Isomorphism of X to itself. [closed]
$begingroup$
Let $(X, +, bullet)$ and $(X,tilde{+}, tilde{bullet})$ be vector spaces over $mathbb{R}$, and dim$(X, +, bullet) =n$ (the set $X$ is the same in both spaces). Show that $(X, +, bullet)$ is isomorphic to $(X,tilde{+}, tilde{bullet})$.
linear-algebra vector-space-isomorphism
asked Dec 26 '18 at 12:49
HelloDarknessHelloDarkness
165
$endgroup$
closed as off-topic by Saad, metamorphy, Davide Giraudo, Eric Wofsey, Paul Frost Dec 26 '18 at 23:24
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, metamorphy, Davide Giraudo, Eric Wofsey, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Let $(X, +, bullet)$ and $(X,tilde{+}, tilde{bullet})$ be vector spaces over $mathbb{R}$, and dim$(X, +, bullet) =n$ (the set $X$ is the same in both spaces). Show that $(X, +, bullet)$ is isomorphic to $(X,tilde{+}, tilde{bullet})$.
linear-algebra vector-space-isomorphism
asked Dec 26 '18 at 12:49
HelloDarknessHelloDarkness
165
$endgroup$
closed as off-topic by Saad, metamorphy, Davide Giraudo, Eric Wofsey, Paul Frost Dec 26 '18 at 23:24
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, metamorphy, Davide Giraudo, Eric Wofsey, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
In this generality it's not true. Are there any connections between the operations?
$endgroup$
– Berci
Dec 26 '18 at 12:55
$begingroup$
Sorry for how I formulated the question, I didn't know if this had to be true, thanks! Could you give me counter example?, I find this really weird haha.
$endgroup$
– HelloDarkness
Dec 26 '18 at 12:59
2
$begingroup$
The key is that we have $|Bbb R^2|=|Bbb R|$, i.e. there's a bijection between $Bbb R^2$ and $Bbb R$, so e.g. we can impose a one dimensional vector space structure on $Bbb R^2$ via this bijection.
$endgroup$
– Berci
Dec 26 '18 at 13:04
add a comment |
$begingroup$
Let $(X, +, bullet)$ and $(X,tilde{+}, tilde{bullet})$ be vector spaces over $mathbb{R}$, and dim$(X, +, bullet) =n$ (the set $X$ is the same in both spaces). Show that $(X, +, bullet)$ is isomorphic to $(X,tilde{+}, tilde{bullet})$.
linear-algebra vector-space-isomorphism
asked Dec 26 '18 at 12:49
HelloDarknessHelloDarkness
165
$endgroup$
Let $(X, +, bullet)$ and $(X,tilde{+}, tilde{bullet})$ be vector spaces over $mathbb{R}$, and dim$(X, +, bullet) =n$ (the set $X$ is the same in both spaces). Show that $(X, +, bullet)$ is isomorphic to $(X,tilde{+}, tilde{bullet})$.
linear-algebra vector-space-isomorphism
linear-algebra vector-space-isomorphism
asked Dec 26 '18 at 12:49
HelloDarknessHelloDarkness
165
asked Dec 26 '18 at 12:49
HelloDarknessHelloDarkness
165
asked Dec 26 '18 at 12:49
HelloDarknessHelloDarkness
165
asked Dec 26 '18 at 12:49
HelloDarknessHelloDarkness
165
asked Dec 26 '18 at 12:49
HelloDarknessHelloDarkness
165
165
closed as off-topic by Saad, metamorphy, Davide Giraudo, Eric Wofsey, Paul Frost Dec 26 '18 at 23:24
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, metamorphy, Davide Giraudo, Eric Wofsey, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, metamorphy, Davide Giraudo, Eric Wofsey, Paul Frost Dec 26 '18 at 23:24
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, metamorphy, Davide Giraudo, Eric Wofsey, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
In this generality it's not true. Are there any connections between the operations?
$endgroup$
– Berci
Dec 26 '18 at 12:55
$begingroup$
Sorry for how I formulated the question, I didn't know if this had to be true, thanks! Could you give me counter example?, I find this really weird haha.
$endgroup$
– HelloDarkness
Dec 26 '18 at 12:59
2
$begingroup$
The key is that we have $|Bbb R^2|=|Bbb R|$, i.e. there's a bijection between $Bbb R^2$ and $Bbb R$, so e.g. we can impose a one dimensional vector space structure on $Bbb R^2$ via this bijection.
$endgroup$
– Berci
Dec 26 '18 at 13:04
add a comment |
1
$begingroup$
In this generality it's not true. Are there any connections between the operations?
