Mixing
Isomorphism Clarification and Identification
Clash Royale CLAN TAG#URR8PPP
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1
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I understand that to claim isomorphism, algebraic properties must be preserved...so....a) closed under multiplication and closed under addition.
However, I am unsure how to apply these condition-testing properties in the context of polynomial/space based questions
For example: Which of the following is isomorphic to a subspace of R^3x4
- P9
- P11
- Upper triangular matrices in R^2x3
- R12
How can I show closed under addition/multiplication in each of these contexts?
linear-algebra linear-transformations vector-space-isomorphism
asked Aug 3 at 13:28
PERTURBATIONFLOW
396
 |Â
show 1 more comment
up vote
1
down vote
favorite
I understand that to claim isomorphism, algebraic properties must be preserved...so....a) closed under multiplication and closed under addition.
However, I am unsure how to apply these condition-testing properties in the context of polynomial/space based questions
For example: Which of the following is isomorphic to a subspace of R^3x4
- P9
- P11
- Upper triangular matrices in R^2x3
- R12
How can I show closed under addition/multiplication in each of these contexts?
linear-algebra linear-transformations vector-space-isomorphism
asked Aug 3 at 13:28
PERTURBATIONFLOW
396
It is a little hard to tell which spaces you mean in your enumeration, can you elaborate on that. Also, do you mean with R^3x4 the space $mathbbR^(3,4)$ of $3$ by $4$ matrices?
– zzuussee
Aug 3 at 13:39
Yes, that is the space. P9 refers to characteristic polynomial with degree 9, and p11 refers to characteristic polynomial with degree 11
– PERTURBATIONFLOW
Aug 3 at 14:50
So, I assume that with characteristic polynomial with degree $9$ you mean the matrices which have a characteristic polynomial of degree $9$, i.e. all $9times 9$ matrices?
– zzuussee
Aug 3 at 14:53
1
Vector spaces are classified by their dimension. For vector spaces $V$ and $W$ there exists an injection $V to W$ if and only if $dim(V) leq dim(W)$. Hence for your problem you just need to figure out the dimension of each space
– leibnewtz
Aug 3 at 15:24
@leibnewtz would all of these choices be subspaces:
– PERTURBATIONFLOW
Aug 4 at 0:59
 |Â
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I understand that to claim isomorphism, algebraic properties must be preserved...so....a) closed under multiplication and closed under addition.
However, I am unsure how to apply these condition-testing properties in the context of polynomial/space based questions
For example: Which of the following is isomorphic to a subspace of R^3x4
- P9
- P11
- Upper triangular matrices in R^2x3
- R12
How can I show closed under addition/multiplication in each of these contexts?
linear-algebra linear-transformations vector-space-isomorphism
asked Aug 3 at 13:28
PERTURBATIONFLOW
396
I understand that to claim isomorphism, algebraic properties must be preserved...so....a) closed under multiplication and closed under addition.
However, I am unsure how to apply these condition-testing properties in the context of polynomial/space based questions
For example: Which of the following is isomorphic to a subspace of R^3x4
- P9
- P11
- Upper triangular matrices in R^2x3
- R12
How can I show closed under addition/multiplication in each of these contexts?
linear-algebra linear-transformations vector-space-isomorphism
asked Aug 3 at 13:28
PERTURBATIONFLOW
396
asked Aug 3 at 13:28
PERTURBATIONFLOW
396
asked Aug 3 at 13:28
PERTURBATIONFLOW
396
asked Aug 3 at 13:28
PERTURBATIONFLOW
396
396
It is a little hard to tell which spaces you mean in your enumeration, can you elaborate on that. Also, do you mean with R^3x4 the space $mathbbR^(3,4)$ of $3$ by $4$ matrices?
