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Is this $0v=0$ proof correct?

5

0

$begingroup$

I'm taking my first linear algebra course at university and recently I've been introduced to vector spaces. Now, the teacher has asked us to prove that $0v=0$ for any $v$ using only the definition of a vector space. I've came up with a proof, but I'm unsure whether it's correct or not, as I'm not familiar with proofs in Math.

$$
begin{align}(1+1)v &= 2v \
(1+1)v-2v &= 2v-2v\
[(1+1)-2]v &= 2(v-v) \
0v &= 0
end{align}
$$

If this is correct I can go on proving other properties of vector spaces. Any feedback appreciated!

EDIT: Ok, taking into account José's answer, I came up with this one
$$
(1+0)v=v \
1v+0v=v \
-v+v+0v=-v+v \
0v=0
$$

linear-algebra proof-verification vector-spaces

share|cite|improve this question

edited Apr 27 '18 at 21:27

GNUSupporter 8964民主女神 地下教會

14.2k82652

asked Apr 27 '18 at 20:33

twkmztwkmz

545

$endgroup$

• 
  
  
  

  
  9
  

  
  
  
  
  
  

  
  $begingroup$
  You are missing a $v$ on the left-hand side of the third line. But otherwise it looks fine. A simpler proof is $vec{0} = v - v = (1-1) v = 0v$.
  $endgroup$
  – angryavian
  Apr 27 '18 at 20:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Doesnt definition of vector space define scalar multiplication? So shouldn't 0v=v be true by definition?
  $endgroup$
  – Goldname
  Apr 27 '18 at 21:20
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Goldname The definition of vector space says that $kv in V$ where $V$ is the vector space, but the thing I know about the result of a scalar times a vector is $1v=v$, everything else I have to prove myself
  $endgroup$
  – twkmz
  Apr 27 '18 at 21:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @angryavian proving that $(-1)v=-v$ is a separate assignment which I haven't done yet, so I would prefer to avoid using it
  $endgroup$
  – twkmz
  Apr 27 '18 at 21:24
  

  
  
  
  

add a comment | 

5

0

$begingroup$

I'm taking my first linear algebra course at university and recently I've been introduced to vector spaces. Now, the teacher has asked us to prove that $0v=0$ for any $v$ using only the definition of a vector space. I've came up with a proof, but I'm unsure whether it's correct or not, as I'm not familiar with proofs in Math.

$$
begin{align}(1+1)v &= 2v \
(1+1)v-2v &= 2v-2v\
[(1+1)-2]v &= 2(v-v) \
0v &= 0
end{align}
$$

If this is correct I can go on proving other properties of vector spaces. Any feedback appreciated!

EDIT: Ok, taking into account José's answer, I came up with this one
$$
(1+0)v=v \
1v+0v=v \
-v+v+0v=-v+v \
0v=0
$$

linear-algebra proof-verification vector-spaces

share|cite|improve this question

edited Apr 27 '18 at 21:27

GNUSupporter 8964民主女神 地下教會

14.2k82652

asked Apr 27 '18 at 20:33

twkmztwkmz

545

$endgroup$

• 
  
  
  

  
  9
  

  
  
  
  
  
  

  
  $begingroup$
  You are missing a $v$ on the left-hand side of the third line. But otherwise it looks fine. A simpler proof is $vec{0} = v - v = (1-1) v = 0v$.
  $endgroup$
  – angryavian
  Apr 27 '18 at 20:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Doesnt definition of vector space define scalar multiplication? So shouldn't 0v=v be true by definition?
  $endgroup$
  – Goldname
  Apr 27 '18 at 21:20
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Goldname The definition of vector space says that $kv in V$ where $V$ is the vector space, but the thing I know about the result of a scalar times a vector is $1v=v$, everything else I have to prove myself
  $endgroup$
  – twkmz
  Apr 27 '18 at 21:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @angryavian proving that $(-1)v=-v$ is a separate assignment which I haven't done yet, so I would prefer to avoid using it
  $endgroup$
  – twkmz
  Apr 27 '18 at 21:24
  

  
  
  
  

add a comment | 

$begingroup$

I'm taking my first linear algebra course at university and recently I've been introduced to vector spaces. Now, the teacher has asked us to prove that $0v=0$ for any $v$ using only the definition of a vector space. I've came up with a proof, but I'm unsure whether it's correct or not, as I'm not familiar with proofs in Math.

$$
begin{align}(1+1)v &= 2v \
(1+1)v-2v &= 2v-2v\
[(1+1)-2]v &= 2(v-v) \
0v &= 0
end{align}
$$

If this is correct I can go on proving other properties of vector spaces. Any feedback appreciated!

