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Why is a space of functions equal to the n-fold product of field F?
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The space of functions whose domain is $S$ and their values lie in $F$ (a field) is shown by:
$$mathbfFunc(S,F) = f: S to F (1)$$
In linear algebra by Peter Petersen on page 11 example 1.4.6 is written that if $S = 1,2,...,n$ then:
$$mathbfFunc(1,2,...,n,F) = F^n (2)$$
The book continues that the vectors in $F^n$ can be also thought of functions.
I cannot understand equation 2 and the statement mentioned above. Why is the space of functions equal to the vector space $F^n$?
For example, if $n = 1$, then we have:
$$mathbfFunc(1,F) = F$$
How does the space of all functions which their input is 1 equal to the field F?
linear-algebra functions vector-spaces
asked 16 hours ago
MOON
1237
add a comment |Â
up vote
0
down vote
favorite
The space of functions whose domain is $S$ and their values lie in $F$ (a field) is shown by:
$$mathbfFunc(S,F) = f: S to F (1)$$
In linear algebra by Peter Petersen on page 11 example 1.4.6 is written that if $S = 1,2,...,n$ then:
$$mathbfFunc(1,2,...,n,F) = F^n (2)$$
The book continues that the vectors in $F^n$ can be also thought of functions.
I cannot understand equation 2 and the statement mentioned above. Why is the space of functions equal to the vector space $F^n$?
For example, if $n = 1$, then we have:
$$mathbfFunc(1,F) = F$$
How does the space of all functions which their input is 1 equal to the field F?
linear-algebra functions vector-spaces
asked 16 hours ago
MOON
1237
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The space of functions whose domain is $S$ and their values lie in $F$ (a field) is shown by:
$$mathbfFunc(S,F) = f: S to F (1)$$
In linear algebra by Peter Petersen on page 11 example 1.4.6 is written that if $S = 1,2,...,n$ then:
$$mathbfFunc(1,2,...,n,F) = F^n (2)$$
The book continues that the vectors in $F^n$ can be also thought of functions.
I cannot understand equation 2 and the statement mentioned above. Why is the space of functions equal to the vector space $F^n$?
For example, if $n = 1$, then we have:
$$mathbfFunc(1,F) = F$$
How does the space of all functions which their input is 1 equal to the field F?
linear-algebra functions vector-spaces
asked 16 hours ago
MOON
1237
The space of functions whose domain is $S$ and their values lie in $F$ (a field) is shown by:
$$mathbfFunc(S,F) = f: S to F (1)$$
In linear algebra by Peter Petersen on page 11 example 1.4.6 is written that if $S = 1,2,...,n$ then:
$$mathbfFunc(1,2,...,n,F) = F^n (2)$$
The book continues that the vectors in $F^n$ can be also thought of functions.
I cannot understand equation 2 and the statement mentioned above. Why is the space of functions equal to the vector space $F^n$?
For example, if $n = 1$, then we have:
$$mathbfFunc(1,F) = F$$
How does the space of all functions which their input is 1 equal to the field F?
linear-algebra functions vector-spaces
asked 16 hours ago
MOON
1237
edited 16 hours ago
asked 16 hours ago
MOON
1237
asked 16 hours ago
MOON
1237
asked 16 hours ago
MOON
1237
1237
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
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up vote
5
down vote
accepted
There is a natural bijection
$$mathbfFunc(1,2,...,n,F) xrightarrowsim F^n$$
given by sending an element $fin mathbfFunc(1,2,...,n,F)$ to the $n$-uple $(f(1),f(2),ldots ,f(n))$. This function is bijective. To see this, you can either show that it is both injective and surjective, either exhibit its inverse. To me, the latter is the most direct way here.
The inverse is clearly given by the map that sends an $n$-uple $(y_1,ldots ,y_n)in F^n$ to the function $fin mathbfFunc(1,2,...,n,F)$ defined by the relations $forall iin1,2,...,n, f(i)=y_i$.
