Mixing
Vector space- linear algebra
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
I don't get how the book solved this problem, because axioms $4$, $5$ and $7$ are about adding not multiplying.
I really need to understand this. Thank you
linear-algebra vector-spaces axioms
edited Jul 28 at 16:41
Daniel Buck
2,2841623
asked Jul 28 at 15:28
user580158
31
add a comment |Â
up vote
0
down vote
favorite
I don't get how the book solved this problem, because axioms $4$, $5$ and $7$ are about adding not multiplying.
I really need to understand this. Thank you
linear-algebra vector-spaces axioms
edited Jul 28 at 16:41
Daniel Buck
2,2841623
asked Jul 28 at 15:28
user580158
31
1
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
– José Carlos Santos
Jul 28 at 15:33
The axioms are actually not about adding; they are about two operations, one of which is normally written as addition.
– md2perpe
Jul 28 at 17:30
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I don't get how the book solved this problem, because axioms $4$, $5$ and $7$ are about adding not multiplying.
I really need to understand this. Thank you
linear-algebra vector-spaces axioms
edited Jul 28 at 16:41
Daniel Buck
2,2841623
asked Jul 28 at 15:28
user580158
31
I don't get how the book solved this problem, because axioms $4$, $5$ and $7$ are about adding not multiplying.
I really need to understand this. Thank you
linear-algebra vector-spaces axioms
edited Jul 28 at 16:41
Daniel Buck
2,2841623
asked Jul 28 at 15:28
user580158
31
edited Jul 28 at 16:41
Daniel Buck
2,2841623
edited Jul 28 at 16:41
Daniel Buck
2,2841623
edited Jul 28 at 16:41
Daniel Buck
2,2841623
2,2841623
asked Jul 28 at 15:28
user580158
31
asked Jul 28 at 15:28
user580158
31
asked Jul 28 at 15:28
user580158
31
31
1
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
– José Carlos Santos
Jul 28 at 15:33
The axioms are actually not about adding; they are about two operations, one of which is normally written as addition.
– md2perpe
Jul 28 at 17:30
add a comment |Â
1
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
– José Carlos Santos
Jul 28 at 15:33
The axioms are actually not about adding; they are about two operations, one of which is normally written as addition.
– md2perpe
Jul 28 at 17:30
1
1
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
– José Carlos Santos
Jul 28 at 15:33
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
– José Carlos Santos
Jul 28 at 15:33
The axioms are actually not about adding; they are about two operations, one of which is normally written as addition.
– md2perpe
Jul 28 at 17:30
The axioms are actually not about adding; they are about two operations, one of which is normally written as addition.
– md2perpe
Jul 28 at 17:30
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
accepted
The key phrase is "vector addition is numerical multiplication"
That is they define the operation $+_V$ as $u+_Vv=ucdot v$. I explicitly write out the subscript $V$ for the vector addition so as to highlight the addition is not the convention addition that you are familiar with.
We can check that $u+_V1 = u cdot 1=u$.
I will leave the rest as an exercise.
answered Jul 28 at 15:33
Siong Thye Goh
77k134794
yup, the zero vector $0_V= 1$.
– Siong Thye Goh
Jul 28 at 15:47
add a comment |Â
up vote
2
down vote
I certainly understand that it looks confusing when the book writes $u+v=uv$ and $ku=u^k,$ so let us instead introduce two operators $oplus : V times V to V$ and $odot : mathbb R times V to V$ satisfying some axioms like
$(u oplus v) oplus w = u oplus (v oplus w)$
and
$k otimes (u oplus v) = (k odot u) oplus (k odot v).$
I won't list all axioms since I don't want to confuse the enumeration with the one in you course literature.