$endgroup$
– Berci
Dec 26 '18 at 12:55
$begingroup$
Sorry for how I formulated the question, I didn't know if this had to be true, thanks! Could you give me counter example?, I find this really weird haha.
$endgroup$
– HelloDarkness
Dec 26 '18 at 12:59
2
$begingroup$
The key is that we have $|Bbb R^2|=|Bbb R|$, i.e. there's a bijection between $Bbb R^2$ and $Bbb R$, so e.g. we can impose a one dimensional vector space structure on $Bbb R^2$ via this bijection.
$endgroup$
– Berci
Dec 26 '18 at 13:04
1
1
$begingroup$
In this generality it's not true. Are there any connections between the operations?
$endgroup$
– Berci
Dec 26 '18 at 12:55
$begingroup$
In this generality it's not true. Are there any connections between the operations?
$endgroup$
– Berci
Dec 26 '18 at 12:55
$begingroup$
Sorry for how I formulated the question, I didn't know if this had to be true, thanks! Could you give me counter example?, I find this really weird haha.
$endgroup$
– HelloDarkness
Dec 26 '18 at 12:59
$begingroup$
Sorry for how I formulated the question, I didn't know if this had to be true, thanks! Could you give me counter example?, I find this really weird haha.
$endgroup$
– HelloDarkness
Dec 26 '18 at 12:59
2
2
$begingroup$
The key is that we have $|Bbb R^2|=|Bbb R|$, i.e. there's a bijection between $Bbb R^2$ and $Bbb R$, so e.g. we can impose a one dimensional vector space structure on $Bbb R^2$ via this bijection.
$endgroup$
– Berci
Dec 26 '18 at 13:04
$begingroup$
The key is that we have $|Bbb R^2|=|Bbb R|$, i.e. there's a bijection between $Bbb R^2$ and $Bbb R$, so e.g. we can impose a one dimensional vector space structure on $Bbb R^2$ via this bijection.
$endgroup$
– Berci
Dec 26 '18 at 13:04
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Since two finite dimensional spaces (over the same field) are isomorphic if and only if they have the same dimension, it follows that:
$$(X, +, bullet)cong(X,tilde{+}, tilde{bullet})$$
if and only if $$dim (X,tilde{+}, tilde{bullet}) = n$$
answered Dec 26 '18 at 13:04
Math_QEDMath_QED
7,58431452
$endgroup$
add a comment |
$begingroup$
The fact that the set is the same is a distraction.
Choose a basis $(v_1, dots,v_n)$ for $(X,+,cdot)$. Choose a basis $w_1, dots w_n$ for $(X, tilde{+}, tilde{cdot})$.
Define $T:X to X$ by $T(v_i)=w_i$ and extend the definition to $X$ linearly ($T(alpha_1 cdot v_i+ alpha_2 cdot v_j)=alpha_1 tilde{cdot}T(v_i)tilde{+}alpha_2tilde{cdot}T(v_j)$). This map will be surjective as its onto the basis elements of $(X, tilde{+}, tilde{cdot})$. By the rank nullity theorem it will be invective.
(Check that this defines a linear map)
(This answer assumes that the dimensions are the same)
answered Dec 26 '18 at 13:01
GalGal
92
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since two finite dimensional spaces (over the same field) are isomorphic if and only if they have the same dimension, it follows that:
$$(X, +, bullet)cong(X,tilde{+}, tilde{bullet})$$
if and only if $$dim (X,tilde{+}, tilde{bullet}) = n$$
answered Dec 26 '18 at 13:04
Math_QEDMath_QED
7,58431452
$endgroup$
add a comment |
$begingroup$
Since two finite dimensional spaces (over the same field) are isomorphic if and only if they have the same dimension, it follows that:
$$(X, +, bullet)cong(X,tilde{+}, tilde{bullet})$$
if and only if $$dim (X,tilde{+}, tilde{bullet}) = n$$
answered Dec 26 '18 at 13:04
Math_QEDMath_QED
7,58431452
$endgroup$
add a comment |
$begingroup$
Since two finite dimensional spaces (over the same field) are isomorphic if and only if they have the same dimension, it follows that:
$$(X, +, bullet)cong(X,tilde{+}, tilde{bullet})$$
if and only if $$dim (X,tilde{+}, tilde{bullet}) = n$$
answered Dec 26 '18 at 13:04
Math_QEDMath_QED
7,58431452
$endgroup$
Since two finite dimensional spaces (over the same field) are isomorphic if and only if they have the same dimension, it follows that:
$$(X, +, bullet)cong(X,tilde{+}, tilde{bullet})$$
if and only if $$dim (X,tilde{+}, tilde{bullet}) = n$$
answered Dec 26 '18 at 13:04
Math_QEDMath_QED
7,58431452
answered Dec 26 '18 at 13:04
Math_QEDMath_QED
7,58431452
answered Dec 26 '18 at 13:04
Math_QEDMath_QED
7,58431452
answered Dec 26 '18 at 13:04
Math_QEDMath_QED
7,58431452
7,58431452
add a comment |
add a comment |
$begingroup$
The fact that the set is the same is a distraction.
