Mixing
Free vector space of a finite set is isomorphic to its function space
$begingroup$
Let $X$ be a finite set and let $F(X, mathbb{C})$ be the space of functions $f: X rightarrow mathbb{C}$. How can I show $mathbb{C}X$, the free vector space generated by $X$ is isomorphic to the above function space? Isn't the function space infinite dimensional, so how can they be isomorphic?
linear-algebra functional-analysis
asked Jan 16 at 3:21
David FengDavid Feng
979
$endgroup$
add a comment |
$begingroup$
Let $X$ be a finite set and let $F(X, mathbb{C})$ be the space of functions $f: X rightarrow mathbb{C}$. How can I show $mathbb{C}X$, the free vector space generated by $X$ is isomorphic to the above function space? Isn't the function space infinite dimensional, so how can they be isomorphic?
linear-algebra functional-analysis
asked Jan 16 at 3:21
David FengDavid Feng
979
$endgroup$
$begingroup$
$X$ is finite, so it is not infinite dimensional.
$endgroup$
– Randall
Jan 16 at 3:22
1
$begingroup$
An element of $mathbb{C}X$ is a formal sum $sum_{k=1}^n z_k x_k$ where $X={x_1, ldots, x_n}$. Do you see how to use this to build a function $f: X to mathbb{C}$?
$endgroup$
– Randall
Jan 16 at 3:24
add a comment |
$begingroup$
Let $X$ be a finite set and let $F(X, mathbb{C})$ be the space of functions $f: X rightarrow mathbb{C}$. How can I show $mathbb{C}X$, the free vector space generated by $X$ is isomorphic to the above function space? Isn't the function space infinite dimensional, so how can they be isomorphic?
linear-algebra functional-analysis
asked Jan 16 at 3:21
David FengDavid Feng
979
$endgroup$
Let $X$ be a finite set and let $F(X, mathbb{C})$ be the space of functions $f: X rightarrow mathbb{C}$. How can I show $mathbb{C}X$, the free vector space generated by $X$ is isomorphic to the above function space? Isn't the function space infinite dimensional, so how can they be isomorphic?
linear-algebra functional-analysis
linear-algebra functional-analysis
asked Jan 16 at 3:21
David FengDavid Feng
979
asked Jan 16 at 3:21
David FengDavid Feng
979
asked Jan 16 at 3:21
David FengDavid Feng
979
asked Jan 16 at 3:21
David FengDavid Feng
979
asked Jan 16 at 3:21
David FengDavid Feng
979
979
$begingroup$
$X$ is finite, so it is not infinite dimensional.
$endgroup$
– Randall
Jan 16 at 3:22
1
$begingroup$
An element of $mathbb{C}X$ is a formal sum $sum_{k=1}^n z_k x_k$ where $X={x_1, ldots, x_n}$. Do you see how to use this to build a function $f: X to mathbb{C}$?
$endgroup$
– Randall
Jan 16 at 3:24
add a comment |
$begingroup$
$X$ is finite, so it is not infinite dimensional.
$endgroup$
– Randall
Jan 16 at 3:22
1
$begingroup$
An element of $mathbb{C}X$ is a formal sum $sum_{k=1}^n z_k x_k$ where $X={x_1, ldots, x_n}$. Do you see how to use this to build a function $f: X to mathbb{C}$?
$endgroup$
– Randall
Jan 16 at 3:24
$begingroup$
$X$ is finite, so it is not infinite dimensional.
$endgroup$
– Randall
Jan 16 at 3:22
$begingroup$
$X$ is finite, so it is not infinite dimensional.
$endgroup$
– Randall
Jan 16 at 3:22
1
1
$begingroup$
An element of $mathbb{C}X$ is a formal sum $sum_{k=1}^n z_k x_k$ where $X={x_1, ldots, x_n}$. Do you see how to use this to build a function $f: X to mathbb{C}$?
$endgroup$
– Randall
Jan 16 at 3:24
$begingroup$
An element of $mathbb{C}X$ is a formal sum $sum_{k=1}^n z_k x_k$ where $X={x_1, ldots, x_n}$. Do you see how to use this to build a function $f: X to mathbb{C}$?
$endgroup$
– Randall
Jan 16 at 3:24
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Consider the map
$$varphi : F(X,mathbb{C}) to mathbb{C}X\
f mapsto sum_{xin X} f(x)x$$
we want to show tht this is a bijection for $X$ finite. First consider surjectivity:
let $y = sum_{xin X} lambda_x x$ then this has the preimage $$z: X to mathbb{C} \ x mapsto lambda_x.$$
Now for injectivity:
since the set ${x}_{xin X} subset mathbb{C}X$ is linear independent by definition, we have that $0= varphi(f)=sum_{xin X} f(x)x$ if and only if $f(x)=0$ for all $x in X$, but this ust means that $f$ is the zero map, and hence we have that $ker (varphi) =0$ which proves injectivity.