NB: Note that when writes $mathbfFunc(1,2,...,n,F) = F^n$ with an equal sign, this is actually a usual abuse of notation. Both sets clearly are not equal. Their respective objects are of different nature. However, this equal sign is meant to signify "both objects can be identified in a natural manner". The good notion of "to be identified" depends on the respective algebraic structure of both sets. Actually, here we may also see both sets as $F$-vector spaces. It happens that not only we have a natural bijection between them as sets, but this bijection turns out to be a natural isomorphism between them as $F$-vector spaces. That is, it respects the sum and the extern product by elements of $F$, as it can be easily checked.
answered 16 hours ago
Suzet
2,003324
1
Back in a prior geological era, I was taught to designate the set of all functions from $X$ to $Y$ as $Y^X$. Anyway, plus one.
– Lubin
16 hours ago
I was taught this way too @Lubin (which was actually 4 years ago).
– Suzet
16 hours ago
@Lubin It is still used nowadays. And I like this notation a lot. I also prefer writing $2^X$ to $mathcalP(X)$ for the power set of a set $X$.
– Batominovski
16 hours ago
@Suzet Is the following right? We can think of $F^n$ as a vector space over F whose elements are n-tuples and there is a one-to-one correspondence between these n-tuples and the functions (vectors) in the space of functions.
– MOON
16 hours ago
Oh Lord, are you telling me I’m up to date? Horrors!
– Lubin
16 hours ago
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
There is a natural bijection
$$mathbfFunc(1,2,...,n,F) xrightarrowsim F^n$$
given by sending an element $fin mathbfFunc(1,2,...,n,F)$ to the $n$-uple $(f(1),f(2),ldots ,f(n))$. This function is bijective. To see this, you can either show that it is both injective and surjective, either exhibit its inverse. To me, the latter is the most direct way here.
The inverse is clearly given by the map that sends an $n$-uple $(y_1,ldots ,y_n)in F^n$ to the function $fin mathbfFunc(1,2,...,n,F)$ defined by the relations $forall iin1,2,...,n, f(i)=y_i$.
NB: Note that when writes $mathbfFunc(1,2,...,n,F) = F^n$ with an equal sign, this is actually a usual abuse of notation. Both sets clearly are not equal. Their respective objects are of different nature. However, this equal sign is meant to signify "both objects can be identified in a natural manner". The good notion of "to be identified" depends on the respective algebraic structure of both sets. Actually, here we may also see both sets as $F$-vector spaces. It happens that not only we have a natural bijection between them as sets, but this bijection turns out to be a natural isomorphism between them as $F$-vector spaces. That is, it respects the sum and the extern product by elements of $F$, as it can be easily checked.
answered 16 hours ago
Suzet
2,003324
1
Back in a prior geological era, I was taught to designate the set of all functions from $X$ to $Y$ as $Y^X$. Anyway, plus one.
– Lubin
16 hours ago
I was taught this way too @Lubin (which was actually 4 years ago).
– Suzet
16 hours ago
@Lubin It is still used nowadays. And I like this notation a lot. I also prefer writing $2^X$ to $mathcalP(X)$ for the power set of a set $X$.
– Batominovski
16 hours ago
@Suzet Is the following right? We can think of $F^n$ as a vector space over F whose elements are n-tuples and there is a one-to-one correspondence between these n-tuples and the functions (vectors) in the space of functions.
– MOON
16 hours ago
Oh Lord, are you telling me I’m up to date? Horrors!
– Lubin
16 hours ago
 |Â
show 1 more comment
up vote
5
down vote
accepted
There is a natural bijection
$$mathbfFunc(1,2,...,n,F) xrightarrowsim F^n$$
given by sending an element $fin mathbfFunc(1,2,...,n,F)$ to the $n$-uple $(f(1),f(2),ldots ,f(n))$. This function is bijective. To see this, you can either show that it is both injective and surjective, either exhibit its inverse. To me, the latter is the most direct way here.
