim?
Consider $V=mathbb{R}^2 $ and $W_3= {(x,0) : xin mathbb{R}}$, $W_1= {(0,y) : yin mathbb{R}}$ and $W_2= {(t,t) : tin mathbb{R}}$
You should be able to verify that $W_1+W_3 = W_2 + W_3$
What does that tell you about your claim?
edited Nov 25 at 14:35
answered Nov 25 at 13:49
Thomas
16.7k21631
16.7k21631
Does it say that the claim is false?
– oypus
Nov 25 at 13:53
add a comment |
Does it say that the claim is false?
– oypus
Nov 25 at 13:53
Does it say that the claim is false?
– oypus
Nov 25 at 13:53
Does it say that the claim is false?
– oypus
Nov 25 at 13:53
add a comment |
What does your claim say if $W_3$ is the whole space $V$?
You can't subtract subspaces.
Thank you, but that only answers one component of my question, may you elaborate on you answer.
– oypus
Nov 25 at 13:43
1
@oypus This answer answers your question completely.
– Thomas
Nov 25 at 13:50
3
SInce your claim is false, there is no way to "demonstrate the same thing in a more rigorous way".Your notation is fine. Better notation would not make the claim true.
– Ethan Bolker
Nov 25 at 13:51
add a comment |
What does your claim say if $W_3$ is the whole space $V$?
You can't subtract subspaces.
Thank you, but that only answers one component of my question, may you elaborate on you answer.
– oypus
Nov 25 at 13:43
1
@oypus This answer answers your question completely.
– Thomas
Nov 25 at 13:50
3
SInce your claim is false, there is no way to "demonstrate the same thing in a more rigorous way".Your notation is fine. Better notation would not make the claim true.
– Ethan Bolker
Nov 25 at 13:51
add a comment |
What does your claim say if $W_3$ is the whole space $V$?
You can't subtract subspaces.
What does your claim say if $W_3$ is the whole space $V$?
You can't subtract subspaces.
answered Nov 25 at 13:39
Ethan Bolker
40.9k546108
40.9k546108
Thank you, but that only answers one component of my question, may you elaborate on you answer.
– oypus
Nov 25 at 13:43
1
@oypus This answer answers your question completely.
– Thomas
Nov 25 at 13:50
3
SInce your claim is false, there is no way to "demonstrate the same thing in a more rigorous way".Your notation is fine. Better notation would not make the claim true.
– Ethan Bolker
Nov 25 at 13:51
add a comment |
Thank you, but that only answers one component of my question, may you elaborate on you answer.
– oypus
Nov 25 at 13:43
1
@oypus This answer answers your question completely.
– Thomas
Nov 25 at 13:50
3
SInce your claim is false, there is no way to "demonstrate the same thing in a more rigorous way".Your notation is fine. Better notation would not make the claim true.
– Ethan Bolker
Nov 25 at 13:51
Thank you, but that only answers one component of my question, may you elaborate on you answer.
– oypus
Nov 25 at 13:43
Thank you, but that only answers one component of my question, may you elaborate on you answer.
– oypus
Nov 25 at 13:43
1
1
@oypus This answer answers your question completely.
– Thomas
Nov 25 at 13:50
@oypus This answer answers your question completely.
– Thomas
Nov 25 at 13:50
3
3
SInce your claim is false, there is no way to "demonstrate the same thing in a more rigorous way".Your notation is fine. Better notation would not make the claim true.
– Ethan Bolker
Nov 25 at 13:51
SInce your claim is false, there is no way to "demonstrate the same thing in a more rigorous way".Your notation is fine. Better notation would not make the claim true.
– Ethan Bolker
Nov 25 at 13:51
add a comment |
There is a very simple counterexample. Since the statement is about arbitrary subspaces $W_1$, $W_2$ and $W_3$ only subject to the condition that $W_1+W_3=W_2+W_3$, it should in particular hold for $W_1={0}$, $W_2=W_3=V$. Then from
$$
{0}+V=V+V
$$
(which is true), you'd conclude that
$$
{0}=V
$$
Now any non trivial vector space is a counterexample.
From a slightly higher point of view, the set $mathscr{L}(V)$ of subspaces of $V$ is a commutative monoid under the $+$ operation, because it is associative and has the neutral element ${0}$, because ${0}+W=W$, for every $Winmathscr{L}(V)$.
However, this monoid cannot be cancellative (that is $W_1+W_3=W_2+W_3$ implies $W_1=W_2$) for several reasons, the most important one being that it has an absorbing element, namely $V$:
$$
W+V=V
$$
for every $Winmathscr{L}(V)$. An absorbing element cannot have a symmetric element, unless the monoid is trivial.
Also, if $U$ is a subspace of $V$ and $mathscr{L}(V;subseteq U)$ denotes the set of subspaces of $V$ contained in $U$, we have $mathscr{L}(V;subseteq U)=mathscr{L}(U)$, which is thus a submonoid of $mathscr{L}(V)$. Since every $Uinmathscr{L}(V)$ is the absorbing element in a submonoid, no element can have a symmetric element, except for ${0}$.