So since $varphi$ is injective and surjective, it is bijective. Also it is linear, which is an easy calculation, respectively as $varphi$ is the finite sum of evalutions clear.
And so $varphi$ is an iso
answered Jan 16 at 8:13
EnkiduEnkidu
1,36719
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Consider the map
$$varphi : F(X,mathbb{C}) to mathbb{C}X\
f mapsto sum_{xin X} f(x)x$$
we want to show tht this is a bijection for $X$ finite. First consider surjectivity:
let $y = sum_{xin X} lambda_x x$ then this has the preimage $$z: X to mathbb{C} \ x mapsto lambda_x.$$
Now for injectivity:
since the set ${x}_{xin X} subset mathbb{C}X$ is linear independent by definition, we have that $0= varphi(f)=sum_{xin X} f(x)x$ if and only if $f(x)=0$ for all $x in X$, but this ust means that $f$ is the zero map, and hence we have that $ker (varphi) =0$ which proves injectivity.
So since $varphi$ is injective and surjective, it is bijective. Also it is linear, which is an easy calculation, respectively as $varphi$ is the finite sum of evalutions clear.
And so $varphi$ is an iso
answered Jan 16 at 8:13
EnkiduEnkidu
1,36719
$endgroup$
add a comment |
$begingroup$
Consider the map
$$varphi : F(X,mathbb{C}) to mathbb{C}X\
f mapsto sum_{xin X} f(x)x$$
we want to show tht this is a bijection for $X$ finite. First consider surjectivity:
let $y = sum_{xin X} lambda_x x$ then this has the preimage $$z: X to mathbb{C} \ x mapsto lambda_x.$$
Now for injectivity:
since the set ${x}_{xin X} subset mathbb{C}X$ is linear independent by definition, we have that $0= varphi(f)=sum_{xin X} f(x)x$ if and only if $f(x)=0$ for all $x in X$, but this ust means that $f$ is the zero map, and hence we have that $ker (varphi) =0$ which proves injectivity.
So since $varphi$ is injective and surjective, it is bijective. Also it is linear, which is an easy calculation, respectively as $varphi$ is the finite sum of evalutions clear.
And so $varphi$ is an iso
answered Jan 16 at 8:13
EnkiduEnkidu
1,36719
$endgroup$
add a comment |
$begingroup$
Consider the map
$$varphi : F(X,mathbb{C}) to mathbb{C}X\
f mapsto sum_{xin X} f(x)x$$
we want to show tht this is a bijection for $X$ finite. First consider surjectivity:
let $y = sum_{xin X} lambda_x x$ then this has the preimage $$z: X to mathbb{C} \ x mapsto lambda_x.$$
Now for injectivity:
since the set ${x}_{xin X} subset mathbb{C}X$ is linear independent by definition, we have that $0= varphi(f)=sum_{xin X} f(x)x$ if and only if $f(x)=0$ for all $x in X$, but this ust means that $f$ is the zero map, and hence we have that $ker (varphi) =0$ which proves injectivity.
So since $varphi$ is injective and surjective, it is bijective. Also it is linear, which is an easy calculation, respectively as $varphi$ is the finite sum of evalutions clear.
And so $varphi$ is an iso
answered Jan 16 at 8:13
EnkiduEnkidu
1,36719
$endgroup$
Consider the map
$$varphi : F(X,mathbb{C}) to mathbb{C}X\
f mapsto sum_{xin X} f(x)x$$
we want to show tht this is a bijection for $X$ finite. First consider surjectivity:
let $y = sum_{xin X} lambda_x x$ then this has the preimage $$z: X to mathbb{C} \ x mapsto lambda_x.$$
Now for injectivity:
since the set ${x}_{xin X} subset mathbb{C}X$ is linear independent by definition, we have that $0= varphi(f)=sum_{xin X} f(x)x$ if and only if $f(x)=0$ for all $x in X$, but this ust means that $f$ is the zero map, and hence we have that $ker (varphi) =0$ which proves injectivity.
So since $varphi$ is injective and surjective, it is bijective. Also it is linear, which is an easy calculation, respectively as $varphi$ is the finite sum of evalutions clear.
And so $varphi$ is an iso
answered Jan 16 at 8:13
EnkiduEnkidu
1,36719
answered Jan 16 at 8:13
EnkiduEnkidu
1,36719
answered Jan 16 at 8:13
EnkiduEnkidu
1,36719
answered Jan 16 at 8:13
EnkiduEnkidu
1,36719
1,36719
add a comment |
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$begingroup$
$X$ is finite, so it is not infinite dimensional.
$endgroup$
– Randall
Jan 16 at 3:22
1
$begingroup$
An element of $mathbb{C}X$ is a formal sum $sum_{k=1}^n z_k x_k$ where $X={x_1, ldots, x_n}$. Do you see how to use this to build a function $f: X to mathbb{C}$?
$endgroup$
– Randall
Jan 16 at 3:24