– zzuussee
Aug 3 at 13:39
Yes, that is the space. P9 refers to characteristic polynomial with degree 9, and p11 refers to characteristic polynomial with degree 11
– PERTURBATIONFLOW
Aug 3 at 14:50
So, I assume that with characteristic polynomial with degree $9$ you mean the matrices which have a characteristic polynomial of degree $9$, i.e. all $9times 9$ matrices?
– zzuussee
Aug 3 at 14:53
1
Vector spaces are classified by their dimension. For vector spaces $V$ and $W$ there exists an injection $V to W$ if and only if $dim(V) leq dim(W)$. Hence for your problem you just need to figure out the dimension of each space
– leibnewtz
Aug 3 at 15:24
@leibnewtz would all of these choices be subspaces:
– PERTURBATIONFLOW
Aug 4 at 0:59
 |Â
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It is a little hard to tell which spaces you mean in your enumeration, can you elaborate on that. Also, do you mean with R^3x4 the space $mathbbR^(3,4)$ of $3$ by $4$ matrices?
– zzuussee
Aug 3 at 13:39
Yes, that is the space. P9 refers to characteristic polynomial with degree 9, and p11 refers to characteristic polynomial with degree 11
– PERTURBATIONFLOW
Aug 3 at 14:50
So, I assume that with characteristic polynomial with degree $9$ you mean the matrices which have a characteristic polynomial of degree $9$, i.e. all $9times 9$ matrices?
– zzuussee
Aug 3 at 14:53
1
Vector spaces are classified by their dimension. For vector spaces $V$ and $W$ there exists an injection $V to W$ if and only if $dim(V) leq dim(W)$. Hence for your problem you just need to figure out the dimension of each space
– leibnewtz
Aug 3 at 15:24
@leibnewtz would all of these choices be subspaces:
– PERTURBATIONFLOW
Aug 4 at 0:59
It is a little hard to tell which spaces you mean in your enumeration, can you elaborate on that. Also, do you mean with R^3x4 the space $mathbbR^(3,4)$ of $3$ by $4$ matrices?
– zzuussee
Aug 3 at 13:39
It is a little hard to tell which spaces you mean in your enumeration, can you elaborate on that. Also, do you mean with R^3x4 the space $mathbbR^(3,4)$ of $3$ by $4$ matrices?
– zzuussee
Aug 3 at 13:39
Yes, that is the space. P9 refers to characteristic polynomial with degree 9, and p11 refers to characteristic polynomial with degree 11
– PERTURBATIONFLOW
Aug 3 at 14:50
Yes, that is the space. P9 refers to characteristic polynomial with degree 9, and p11 refers to characteristic polynomial with degree 11
– PERTURBATIONFLOW
Aug 3 at 14:50
So, I assume that with characteristic polynomial with degree $9$ you mean the matrices which have a characteristic polynomial of degree $9$, i.e. all $9times 9$ matrices?
– zzuussee
Aug 3 at 14:53
So, I assume that with characteristic polynomial with degree $9$ you mean the matrices which have a characteristic polynomial of degree $9$, i.e. all $9times 9$ matrices?
– zzuussee
Aug 3 at 14:53
1
1
Vector spaces are classified by their dimension. For vector spaces $V$ and $W$ there exists an injection $V to W$ if and only if $dim(V) leq dim(W)$. Hence for your problem you just need to figure out the dimension of each space
– leibnewtz
Aug 3 at 15:24
Vector spaces are classified by their dimension. For vector spaces $V$ and $W$ there exists an injection $V to W$ if and only if $dim(V) leq dim(W)$. Hence for your problem you just need to figure out the dimension of each space
– leibnewtz
Aug 3 at 15:24
@leibnewtz would all of these choices be subspaces:
– PERTURBATIONFLOW
Aug 4 at 0:59
@leibnewtz would all of these choices be subspaces:
– PERTURBATIONFLOW
Aug 4 at 0:59
 |Â
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2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
The simpler way to find out if two finite dimensional vector spaces are isomorphic, as others said, is to find out what if the dimension of the vector space that you're studying. So we can state a little "theorem"
Two finite dimensional vector spaces $V,W$ are isomorphic iff $$dim(V)=dim(W)$$ Here are some references.