Note. By “symmetric element” of an element $x$ in a monoid $(M,*,e)$ I mean an element $y$ such that $x*y=y*x=e$.
Correct me if I'm wrong but essentially what you are saying is that since addition isn't always 'symmetric' i.e. W+V=V we cannot use inverse operations like subtraction without a prior knowledge of the sets we are dealing with, because inverse operations imply or assume symmetry.
– oypus
Nov 25 at 14:13
Is "symmetric element" the same as "inverse element", or is there more to what you're saying?
– Henning Makholm
Nov 25 at 14:20
@HenningMakholm Yes, I used “symmetric” to be agnostic about the operation being denoted additively or multiplicatively. I added a note.
– egreg
Nov 25 at 14:24
1
@egreg: Okay. (However, in my experience "inverse" too is used for both additively and multiplicatively notated monoids).
– Henning Makholm
Nov 25 at 14:29
add a comment |
There is a very simple counterexample. Since the statement is about arbitrary subspaces $W_1$, $W_2$ and $W_3$ only subject to the condition that $W_1+W_3=W_2+W_3$, it should in particular hold for $W_1={0}$, $W_2=W_3=V$. Then from
$$
{0}+V=V+V
$$
(which is true), you'd conclude that
$$
{0}=V
$$
Now any non trivial vector space is a counterexample.
From a slightly higher point of view, the set $mathscr{L}(V)$ of subspaces of $V$ is a commutative monoid under the $+$ operation, because it is associative and has the neutral element ${0}$, because ${0}+W=W$, for every $Winmathscr{L}(V)$.
However, this monoid cannot be cancellative (that is $W_1+W_3=W_2+W_3$ implies $W_1=W_2$) for several reasons, the most important one being that it has an absorbing element, namely $V$:
$$
W+V=V
$$
for every $Winmathscr{L}(V)$. An absorbing element cannot have a symmetric element, unless the monoid is trivial.
Also, if $U$ is a subspace of $V$ and $mathscr{L}(V;subseteq U)$ denotes the set of subspaces of $V$ contained in $U$, we have $mathscr{L}(V;subseteq U)=mathscr{L}(U)$, which is thus a submonoid of $mathscr{L}(V)$. Since every $Uinmathscr{L}(V)$ is the absorbing element in a submonoid, no element can have a symmetric element, except for ${0}$.
Note. By “symmetric element” of an element $x$ in a monoid $(M,*,e)$ I mean an element $y$ such that $x*y=y*x=e$.
Correct me if I'm wrong but essentially what you are saying is that since addition isn't always 'symmetric' i.e. W+V=V we cannot use inverse operations like subtraction without a prior knowledge of the sets we are dealing with, because inverse operations imply or assume symmetry.
– oypus
Nov 25 at 14:13
Is "symmetric element" the same as "inverse element", or is there more to what you're saying?
– Henning Makholm
Nov 25 at 14:20
@HenningMakholm Yes, I used “symmetric” to be agnostic about the operation being denoted additively or multiplicatively. I added a note.
– egreg
Nov 25 at 14:24
1
@egreg: Okay. (However, in my experience "inverse" too is used for both additively and multiplicatively notated monoids).
– Henning Makholm
Nov 25 at 14:29
add a comment |
There is a very simple counterexample. Since the statement is about arbitrary subspaces $W_1$, $W_2$ and $W_3$ only subject to the condition that $W_1+W_3=W_2+W_3$, it should in particular hold for $W_1={0}$, $W_2=W_3=V$. Then from
$$
{0}+V=V+V
$$
(which is true), you'd conclude that
$$
{0}=V
$$
Now any non trivial vector space is a counterexample.
From a slightly higher point of view, the set $mathscr{L}(V)$ of subspaces of $V$ is a commutative monoid under the $+$ operation, because it is associative and has the neutral element ${0}$, because ${0}+W=W$, for every $Winmathscr{L}(V)$.
However, this monoid cannot be cancellative (that is $W_1+W_3=W_2+W_3$ implies $W_1=W_2$) for several reasons, the most important one being that it has an absorbing element, namely $V$:
$$
W+V=V
$$
for every $Winmathscr{L}(V)$. An absorbing element cannot have a symmetric element, unless the monoid is trivial.
Also, if $U$ is a subspace of $V$ and $mathscr{L}(V;subseteq U)$ denotes the set of subspaces of $V$ contained in $U$, we have $mathscr{L}(V;subseteq U)=mathscr{L}(U)$, which is thus a submonoid of $mathscr{L}(V)$. Since every $Uinmathscr{L}(V)$ is the absorbing element in a submonoid, no element can have a symmetric element, except for ${0}$.
Note. By “symmetric element” of an element $x$ in a monoid $(M,*,e)$ I mean an element $y$ such that $x*y=y*x=e$.