In the example given, they take $V = (0, infty)$ and define $oplus$ and $odot$ by $u oplus v = uv$ and $k odot u = u^k.$ Then, for example, they show the second axiom that I listed:
$$beginalign
k odot (u oplus v)
& = text by def. of $odot$ = (u oplus v)^k \
& = text by def. of $oplus$ = (uv)^k \
& = text by one of the power laws = u^k v^k \
& = text by def. of $oplus$ = u^k oplus v^k \
& = text by def. of $odot$ = (k odot u) oplus (k odot v)
endalign$$
It is very uncommon though with vector spaces where the operations are not derived from ordinary addition and multiplication with a real or complex number. Therefore the operations are normally written as just $+$ and juxtaposition (i.e. $ku$ without an explicit operation).
answered Jul 28 at 18:14
md2perpe
5,79511022
add a comment |Â
up vote
0
down vote
Here $V$ is the set $mathbbR_>0$, so the vector $mathbfv$ is just the real number $v$ (although it is still a vector in $V$).
Now define the operations on $V$:
$$u+v=uv quadtextbfVector addition is numerical multiplicationtag1$$
$$ku=u^k quadtextbfScalar multiplication is numerical exponentiationtag2$$
Now with operations $(1)$ and $(2)$ set up you have to try each axiom in turn to see if it satisfies them.
For Axiom $4$ we want the definition of the zero vector, i.e., the element $i$ that makes $u+i=u$. It's obviously not $i=0$, as $0notin V$ (if it were it would give: $u+0=ucdot0=0$, and so $u=0$, contradicting the fact $u$ is a positive real). In a sense you can work backwards here, since you know $ucdot1=u$, where $u$ and $1$ are seen as numbers to be multiplied. Now looking at $(1)$ you see the addition of two vectors gives a numerical product so $u+1=ucdot1$. But then $u+1=u$, and so the number $1$ acts as the zero-vector in this space.
For Axiom $5$ the negative of a vector $u$ is its reciprocal ($-u=1/u$)
$$u+frac1u=uleft(frac1uright)=1$$
The point here is we want define negative vectors, so we can have additive inverses. We know when we subtract a vector from itself we get the zero-vector. So since $u+(-u)=1$, we need to solve $u+(-u)=u(-u)=1$, and $-u=1/u$ does this.
For Axiom $7$ we need to show $k(u+v)=ku+kv$, so we can have the property of distributivity when multiplying. We have by $(1)$ that $u+v=uv$, so $k(u+v)=k(uv)$. Now use $(2)$ to give $k(uv)=(uv)^k=u^kv^k$, where some power law is invoked. This gives $u^kv^k=u^k+v^k$ by $(1)$, with $u^k+v^k=ku+kv$ by $(2)$, and so $k(u+v)=ku+kv$ as required.
Now do the rest.
answered Jul 28 at 19:30
Daniel Buck
2,2841623
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
The key phrase is "vector addition is numerical multiplication"
That is they define the operation $+_V$ as $u+_Vv=ucdot v$. I explicitly write out the subscript $V$ for the vector addition so as to highlight the addition is not the convention addition that you are familiar with.
We can check that $u+_V1 = u cdot 1=u$.
I will leave the rest as an exercise.
answered Jul 28 at 15:33
Siong Thye Goh
77k134794
yup, the zero vector $0_V= 1$.
– Siong Thye Goh
Jul 28 at 15:47
add a comment |Â
up vote
3
down vote
accepted
The key phrase is "vector addition is numerical multiplication"
That is they define the operation $+_V$ as $u+_Vv=ucdot v$. I explicitly write out the subscript $V$ for the vector addition so as to highlight the addition is not the convention addition that you are familiar with.
We can check that $u+_V1 = u cdot 1=u$.
I will leave the rest as an exercise.
answered Jul 28 at 15:33
Siong Thye Goh
77k134794
yup, the zero vector $0_V= 1$.
– Siong Thye Goh
Jul 28 at 15:47
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
The key phrase is "vector addition is numerical multiplication"
That is they define the operation $+_V$ as $u+_Vv=ucdot v$. I explicitly write out the subscript $V$ for the vector addition so as to highlight the addition is not the convention addition that you are familiar with.