EDIT: Ok, taking into account José's answer, I came up with this one
$$
(1+0)v=v \
1v+0v=v \
-v+v+0v=-v+v \
0v=0
$$

linear-algebra proof-verification vector-spaces

share|cite|improve this question

edited Apr 27 '18 at 21:27

GNUSupporter 8964民主女神 地下教會

14.2k82652

asked Apr 27 '18 at 20:33

twkmztwkmz

545

$endgroup$

share|cite|improve this question

edited Apr 27 '18 at 21:27

GNUSupporter 8964民主女神 地下教會

14.2k82652

asked Apr 27 '18 at 20:33

twkmztwkmz

545

share|cite|improve this question

edited Apr 27 '18 at 21:27

GNUSupporter 8964民主女神 地下教會

14.2k82652

asked Apr 27 '18 at 20:33

twkmztwkmz

545

edited Apr 27 '18 at 21:27

GNUSupporter 8964民主女神 地下教會

14.2k82652

edited Apr 27 '18 at 21:27

GNUSupporter 8964民主女神 地下教會

14.2k82652

asked Apr 27 '18 at 20:33

twkmztwkmz

545

asked Apr 27 '18 at 20:33

twkmztwkmz

545

3 Answers
3

active

oldest

votes

1

$begingroup$

Your proof has some problems. What is $-2v$? Is it $-(2v)$ or is it $(-2)v$? If it is $-(2v)$, then you could jump directly from $2v-2v$ to $0$, but then how do you know that $(1+1)v-2v=bigl((1+1)-2bigr)v$? And if $-2v$ is $(-2)v$, then how do you know that $2v-2v=2(v-v)$?

share|cite|improve this answer

answered Apr 27 '18 at 20:55

José Carlos SantosJosé Carlos Santos

176k24137247

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah, you're right, I was trying to prove that $(-1)v=-v$ as a separate exercise, so I was in fact using properties that I hadn't yet proved
  $endgroup$
  – twkmz
  Apr 27 '18 at 21:16
  

  
  
  
  

add a comment | 

1

$begingroup$

You can write a proof that doesn't require $(-1)v=-v$:
$$
0v=(0+0)v\
0v=0v+0v\
0v-(0v)=0v+0v-(0v)\
0=0v
$$
Note that $x-y$ actually means $x+(-y)$; with $-(0v)$ I denote the negative of $0v$.

Where does your attempt go wrong? With the same convention as before, you can do $(1+1)v-(2v)=2v-(2v)$. The right hand side is $0$ by definition of $-(2v)$, but you cannot go from $(1+1)v-(2v)$ to $((1+1)-2)v$ unless you prove that
$$
-(2v)=(-2)v
$$
which is true but unfortunately depends on $0v=0$. Indeed, for any scalar $a$,
$$
av+(-a)v=(a+(-a))v=0v=0
$$
so $(-a)v$ summed to $av$ is $0$, which means $(-a)v$ is the negative of $av$:
$$
(-a)v=-(av)
$$
Note that in all of this the property $1v=v$ has not been used. With it we can also state
$$
(-1)v=-(1v)=-v
$$

share|cite|improve this answer

answered Apr 27 '18 at 22:06

egregegreg

186k1486209

$endgroup$

add a comment | 

-1

$begingroup$

Here another method :
where $θ$ is the zero vector and $a$ is a scaler
We know that $av$ is a scaled vector and $v + θ = v$
So : $$av + θ = av $$
Hence :$$θ = (a-a)v$$
$$θ = 0v$$

share|cite|improve this answer

edited Mar 26 at 11:35

Yanior Weg

3,06021652

answered Mar 26 at 10:45

User1234User1234

1

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  This makes the same mistake that the OP did, namely assume without argument that $-(av)=(-a)v$.
  $endgroup$
  – Henning Makholm
  Mar 26 at 10:52
  