There is a very simple counterexample. Since the statement is about arbitrary subspaces $W_1$, $W_2$ and $W_3$ only subject to the condition that $W_1+W_3=W_2+W_3$, it should in particular hold for $W_1={0}$, $W_2=W_3=V$. Then from
$$
{0}+V=V+V
$$
(which is true), you'd conclude that
$$
{0}=V
$$
Now any non trivial vector space is a counterexample.
From a slightly higher point of view, the set $mathscr{L}(V)$ of subspaces of $V$ is a commutative monoid under the $+$ operation, because it is associative and has the neutral element ${0}$, because ${0}+W=W$, for every $Winmathscr{L}(V)$.
However, this monoid cannot be cancellative (that is $W_1+W_3=W_2+W_3$ implies $W_1=W_2$) for several reasons, the most important one being that it has an absorbing element, namely $V$:
$$
W+V=V
$$
for every $Winmathscr{L}(V)$. An absorbing element cannot have a symmetric element, unless the monoid is trivial.
Also, if $U$ is a subspace of $V$ and $mathscr{L}(V;subseteq U)$ denotes the set of subspaces of $V$ contained in $U$, we have $mathscr{L}(V;subseteq U)=mathscr{L}(U)$, which is thus a submonoid of $mathscr{L}(V)$. Since every $Uinmathscr{L}(V)$ is the absorbing element in a submonoid, no element can have a symmetric element, except for ${0}$.
Note. By “symmetric element” of an element $x$ in a monoid $(M,*,e)$ I mean an element $y$ such that $x*y=y*x=e$.
edited Nov 25 at 14:25
answered Nov 25 at 14:09
egreg
177k1484200
177k1484200
Correct me if I'm wrong but essentially what you are saying is that since addition isn't always 'symmetric' i.e. W+V=V we cannot use inverse operations like subtraction without a prior knowledge of the sets we are dealing with, because inverse operations imply or assume symmetry.
– oypus
Nov 25 at 14:13
Is "symmetric element" the same as "inverse element", or is there more to what you're saying?
– Henning Makholm
Nov 25 at 14:20
@HenningMakholm Yes, I used “symmetric” to be agnostic about the operation being denoted additively or multiplicatively. I added a note.
– egreg
Nov 25 at 14:24
1
@egreg: Okay. (However, in my experience "inverse" too is used for both additively and multiplicatively notated monoids).
– Henning Makholm
Nov 25 at 14:29
add a comment |
Correct me if I'm wrong but essentially what you are saying is that since addition isn't always 'symmetric' i.e. W+V=V we cannot use inverse operations like subtraction without a prior knowledge of the sets we are dealing with, because inverse operations imply or assume symmetry.
– oypus
Nov 25 at 14:13
Is "symmetric element" the same as "inverse element", or is there more to what you're saying?
– Henning Makholm
Nov 25 at 14:20
@HenningMakholm Yes, I used “symmetric” to be agnostic about the operation being denoted additively or multiplicatively. I added a note.
– egreg
Nov 25 at 14:24
1
@egreg: Okay. (However, in my experience "inverse" too is used for both additively and multiplicatively notated monoids).
– Henning Makholm
Nov 25 at 14:29
Correct me if I'm wrong but essentially what you are saying is that since addition isn't always 'symmetric' i.e. W+V=V we cannot use inverse operations like subtraction without a prior knowledge of the sets we are dealing with, because inverse operations imply or assume symmetry.
– oypus
Nov 25 at 14:13
Correct me if I'm wrong but essentially what you are saying is that since addition isn't always 'symmetric' i.e. W+V=V we cannot use inverse operations like subtraction without a prior knowledge of the sets we are dealing with, because inverse operations imply or assume symmetry.
– oypus
Nov 25 at 14:13
Is "symmetric element" the same as "inverse element", or is there more to what you're saying?
– Henning Makholm
Nov 25 at 14:20
Is "symmetric element" the same as "inverse element", or is there more to what you're saying?
– Henning Makholm
Nov 25 at 14:20
@HenningMakholm Yes, I used “symmetric” to be agnostic about the operation being denoted additively or multiplicatively. I added a note.
– egreg
Nov 25 at 14:24
@HenningMakholm Yes, I used “symmetric” to be agnostic about the operation being denoted additively or multiplicatively. I added a note.
– egreg
Nov 25 at 14:24
1
1
@egreg: Okay. (However, in my experience "inverse" too is used for both additively and multiplicatively notated monoids).
– Henning Makholm
Nov 25 at 14:29
@egreg: Okay. (However, in my experience "inverse" too is used for both additively and multiplicatively notated monoids).
– Henning Makholm
Nov 25 at 14:29
add a comment |
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Subtraction between subspaces of a vector space is not defined.
– Batominovski
Nov 25 at 13:52
2
Your claim is clearly false: ${0}+V=V+V$, for every vector space $V$. If your claim is true, it implies that every vector space just consists of the zero vector. Your attempt at a proof is essentially the same as concluding that $1=2$ from the fact that $1cdot0=2cdot0$ and multiplying both sides by $0^{-1}$.
– egreg
Nov 25 at 13:54