After stating this we can easily find out which of your vector spaces is isomorphic to $mathbbR^3times 4$. First of all we have that $$dim(mathbbR^3times 4) = 12$$ and it's easy to see because to write down a $3$ by $4$ matrix you need $12$ indipendent components.
Now let's see the other dimensions $$beginalign
P_9 &&rightarrow&&dim(P_9)=9tag1\
P_11 &&rightarrow&&dim(P_11)=11tag2\
textupper triangular mathbbR^2times 3&&rightarrow&&dim(textupper triangular mathbbR^2times 3)=5tag3\
mathbbR^12&&rightarrow&&dim(mathbbR^12)=12tag4
endalign$$
For the first and second one the dimension is trivial to find: a polynomial of degree less and equal to $n$ has $n$ coefficients, so the vector space is $n$-dimensional. The third one is more tricky: one would think that the dimension would be $6$ but an upper triangular $2$ by $3$ matrix is of the form $$left(beginmatrixa_11&a_12&a_13\0&a_22&a_23endmatrixright)$$
so the element $a_21=0$ always. The only independent components ar the $5$ left. The fifth is just $mathbbR^n$ so it's dimension is $n$.
Clearly the only isomorphic space to $mathbbR^3times2$ in this list is $mathbbR^12$. If you want to find an isomorphism, which is not requested if the only thing you want is to find if to vector spaces are isomorphic, you could use this isomorphism $$phi:mathbbR^3times4rightarrow mathbbR^12 \
left(beginmatrixa_11&a_12&a_13&a_14\a_21&a_22&a_23&a_24\a_31&a_32&a_33&a_34endmatrixright)mapsto (a_11,a_12,a_13,a_14,a_21,a_22,a_23,a_24,a_31,a_32,a_33,a_34)$$ i.e. you take all the entries in the matrix and map them to a vector that has as components the $a_ij$ of the matrix.
answered 2 days ago
Davide Morgante
1,630220
Can you elaborate further as to why P9 and P11 are not isomorphic. I simply thought because their dimensions are smaller, they can be isomorphic
– PERTURBATIONFLOW
yesterday
If being isomorphic means to have an isomorphism between two spaces, necessary the two spces have to have the same dimension. An isomorphism is a map with a one-to-one correspondence: if you have $n$ elements in a set and $m$ in another, with $m>n$, you cannot construct an isomorphism because, at most, $n$ of the $m$ elements can be mapped from one set to another
– Davide Morgante
yesterday
I'm from a phone now and I cannot give you a thorough explanation! When I'll be on pc I'll give you a better answer
– Davide Morgante
yesterday
I see. I think I'm thinking in terms of dimensions. For example, if I had a linear transformation T from V to R^2x2 with the KernelT =0. Then I know for sure that dimension V is less than or equal to 4. Or is that the wrong assumption?
– PERTURBATIONFLOW
yesterday
What you're saying is true but in this case the linear map you're searching is an isomorphism which let's you know that the two spaces on which it operates are isomorphic (so they have the same dimension)
– Davide Morgante
yesterday
 |Â
show 2 more comments
up vote
1
down vote
Leibnewtz is right for finite dimensional vector spaces. You basically just need to count the number of degrees of freedom in your space to determine the dimension. For example, an element of $R^(3,4)$ contains 12 degrees of freedom, so it has dimension $12$. So it is obvious that $R^12$ is isomorphic to $R^(3,4)$. You could also easily construct the isomorphism between these spaces, for example, by stacking the columns of an element of $R^(3,4)$ into a single long column vector, and check that this map preserves algebraic properties.
answered Aug 3 at 15:49
amarney
978215
Is it false for infinite dimensional spaces?