  
  

  
  
  
  

add a comment | 

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1

$begingroup$

Your proof has some problems. What is $-2v$? Is it $-(2v)$ or is it $(-2)v$? If it is $-(2v)$, then you could jump directly from $2v-2v$ to $0$, but then how do you know that $(1+1)v-2v=bigl((1+1)-2bigr)v$? And if $-2v$ is $(-2)v$, then how do you know that $2v-2v=2(v-v)$?

share|cite|improve this answer

answered Apr 27 '18 at 20:55

José Carlos SantosJosé Carlos Santos

176k24137247

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah, you're right, I was trying to prove that $(-1)v=-v$ as a separate exercise, so I was in fact using properties that I hadn't yet proved
  $endgroup$
  – twkmz
  Apr 27 '18 at 21:16
  

  
  
  
  

add a comment | 

1

$begingroup$

Your proof has some problems. What is $-2v$? Is it $-(2v)$ or is it $(-2)v$? If it is $-(2v)$, then you could jump directly from $2v-2v$ to $0$, but then how do you know that $(1+1)v-2v=bigl((1+1)-2bigr)v$? And if $-2v$ is $(-2)v$, then how do you know that $2v-2v=2(v-v)$?

share|cite|improve this answer

answered Apr 27 '18 at 20:55

José Carlos SantosJosé Carlos Santos

176k24137247

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah, you're right, I was trying to prove that $(-1)v=-v$ as a separate exercise, so I was in fact using properties that I hadn't yet proved
  $endgroup$
  – twkmz
  Apr 27 '18 at 21:16
  

  
  
  
  

add a comment | 

$begingroup$

Your proof has some problems. What is $-2v$? Is it $-(2v)$ or is it $(-2)v$? If it is $-(2v)$, then you could jump directly from $2v-2v$ to $0$, but then how do you know that $(1+1)v-2v=bigl((1+1)-2bigr)v$? And if $-2v$ is $(-2)v$, then how do you know that $2v-2v=2(v-v)$?

share|cite|improve this answer

answered Apr 27 '18 at 20:55

José Carlos SantosJosé Carlos Santos

176k24137247

$endgroup$

share|cite|improve this answer

answered Apr 27 '18 at 20:55

José Carlos SantosJosé Carlos Santos

176k24137247

answered Apr 27 '18 at 20:55

José Carlos SantosJosé Carlos Santos

176k24137247

answered Apr 27 '18 at 20:55

José Carlos SantosJosé Carlos Santos

176k24137247

1

$begingroup$

You can write a proof that doesn't require $(-1)v=-v$:
$$
0v=(0+0)v\
0v=0v+0v\
0v-(0v)=0v+0v-(0v)\
0=0v
$$
Note that $x-y$ actually means $x+(-y)$; with $-(0v)$ I denote the negative of $0v$.

Where does your attempt go wrong? With the same convention as before, you can do $(1+1)v-(2v)=2v-(2v)$. The right hand side is $0$ by definition of $-(2v)$, but you cannot go from $(1+1)v-(2v)$ to $((1+1)-2)v$ unless you prove that
$$
-(2v)=(-2)v
$$
which is true but unfortunately depends on $0v=0$. Indeed, for any scalar $a$,
$$
av+(-a)v=(a+(-a))v=0v=0
$$
so $(-a)v$ summed to $av$ is $0$, which means $(-a)v$ is the negative of $av$:
$$
(-a)v=-(av)
$$
Note that in all of this the property $1v=v$ has not been used. With it we can also state
$$
(-1)v=-(1v)=-v
$$

share|cite|improve this answer

answered Apr 27 '18 at 22:06

egregegreg

186k1486209

$endgroup$

add a comment | 

1

$begingroup$

You can write a proof that doesn't require $(-1)v=-v$:
$$
0v=(0+0)v\
0v=0v+0v\
0v-(0v)=0v+0v-(0v)\
0=0v
$$
Note that $x-y$ actually means $x+(-y)$; with $-(0v)$ I denote the negative of $0v$.