We can check that $u+_V1 = u cdot 1=u$.
I will leave the rest as an exercise.
answered Jul 28 at 15:33
Siong Thye Goh
77k134794
The key phrase is "vector addition is numerical multiplication"
That is they define the operation $+_V$ as $u+_Vv=ucdot v$. I explicitly write out the subscript $V$ for the vector addition so as to highlight the addition is not the convention addition that you are familiar with.
We can check that $u+_V1 = u cdot 1=u$.
I will leave the rest as an exercise.
answered Jul 28 at 15:33
Siong Thye Goh
77k134794
answered Jul 28 at 15:33
Siong Thye Goh
77k134794
answered Jul 28 at 15:33
Siong Thye Goh
77k134794
answered Jul 28 at 15:33
Siong Thye Goh
77k134794
77k134794
yup, the zero vector $0_V= 1$.
– Siong Thye Goh
Jul 28 at 15:47
add a comment |Â
yup, the zero vector $0_V= 1$.
– Siong Thye Goh
Jul 28 at 15:47
yup, the zero vector $0_V= 1$.
– Siong Thye Goh
Jul 28 at 15:47
yup, the zero vector $0_V= 1$.
– Siong Thye Goh
Jul 28 at 15:47
add a comment |Â
up vote
2
down vote
I certainly understand that it looks confusing when the book writes $u+v=uv$ and $ku=u^k,$ so let us instead introduce two operators $oplus : V times V to V$ and $odot : mathbb R times V to V$ satisfying some axioms like
$(u oplus v) oplus w = u oplus (v oplus w)$
and
$k otimes (u oplus v) = (k odot u) oplus (k odot v).$
I won't list all axioms since I don't want to confuse the enumeration with the one in you course literature.
In the example given, they take $V = (0, infty)$ and define $oplus$ and $odot$ by $u oplus v = uv$ and $k odot u = u^k.$ Then, for example, they show the second axiom that I listed:
$$beginalign
k odot (u oplus v)
& = text by def. of $odot$ = (u oplus v)^k \
& = text by def. of $oplus$ = (uv)^k \
& = text by one of the power laws = u^k v^k \
& = text by def. of $oplus$ = u^k oplus v^k \
& = text by def. of $odot$ = (k odot u) oplus (k odot v)
endalign$$
It is very uncommon though with vector spaces where the operations are not derived from ordinary addition and multiplication with a real or complex number. Therefore the operations are normally written as just $+$ and juxtaposition (i.e. $ku$ without an explicit operation).
answered Jul 28 at 18:14
md2perpe
5,79511022
add a comment |Â
up vote
2
down vote
I certainly understand that it looks confusing when the book writes $u+v=uv$ and $ku=u^k,$ so let us instead introduce two operators $oplus : V times V to V$ and $odot : mathbb R times V to V$ satisfying some axioms like
$(u oplus v) oplus w = u oplus (v oplus w)$
and
$k otimes (u oplus v) = (k odot u) oplus (k odot v).$
I won't list all axioms since I don't want to confuse the enumeration with the one in you course literature.
In the example given, they take $V = (0, infty)$ and define $oplus$ and $odot$ by $u oplus v = uv$ and $k odot u = u^k.$ Then, for example, they show the second axiom that I listed:
$$beginalign
k odot (u oplus v)
& = text by def. of $odot$ = (u oplus v)^k \
& = text by def. of $oplus$ = (uv)^k \
& = text by one of the power laws = u^k v^k \
& = text by def. of $oplus$ = u^k oplus v^k \
& = text by def. of $odot$ = (k odot u) oplus (k odot v)
endalign$$
It is very uncommon though with vector spaces where the operations are not derived from ordinary addition and multiplication with a real or complex number. Therefore the operations are normally written as just $+$ and juxtaposition (i.e. $ku$ without an explicit operation).