– leibnewtz
Aug 3 at 23:53
would all of these choices be subspaces: We have 9 <= 12 We have 11 <= 12 We have 6 <=12 -> 6 from 2X3 We have 12<=12
– PERTURBATIONFLOW
Aug 4 at 1:33
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The simpler way to find out if two finite dimensional vector spaces are isomorphic, as others said, is to find out what if the dimension of the vector space that you're studying. So we can state a little "theorem"
Two finite dimensional vector spaces $V,W$ are isomorphic iff $$dim(V)=dim(W)$$ Here are some references.
After stating this we can easily find out which of your vector spaces is isomorphic to $mathbbR^3times 4$. First of all we have that $$dim(mathbbR^3times 4) = 12$$ and it's easy to see because to write down a $3$ by $4$ matrix you need $12$ indipendent components.
Now let's see the other dimensions $$beginalign
P_9 &&rightarrow&&dim(P_9)=9tag1\
P_11 &&rightarrow&&dim(P_11)=11tag2\
textupper triangular mathbbR^2times 3&&rightarrow&&dim(textupper triangular mathbbR^2times 3)=5tag3\
mathbbR^12&&rightarrow&&dim(mathbbR^12)=12tag4
endalign$$
For the first and second one the dimension is trivial to find: a polynomial of degree less and equal to $n$ has $n$ coefficients, so the vector space is $n$-dimensional. The third one is more tricky: one would think that the dimension would be $6$ but an upper triangular $2$ by $3$ matrix is of the form $$left(beginmatrixa_11&a_12&a_13\0&a_22&a_23endmatrixright)$$
so the element $a_21=0$ always. The only independent components ar the $5$ left. The fifth is just $mathbbR^n$ so it's dimension is $n$.
Clearly the only isomorphic space to $mathbbR^3times2$ in this list is $mathbbR^12$. If you want to find an isomorphism, which is not requested if the only thing you want is to find if to vector spaces are isomorphic, you could use this isomorphism $$phi:mathbbR^3times4rightarrow mathbbR^12 \
left(beginmatrixa_11&a_12&a_13&a_14\a_21&a_22&a_23&a_24\a_31&a_32&a_33&a_34endmatrixright)mapsto (a_11,a_12,a_13,a_14,a_21,a_22,a_23,a_24,a_31,a_32,a_33,a_34)$$ i.e. you take all the entries in the matrix and map them to a vector that has as components the $a_ij$ of the matrix.
answered 2 days ago
Davide Morgante
1,630220
Can you elaborate further as to why P9 and P11 are not isomorphic. I simply thought because their dimensions are smaller, they can be isomorphic
– PERTURBATIONFLOW
yesterday
If being isomorphic means to have an isomorphism between two spaces, necessary the two spces have to have the same dimension. An isomorphism is a map with a one-to-one correspondence: if you have $n$ elements in a set and $m$ in another, with $m>n$, you cannot construct an isomorphism because, at most, $n$ of the $m$ elements can be mapped from one set to another
– Davide Morgante
yesterday
I'm from a phone now and I cannot give you a thorough explanation! When I'll be on pc I'll give you a better answer
– Davide Morgante
yesterday
I see. I think I'm thinking in terms of dimensions. For example, if I had a linear transformation T from V to R^2x2 with the KernelT =0. Then I know for sure that dimension V is less than or equal to 4. Or is that the wrong assumption?
– PERTURBATIONFLOW
yesterday
What you're saying is true but in this case the linear map you're searching is an isomorphism which let's you know that the two spaces on which it operates are isomorphic (so they have the same dimension)
– Davide Morgante
yesterday
 |Â
show 2 more comments
up vote
1
down vote
accepted
The simpler way to find out if two finite dimensional vector spaces are isomorphic, as others said, is to find out what if the dimension of the vector space that you're studying. So we can state a little "theorem"
Two finite dimensional vector spaces $V,W$ are isomorphic iff $$dim(V)=dim(W)$$ Here are some references.