Where does your attempt go wrong? With the same convention as before, you can do $(1+1)v-(2v)=2v-(2v)$. The right hand side is $0$ by definition of $-(2v)$, but you cannot go from $(1+1)v-(2v)$ to $((1+1)-2)v$ unless you prove that
$$
-(2v)=(-2)v
$$
which is true but unfortunately depends on $0v=0$. Indeed, for any scalar $a$,
$$
av+(-a)v=(a+(-a))v=0v=0
$$
so $(-a)v$ summed to $av$ is $0$, which means $(-a)v$ is the negative of $av$:
$$
(-a)v=-(av)
$$
Note that in all of this the property $1v=v$ has not been used. With it we can also state
$$
(-1)v=-(1v)=-v
$$

share|cite|improve this answer

answered Apr 27 '18 at 22:06

egregegreg

186k1486209

$endgroup$

add a comment | 

$begingroup$

You can write a proof that doesn't require $(-1)v=-v$:
$$
0v=(0+0)v\
0v=0v+0v\
0v-(0v)=0v+0v-(0v)\
0=0v
$$
Note that $x-y$ actually means $x+(-y)$; with $-(0v)$ I denote the negative of $0v$.

Where does your attempt go wrong? With the same convention as before, you can do $(1+1)v-(2v)=2v-(2v)$. The right hand side is $0$ by definition of $-(2v)$, but you cannot go from $(1+1)v-(2v)$ to $((1+1)-2)v$ unless you prove that
$$
-(2v)=(-2)v
$$
which is true but unfortunately depends on $0v=0$. Indeed, for any scalar $a$,
$$
av+(-a)v=(a+(-a))v=0v=0
$$
so $(-a)v$ summed to $av$ is $0$, which means $(-a)v$ is the negative of $av$:
$$
(-a)v=-(av)
$$
Note that in all of this the property $1v=v$ has not been used. With it we can also state
$$
(-1)v=-(1v)=-v
$$

share|cite|improve this answer

answered Apr 27 '18 at 22:06

egregegreg

186k1486209

$endgroup$

share|cite|improve this answer

answered Apr 27 '18 at 22:06

egregegreg

186k1486209

answered Apr 27 '18 at 22:06

egregegreg

186k1486209

answered Apr 27 '18 at 22:06

egregegreg

186k1486209

-1

$begingroup$

Here another method :
where $θ$ is the zero vector and $a$ is a scaler
We know that $av$ is a scaled vector and $v + θ = v$
So : $$av + θ = av $$
Hence :$$θ = (a-a)v$$
$$θ = 0v$$

share|cite|improve this answer

edited Mar 26 at 11:35

Yanior Weg

3,06021652

answered Mar 26 at 10:45

User1234User1234

1

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  This makes the same mistake that the OP did, namely assume without argument that $-(av)=(-a)v$.
  $endgroup$
  – Henning Makholm
  Mar 26 at 10:52
  
  
  

  
  
  
  

add a comment | 

-1

$begingroup$

Here another method :
where $θ$ is the zero vector and $a$ is a scaler
We know that $av$ is a scaled vector and $v + θ = v$
So : $$av + θ = av $$
Hence :$$θ = (a-a)v$$
$$θ = 0v$$

share|cite|improve this answer

edited Mar 26 at 11:35

Yanior Weg

3,06021652

answered Mar 26 at 10:45

User1234User1234

1

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  This makes the same mistake that the OP did, namely assume without argument that $-(av)=(-a)v$.
  $endgroup$
  – Henning Makholm
  Mar 26 at 10:52
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Here another method :
where $θ$ is the zero vector and $a$ is a scaler
We know that $av$ is a scaled vector and $v + θ = v$
So : $$av + θ = av $$
Hence :$$θ = (a-a)v$$
$$θ = 0v$$

share|cite|improve this answer

edited Mar 26 at 11:35

Yanior Weg

3,06021652

answered Mar 26 at 10:45

User1234User1234

1

$endgroup$

share|cite|improve this answer

edited Mar 26 at 11:35

Yanior Weg

3,06021652

answered Mar 26 at 10:45

User1234User1234

1

edited Mar 26 at 11:35

Yanior Weg

3,06021652

edited Mar 26 at 11:35

Yanior Weg

3,06021652

answered Mar 26 at 10:45

User1234User1234

1

answered Mar 26 at 10:45

User1234User1234

1