The inverse is clearly given by the map that sends an $n$-uple $(y_1,ldots ,y_n)in F^n$ to the function $fin mathbfFunc(1,2,...,n,F)$ defined by the relations $forall iin1,2,...,n, f(i)=y_i$.
NB: Note that when writes $mathbfFunc(1,2,...,n,F) = F^n$ with an equal sign, this is actually a usual abuse of notation. Both sets clearly are not equal. Their respective objects are of different nature. However, this equal sign is meant to signify "both objects can be identified in a natural manner". The good notion of "to be identified" depends on the respective algebraic structure of both sets. Actually, here we may also see both sets as $F$-vector spaces. It happens that not only we have a natural bijection between them as sets, but this bijection turns out to be a natural isomorphism between them as $F$-vector spaces. That is, it respects the sum and the extern product by elements of $F$, as it can be easily checked.
answered 16 hours ago
Suzet
2,003324
1
Back in a prior geological era, I was taught to designate the set of all functions from $X$ to $Y$ as $Y^X$. Anyway, plus one.
– Lubin
16 hours ago
I was taught this way too @Lubin (which was actually 4 years ago).
– Suzet
16 hours ago
@Lubin It is still used nowadays. And I like this notation a lot. I also prefer writing $2^X$ to $mathcalP(X)$ for the power set of a set $X$.
– Batominovski
16 hours ago
@Suzet Is the following right? We can think of $F^n$ as a vector space over F whose elements are n-tuples and there is a one-to-one correspondence between these n-tuples and the functions (vectors) in the space of functions.
– MOON
16 hours ago
Oh Lord, are you telling me I’m up to date? Horrors!
– Lubin
16 hours ago
 |Â
show 1 more comment
up vote
5
down vote
accepted
up vote
5
down vote
accepted
There is a natural bijection
$$mathbfFunc(1,2,...,n,F) xrightarrowsim F^n$$
given by sending an element $fin mathbfFunc(1,2,...,n,F)$ to the $n$-uple $(f(1),f(2),ldots ,f(n))$. This function is bijective. To see this, you can either show that it is both injective and surjective, either exhibit its inverse. To me, the latter is the most direct way here.
The inverse is clearly given by the map that sends an $n$-uple $(y_1,ldots ,y_n)in F^n$ to the function $fin mathbfFunc(1,2,...,n,F)$ defined by the relations $forall iin1,2,...,n, f(i)=y_i$.
NB: Note that when writes $mathbfFunc(1,2,...,n,F) = F^n$ with an equal sign, this is actually a usual abuse of notation. Both sets clearly are not equal. Their respective objects are of different nature. However, this equal sign is meant to signify "both objects can be identified in a natural manner". The good notion of "to be identified" depends on the respective algebraic structure of both sets. Actually, here we may also see both sets as $F$-vector spaces. It happens that not only we have a natural bijection between them as sets, but this bijection turns out to be a natural isomorphism between them as $F$-vector spaces. That is, it respects the sum and the extern product by elements of $F$, as it can be easily checked.
answered 16 hours ago
Suzet
2,003324
There is a natural bijection
$$mathbfFunc(1,2,...,n,F) xrightarrowsim F^n$$
given by sending an element $fin mathbfFunc(1,2,...,n,F)$ to the $n$-uple $(f(1),f(2),ldots ,f(n))$. This function is bijective. To see this, you can either show that it is both injective and surjective, either exhibit its inverse. To me, the latter is the most direct way here.
The inverse is clearly given by the map that sends an $n$-uple $(y_1,ldots ,y_n)in F^n$ to the function $fin mathbfFunc(1,2,...,n,F)$ defined by the relations $forall iin1,2,...,n, f(i)=y_i$.