After stating this we can easily find out which of your vector spaces is isomorphic to $mathbbR^3times 4$. First of all we have that $$dim(mathbbR^3times 4) = 12$$ and it's easy to see because to write down a $3$ by $4$ matrix you need $12$ indipendent components.
Now let's see the other dimensions $$beginalign
P_9 &&rightarrow&&dim(P_9)=9tag1\
P_11 &&rightarrow&&dim(P_11)=11tag2\
textupper triangular mathbbR^2times 3&&rightarrow&&dim(textupper triangular mathbbR^2times 3)=5tag3\
mathbbR^12&&rightarrow&&dim(mathbbR^12)=12tag4
endalign$$
For the first and second one the dimension is trivial to find: a polynomial of degree less and equal to $n$ has $n$ coefficients, so the vector space is $n$-dimensional. The third one is more tricky: one would think that the dimension would be $6$ but an upper triangular $2$ by $3$ matrix is of the form $$left(beginmatrixa_11&a_12&a_13\0&a_22&a_23endmatrixright)$$
so the element $a_21=0$ always. The only independent components ar the $5$ left. The fifth is just $mathbbR^n$ so it's dimension is $n$.
Clearly the only isomorphic space to $mathbbR^3times2$ in this list is $mathbbR^12$. If you want to find an isomorphism, which is not requested if the only thing you want is to find if to vector spaces are isomorphic, you could use this isomorphism $$phi:mathbbR^3times4rightarrow mathbbR^12 \
left(beginmatrixa_11&a_12&a_13&a_14\a_21&a_22&a_23&a_24\a_31&a_32&a_33&a_34endmatrixright)mapsto (a_11,a_12,a_13,a_14,a_21,a_22,a_23,a_24,a_31,a_32,a_33,a_34)$$ i.e. you take all the entries in the matrix and map them to a vector that has as components the $a_ij$ of the matrix.
answered 2 days ago
Davide Morgante
1,630220
Can you elaborate further as to why P9 and P11 are not isomorphic. I simply thought because their dimensions are smaller, they can be isomorphic
– PERTURBATIONFLOW
yesterday
If being isomorphic means to have an isomorphism between two spaces, necessary the two spces have to have the same dimension. An isomorphism is a map with a one-to-one correspondence: if you have $n$ elements in a set and $m$ in another, with $m>n$, you cannot construct an isomorphism because, at most, $n$ of the $m$ elements can be mapped from one set to another
– Davide Morgante
yesterday
I'm from a phone now and I cannot give you a thorough explanation! When I'll be on pc I'll give you a better answer
– Davide Morgante
yesterday
I see. I think I'm thinking in terms of dimensions. For example, if I had a linear transformation T from V to R^2x2 with the KernelT =0. Then I know for sure that dimension V is less than or equal to 4. Or is that the wrong assumption?
– PERTURBATIONFLOW
yesterday
What you're saying is true but in this case the linear map you're searching is an isomorphism which let's you know that the two spaces on which it operates are isomorphic (so they have the same dimension)
– Davide Morgante
yesterday
 |Â
show 2 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The simpler way to find out if two finite dimensional vector spaces are isomorphic, as others said, is to find out what if the dimension of the vector space that you're studying. So we can state a little "theorem"
Two finite dimensional vector spaces $V,W$ are isomorphic iff $$dim(V)=dim(W)$$ Here are some references.
After stating this we can easily find out which of your vector spaces is isomorphic to $mathbbR^3times 4$. First of all we have that $$dim(mathbbR^3times 4) = 12$$ and it's easy to see because to write down a $3$ by $4$ matrix you need $12$ indipendent components.