Choose a basis $(v_1, dots,v_n)$ for $(X,+,cdot)$. Choose a basis $w_1, dots w_n$ for $(X, tilde{+}, tilde{cdot})$.
Define $T:X to X$ by $T(v_i)=w_i$ and extend the definition to $X$ linearly ($T(alpha_1 cdot v_i+ alpha_2 cdot v_j)=alpha_1 tilde{cdot}T(v_i)tilde{+}alpha_2tilde{cdot}T(v_j)$). This map will be surjective as its onto the basis elements of $(X, tilde{+}, tilde{cdot})$. By the rank nullity theorem it will be invective.
(Check that this defines a linear map)
(This answer assumes that the dimensions are the same)
answered Dec 26 '18 at 13:01
GalGal
92
$endgroup$
add a comment |
$begingroup$
The fact that the set is the same is a distraction.
Choose a basis $(v_1, dots,v_n)$ for $(X,+,cdot)$. Choose a basis $w_1, dots w_n$ for $(X, tilde{+}, tilde{cdot})$.
Define $T:X to X$ by $T(v_i)=w_i$ and extend the definition to $X$ linearly ($T(alpha_1 cdot v_i+ alpha_2 cdot v_j)=alpha_1 tilde{cdot}T(v_i)tilde{+}alpha_2tilde{cdot}T(v_j)$). This map will be surjective as its onto the basis elements of $(X, tilde{+}, tilde{cdot})$. By the rank nullity theorem it will be invective.
(Check that this defines a linear map)
(This answer assumes that the dimensions are the same)
answered Dec 26 '18 at 13:01
GalGal
92
$endgroup$
add a comment |
$begingroup$
The fact that the set is the same is a distraction.
Choose a basis $(v_1, dots,v_n)$ for $(X,+,cdot)$. Choose a basis $w_1, dots w_n$ for $(X, tilde{+}, tilde{cdot})$.
Define $T:X to X$ by $T(v_i)=w_i$ and extend the definition to $X$ linearly ($T(alpha_1 cdot v_i+ alpha_2 cdot v_j)=alpha_1 tilde{cdot}T(v_i)tilde{+}alpha_2tilde{cdot}T(v_j)$). This map will be surjective as its onto the basis elements of $(X, tilde{+}, tilde{cdot})$. By the rank nullity theorem it will be invective.
(Check that this defines a linear map)
(This answer assumes that the dimensions are the same)
answered Dec 26 '18 at 13:01
GalGal
92
$endgroup$
The fact that the set is the same is a distraction.
Choose a basis $(v_1, dots,v_n)$ for $(X,+,cdot)$. Choose a basis $w_1, dots w_n$ for $(X, tilde{+}, tilde{cdot})$.
Define $T:X to X$ by $T(v_i)=w_i$ and extend the definition to $X$ linearly ($T(alpha_1 cdot v_i+ alpha_2 cdot v_j)=alpha_1 tilde{cdot}T(v_i)tilde{+}alpha_2tilde{cdot}T(v_j)$). This map will be surjective as its onto the basis elements of $(X, tilde{+}, tilde{cdot})$. By the rank nullity theorem it will be invective.
(Check that this defines a linear map)
(This answer assumes that the dimensions are the same)
answered Dec 26 '18 at 13:01
GalGal
92
answered Dec 26 '18 at 13:01
GalGal
92
answered Dec 26 '18 at 13:01
GalGal
92
answered Dec 26 '18 at 13:01
GalGal
92
92
add a comment |
add a comment |
1
$begingroup$
In this generality it's not true. Are there any connections between the operations?
$endgroup$
– Berci
Dec 26 '18 at 12:55
$begingroup$
Sorry for how I formulated the question, I didn't know if this had to be true, thanks! Could you give me counter example?, I find this really weird haha.
$endgroup$
– HelloDarkness
Dec 26 '18 at 12:59
2
$begingroup$
The key is that we have $|Bbb R^2|=|Bbb R|$, i.e. there's a bijection between $Bbb R^2$ and $Bbb R$, so e.g. we can impose a one dimensional vector space structure on $Bbb R^2$ via this bijection.
$endgroup$
– Berci
Dec 26 '18 at 13:04