answered Jul 28 at 18:14
md2perpe
5,79511022
add a comment |Â
up vote
2
down vote
up vote
2
down vote
I certainly understand that it looks confusing when the book writes $u+v=uv$ and $ku=u^k,$ so let us instead introduce two operators $oplus : V times V to V$ and $odot : mathbb R times V to V$ satisfying some axioms like
$(u oplus v) oplus w = u oplus (v oplus w)$
and
$k otimes (u oplus v) = (k odot u) oplus (k odot v).$
I won't list all axioms since I don't want to confuse the enumeration with the one in you course literature.
In the example given, they take $V = (0, infty)$ and define $oplus$ and $odot$ by $u oplus v = uv$ and $k odot u = u^k.$ Then, for example, they show the second axiom that I listed:
$$beginalign
k odot (u oplus v)
& = text by def. of $odot$ = (u oplus v)^k \
& = text by def. of $oplus$ = (uv)^k \
& = text by one of the power laws = u^k v^k \
& = text by def. of $oplus$ = u^k oplus v^k \
& = text by def. of $odot$ = (k odot u) oplus (k odot v)
endalign$$
It is very uncommon though with vector spaces where the operations are not derived from ordinary addition and multiplication with a real or complex number. Therefore the operations are normally written as just $+$ and juxtaposition (i.e. $ku$ without an explicit operation).
answered Jul 28 at 18:14
md2perpe
5,79511022
I certainly understand that it looks confusing when the book writes $u+v=uv$ and $ku=u^k,$ so let us instead introduce two operators $oplus : V times V to V$ and $odot : mathbb R times V to V$ satisfying some axioms like
$(u oplus v) oplus w = u oplus (v oplus w)$
and
$k otimes (u oplus v) = (k odot u) oplus (k odot v).$
I won't list all axioms since I don't want to confuse the enumeration with the one in you course literature.
In the example given, they take $V = (0, infty)$ and define $oplus$ and $odot$ by $u oplus v = uv$ and $k odot u = u^k.$ Then, for example, they show the second axiom that I listed:
$$beginalign
k odot (u oplus v)
& = text by def. of $odot$ = (u oplus v)^k \
& = text by def. of $oplus$ = (uv)^k \
& = text by one of the power laws = u^k v^k \
& = text by def. of $oplus$ = u^k oplus v^k \
& = text by def. of $odot$ = (k odot u) oplus (k odot v)
endalign$$
It is very uncommon though with vector spaces where the operations are not derived from ordinary addition and multiplication with a real or complex number. Therefore the operations are normally written as just $+$ and juxtaposition (i.e. $ku$ without an explicit operation).
answered Jul 28 at 18:14
md2perpe
5,79511022
answered Jul 28 at 18:14
md2perpe
5,79511022
answered Jul 28 at 18:14
md2perpe
5,79511022
answered Jul 28 at 18:14
md2perpe
5,79511022
5,79511022
add a comment |Â
add a comment |Â
up vote
0
down vote
Here $V$ is the set $mathbbR_>0$, so the vector $mathbfv$ is just the real number $v$ (although it is still a vector in $V$).
Now define the operations on $V$:
$$u+v=uv quadtextbfVector addition is numerical multiplicationtag1$$
$$ku=u^k quadtextbfScalar multiplication is numerical exponentiationtag2$$
Now with operations $(1)$ and $(2)$ set up you have to try each axiom in turn to see if it satisfies them.
For Axiom $4$ we want the definition of the zero vector, i.e., the element $i$ that makes $u+i=u$. It's obviously not $i=0$, as $0notin V$ (if it were it would give: $u+0=ucdot0=0$, and so $u=0$, contradicting the fact $u$ is a positive real). In a sense you can work backwards here, since you know $ucdot1=u$, where $u$ and $1$ are seen as numbers to be multiplied. Now looking at $(1)$ you see the addition of two vectors gives a numerical product so $u+1=ucdot1$. But then $u+1=u$, and so the number $1$ acts as the zero-vector in this space.