NB: Note that when writes $mathbfFunc(1,2,...,n,F) = F^n$ with an equal sign, this is actually a usual abuse of notation. Both sets clearly are not equal. Their respective objects are of different nature. However, this equal sign is meant to signify "both objects can be identified in a natural manner". The good notion of "to be identified" depends on the respective algebraic structure of both sets. Actually, here we may also see both sets as $F$-vector spaces. It happens that not only we have a natural bijection between them as sets, but this bijection turns out to be a natural isomorphism between them as $F$-vector spaces. That is, it respects the sum and the extern product by elements of $F$, as it can be easily checked.
answered 16 hours ago
Suzet
2,003324
edited 16 hours ago
answered 16 hours ago
Suzet
2,003324
answered 16 hours ago
Suzet
2,003324
answered 16 hours ago
Suzet
2,003324
2,003324
1
Back in a prior geological era, I was taught to designate the set of all functions from $X$ to $Y$ as $Y^X$. Anyway, plus one.
– Lubin
16 hours ago
I was taught this way too @Lubin (which was actually 4 years ago).
– Suzet
16 hours ago
@Lubin It is still used nowadays. And I like this notation a lot. I also prefer writing $2^X$ to $mathcalP(X)$ for the power set of a set $X$.
– Batominovski
16 hours ago
@Suzet Is the following right? We can think of $F^n$ as a vector space over F whose elements are n-tuples and there is a one-to-one correspondence between these n-tuples and the functions (vectors) in the space of functions.
– MOON
16 hours ago
Oh Lord, are you telling me I’m up to date? Horrors!
– Lubin
16 hours ago
 |Â
show 1 more comment
1
Back in a prior geological era, I was taught to designate the set of all functions from $X$ to $Y$ as $Y^X$. Anyway, plus one.
– Lubin
16 hours ago
I was taught this way too @Lubin (which was actually 4 years ago).
– Suzet
16 hours ago
@Lubin It is still used nowadays. And I like this notation a lot. I also prefer writing $2^X$ to $mathcalP(X)$ for the power set of a set $X$.
– Batominovski
16 hours ago
@Suzet Is the following right? We can think of $F^n$ as a vector space over F whose elements are n-tuples and there is a one-to-one correspondence between these n-tuples and the functions (vectors) in the space of functions.
– MOON
16 hours ago
Oh Lord, are you telling me I’m up to date? Horrors!
– Lubin
16 hours ago
1
1
Back in a prior geological era, I was taught to designate the set of all functions from $X$ to $Y$ as $Y^X$. Anyway, plus one.
– Lubin
16 hours ago
Back in a prior geological era, I was taught to designate the set of all functions from $X$ to $Y$ as $Y^X$. Anyway, plus one.
– Lubin
16 hours ago
I was taught this way too @Lubin (which was actually 4 years ago).
– Suzet
16 hours ago
I was taught this way too @Lubin (which was actually 4 years ago).
– Suzet
16 hours ago
@Lubin It is still used nowadays. And I like this notation a lot. I also prefer writing $2^X$ to $mathcalP(X)$ for the power set of a set $X$.
– Batominovski
16 hours ago
@Lubin It is still used nowadays. And I like this notation a lot. I also prefer writing $2^X$ to $mathcalP(X)$ for the power set of a set $X$.
– Batominovski
16 hours ago
@Suzet Is the following right? We can think of $F^n$ as a vector space over F whose elements are n-tuples and there is a one-to-one correspondence between these n-tuples and the functions (vectors) in the space of functions.
– MOON
16 hours ago
@Suzet Is the following right? We can think of $F^n$ as a vector space over F whose elements are n-tuples and there is a one-to-one correspondence between these n-tuples and the functions (vectors) in the space of functions.
– MOON
16 hours ago
Oh Lord, are you telling me I’m up to date? Horrors!
– Lubin
16 hours ago
Oh Lord, are you telling me I’m up to date? Horrors!
– Lubin
16 hours ago
 |Â
show 1 more comment
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