Now let's see the other dimensions $$beginalign
P_9 &&rightarrow&&dim(P_9)=9tag1\
P_11 &&rightarrow&&dim(P_11)=11tag2\
textupper triangular mathbbR^2times 3&&rightarrow&&dim(textupper triangular mathbbR^2times 3)=5tag3\
mathbbR^12&&rightarrow&&dim(mathbbR^12)=12tag4
endalign$$
For the first and second one the dimension is trivial to find: a polynomial of degree less and equal to $n$ has $n$ coefficients, so the vector space is $n$-dimensional. The third one is more tricky: one would think that the dimension would be $6$ but an upper triangular $2$ by $3$ matrix is of the form $$left(beginmatrixa_11&a_12&a_13\0&a_22&a_23endmatrixright)$$
so the element $a_21=0$ always. The only independent components ar the $5$ left. The fifth is just $mathbbR^n$ so it's dimension is $n$.
Clearly the only isomorphic space to $mathbbR^3times2$ in this list is $mathbbR^12$. If you want to find an isomorphism, which is not requested if the only thing you want is to find if to vector spaces are isomorphic, you could use this isomorphism $$phi:mathbbR^3times4rightarrow mathbbR^12 \
left(beginmatrixa_11&a_12&a_13&a_14\a_21&a_22&a_23&a_24\a_31&a_32&a_33&a_34endmatrixright)mapsto (a_11,a_12,a_13,a_14,a_21,a_22,a_23,a_24,a_31,a_32,a_33,a_34)$$ i.e. you take all the entries in the matrix and map them to a vector that has as components the $a_ij$ of the matrix.
answered 2 days ago
Davide Morgante
1,630220
The simpler way to find out if two finite dimensional vector spaces are isomorphic, as others said, is to find out what if the dimension of the vector space that you're studying. So we can state a little "theorem"
Two finite dimensional vector spaces $V,W$ are isomorphic iff $$dim(V)=dim(W)$$ Here are some references.
After stating this we can easily find out which of your vector spaces is isomorphic to $mathbbR^3times 4$. First of all we have that $$dim(mathbbR^3times 4) = 12$$ and it's easy to see because to write down a $3$ by $4$ matrix you need $12$ indipendent components.
Now let's see the other dimensions $$beginalign
P_9 &&rightarrow&&dim(P_9)=9tag1\
P_11 &&rightarrow&&dim(P_11)=11tag2\
textupper triangular mathbbR^2times 3&&rightarrow&&dim(textupper triangular mathbbR^2times 3)=5tag3\
mathbbR^12&&rightarrow&&dim(mathbbR^12)=12tag4
endalign$$
For the first and second one the dimension is trivial to find: a polynomial of degree less and equal to $n$ has $n$ coefficients, so the vector space is $n$-dimensional. The third one is more tricky: one would think that the dimension would be $6$ but an upper triangular $2$ by $3$ matrix is of the form $$left(beginmatrixa_11&a_12&a_13\0&a_22&a_23endmatrixright)$$
so the element $a_21=0$ always. The only independent components ar the $5$ left. The fifth is just $mathbbR^n$ so it's dimension is $n$.
Clearly the only isomorphic space to $mathbbR^3times2$ in this list is $mathbbR^12$. If you want to find an isomorphism, which is not requested if the only thing you want is to find if to vector spaces are isomorphic, you could use this isomorphism $$phi:mathbbR^3times4rightarrow mathbbR^12 \
left(beginmatrixa_11&a_12&a_13&a_14\a_21&a_22&a_23&a_24\a_31&a_32&a_33&a_34endmatrixright)mapsto (a_11,a_12,a_13,a_14,a_21,a_22,a_23,a_24,a_31,a_32,a_33,a_34)$$ i.e. you take all the entries in the matrix and map them to a vector that has as components the $a_ij$ of the matrix.