For Axiom $5$ the negative of a vector $u$ is its reciprocal ($-u=1/u$)
$$u+frac1u=uleft(frac1uright)=1$$
The point here is we want define negative vectors, so we can have additive inverses. We know when we subtract a vector from itself we get the zero-vector. So since $u+(-u)=1$, we need to solve $u+(-u)=u(-u)=1$, and $-u=1/u$ does this.
For Axiom $7$ we need to show $k(u+v)=ku+kv$, so we can have the property of distributivity when multiplying. We have by $(1)$ that $u+v=uv$, so $k(u+v)=k(uv)$. Now use $(2)$ to give $k(uv)=(uv)^k=u^kv^k$, where some power law is invoked. This gives $u^kv^k=u^k+v^k$ by $(1)$, with $u^k+v^k=ku+kv$ by $(2)$, and so $k(u+v)=ku+kv$ as required.
Now do the rest.
answered Jul 28 at 19:30
Daniel Buck
2,2841623
add a comment |Â
up vote
0
down vote
Here $V$ is the set $mathbbR_>0$, so the vector $mathbfv$ is just the real number $v$ (although it is still a vector in $V$).
Now define the operations on $V$:
$$u+v=uv quadtextbfVector addition is numerical multiplicationtag1$$
$$ku=u^k quadtextbfScalar multiplication is numerical exponentiationtag2$$
Now with operations $(1)$ and $(2)$ set up you have to try each axiom in turn to see if it satisfies them.
For Axiom $4$ we want the definition of the zero vector, i.e., the element $i$ that makes $u+i=u$. It's obviously not $i=0$, as $0notin V$ (if it were it would give: $u+0=ucdot0=0$, and so $u=0$, contradicting the fact $u$ is a positive real). In a sense you can work backwards here, since you know $ucdot1=u$, where $u$ and $1$ are seen as numbers to be multiplied. Now looking at $(1)$ you see the addition of two vectors gives a numerical product so $u+1=ucdot1$. But then $u+1=u$, and so the number $1$ acts as the zero-vector in this space.
For Axiom $5$ the negative of a vector $u$ is its reciprocal ($-u=1/u$)
$$u+frac1u=uleft(frac1uright)=1$$
The point here is we want define negative vectors, so we can have additive inverses. We know when we subtract a vector from itself we get the zero-vector. So since $u+(-u)=1$, we need to solve $u+(-u)=u(-u)=1$, and $-u=1/u$ does this.
For Axiom $7$ we need to show $k(u+v)=ku+kv$, so we can have the property of distributivity when multiplying. We have by $(1)$ that $u+v=uv$, so $k(u+v)=k(uv)$. Now use $(2)$ to give $k(uv)=(uv)^k=u^kv^k$, where some power law is invoked. This gives $u^kv^k=u^k+v^k$ by $(1)$, with $u^k+v^k=ku+kv$ by $(2)$, and so $k(u+v)=ku+kv$ as required.
Now do the rest.
answered Jul 28 at 19:30
Daniel Buck
2,2841623
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Here $V$ is the set $mathbbR_>0$, so the vector $mathbfv$ is just the real number $v$ (although it is still a vector in $V$).
Now define the operations on $V$:
$$u+v=uv quadtextbfVector addition is numerical multiplicationtag1$$
$$ku=u^k quadtextbfScalar multiplication is numerical exponentiationtag2$$
Now with operations $(1)$ and $(2)$ set up you have to try each axiom in turn to see if it satisfies them.
For Axiom $4$ we want the definition of the zero vector, i.e., the element $i$ that makes $u+i=u$. It's obviously not $i=0$, as $0notin V$ (if it were it would give: $u+0=ucdot0=0$, and so $u=0$, contradicting the fact $u$ is a positive real). In a sense you can work backwards here, since you know $ucdot1=u$, where $u$ and $1$ are seen as numbers to be multiplied. Now looking at $(1)$ you see the addition of two vectors gives a numerical product so $u+1=ucdot1$. But then $u+1=u$, and so the number $1$ acts as the zero-vector in this space.
For Axiom $5$ the negative of a vector $u$ is its reciprocal ($-u=1/u$)
$$u+frac1u=uleft(frac1uright)=1$$
The point here is we want define negative vectors, so we can have additive inverses. We know when we subtract a vector from itself we get the zero-vector. So since $u+(-u)=1$, we need to solve $u+(-u)=u(-u)=1$, and $-u=1/u$ does this.
For Axiom $7$ we need to show $k(u+v)=ku+kv$, so we can have the property of distributivity when multiplying. We have by $(1)$ that $u+v=uv$, so $k(u+v)=k(uv)$. Now use $(2)$ to give $k(uv)=(uv)^k=u^kv^k$, where some power law is invoked. This gives $u^kv^k=u^k+v^k$ by $(1)$, with $u^k+v^k=ku+kv$ by $(2)$, and so $k(u+v)=ku+kv$ as required.
Now do the rest.
answered Jul 28 at 19:30
Daniel Buck
2,2841623
Here $V$ is the set $mathbbR_>0$, so the vector $mathbfv$ is just the real number $v$ (although it is still a vector in $V$).
Now define the operations on $V$:
$$u+v=uv quadtextbfVector addition is numerical multiplicationtag1$$
$$ku=u^k quadtextbfScalar multiplication is numerical exponentiationtag2$$
Now with operations $(1)$ and $(2)$ set up you have to try each axiom in turn to see if it satisfies them.
For Axiom $4$ we want the definition of the zero vector, i.e., the element $i$ that makes $u+i=u$. It's obviously not $i=0$, as $0notin V$ (if it were it would give: $u+0=ucdot0=0$, and so $u=0$, contradicting the fact $u$ is a positive real). In a sense you can work backwards here, since you know $ucdot1=u$, where $u$ and $1$ are seen as numbers to be multiplied. Now looking at $(1)$ you see the addition of two vectors gives a numerical product so $u+1=ucdot1$. But then $u+1=u$, and so the number $1$ acts as the zero-vector in this space.
For Axiom $5$ the negative of a vector $u$ is its reciprocal ($-u=1/u$)
$$u+frac1u=uleft(frac1uright)=1$$
The point here is we want define negative vectors, so we can have additive inverses. We know when we subtract a vector from itself we get the zero-vector. So since $u+(-u)=1$, we need to solve $u+(-u)=u(-u)=1$, and $-u=1/u$ does this.
For Axiom $7$ we need to show $k(u+v)=ku+kv$, so we can have the property of distributivity when multiplying. We have by $(1)$ that $u+v=uv$, so $k(u+v)=k(uv)$. Now use $(2)$ to give $k(uv)=(uv)^k=u^kv^k$, where some power law is invoked. This gives $u^kv^k=u^k+v^k$ by $(1)$, with $u^k+v^k=ku+kv$ by $(2)$, and so $k(u+v)=ku+kv$ as required.
Now do the rest.
answered Jul 28 at 19:30
Daniel Buck
2,2841623
edited Jul 28 at 21:41
answered Jul 28 at 19:30
Daniel Buck
2,2841623
answered Jul 28 at 19:30
Daniel Buck
2,2841623
answered Jul 28 at 19:30
Daniel Buck
2,2841623
2,2841623
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2865334%2fvector-space-linear-algebra%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
– José Carlos Santos
Jul 28 at 15:33
The axioms are actually not about adding; they are about two operations, one of which is normally written as addition.
– md2perpe
Jul 28 at 17:30