answered 2 days ago
Davide Morgante
1,630220
answered 2 days ago
Davide Morgante
1,630220
answered 2 days ago
Davide Morgante
1,630220
answered 2 days ago
Davide Morgante
1,630220
1,630220
Can you elaborate further as to why P9 and P11 are not isomorphic. I simply thought because their dimensions are smaller, they can be isomorphic
– PERTURBATIONFLOW
yesterday
If being isomorphic means to have an isomorphism between two spaces, necessary the two spces have to have the same dimension. An isomorphism is a map with a one-to-one correspondence: if you have $n$ elements in a set and $m$ in another, with $m>n$, you cannot construct an isomorphism because, at most, $n$ of the $m$ elements can be mapped from one set to another
– Davide Morgante
yesterday
I'm from a phone now and I cannot give you a thorough explanation! When I'll be on pc I'll give you a better answer
– Davide Morgante
yesterday
I see. I think I'm thinking in terms of dimensions. For example, if I had a linear transformation T from V to R^2x2 with the KernelT =0. Then I know for sure that dimension V is less than or equal to 4. Or is that the wrong assumption?
– PERTURBATIONFLOW
yesterday
What you're saying is true but in this case the linear map you're searching is an isomorphism which let's you know that the two spaces on which it operates are isomorphic (so they have the same dimension)
– Davide Morgante
yesterday
 |Â
show 2 more comments
Can you elaborate further as to why P9 and P11 are not isomorphic. I simply thought because their dimensions are smaller, they can be isomorphic
– PERTURBATIONFLOW
yesterday
If being isomorphic means to have an isomorphism between two spaces, necessary the two spces have to have the same dimension. An isomorphism is a map with a one-to-one correspondence: if you have $n$ elements in a set and $m$ in another, with $m>n$, you cannot construct an isomorphism because, at most, $n$ of the $m$ elements can be mapped from one set to another
– Davide Morgante
yesterday
I'm from a phone now and I cannot give you a thorough explanation! When I'll be on pc I'll give you a better answer
– Davide Morgante
yesterday
I see. I think I'm thinking in terms of dimensions. For example, if I had a linear transformation T from V to R^2x2 with the KernelT =0. Then I know for sure that dimension V is less than or equal to 4. Or is that the wrong assumption?
– PERTURBATIONFLOW
yesterday
What you're saying is true but in this case the linear map you're searching is an isomorphism which let's you know that the two spaces on which it operates are isomorphic (so they have the same dimension)
– Davide Morgante
yesterday
Can you elaborate further as to why P9 and P11 are not isomorphic. I simply thought because their dimensions are smaller, they can be isomorphic
– PERTURBATIONFLOW
yesterday
Can you elaborate further as to why P9 and P11 are not isomorphic. I simply thought because their dimensions are smaller, they can be isomorphic
– PERTURBATIONFLOW
yesterday
If being isomorphic means to have an isomorphism between two spaces, necessary the two spces have to have the same dimension. An isomorphism is a map with a one-to-one correspondence: if you have $n$ elements in a set and $m$ in another, with $m>n$, you cannot construct an isomorphism because, at most, $n$ of the $m$ elements can be mapped from one set to another
– Davide Morgante
yesterday
If being isomorphic means to have an isomorphism between two spaces, necessary the two spces have to have the same dimension. An isomorphism is a map with a one-to-one correspondence: if you have $n$ elements in a set and $m$ in another, with $m>n$, you cannot construct an isomorphism because, at most, $n$ of the $m$ elements can be mapped from one set to another
– Davide Morgante
yesterday
I'm from a phone now and I cannot give you a thorough explanation! When I'll be on pc I'll give you a better answer
– Davide Morgante
yesterday
I'm from a phone now and I cannot give you a thorough explanation! When I'll be on pc I'll give you a better answer
– Davide Morgante
yesterday
I see. I think I'm thinking in terms of dimensions. For example, if I had a linear transformation T from V to R^2x2 with the KernelT =0. Then I know for sure that dimension V is less than or equal to 4. Or is that the wrong assumption?
– PERTURBATIONFLOW
yesterday
I see. I think I'm thinking in terms of dimensions. For example, if I had a linear transformation T from V to R^2x2 with the KernelT =0. Then I know for sure that dimension V is less than or equal to 4. Or is that the wrong assumption?
– PERTURBATIONFLOW
yesterday
What you're saying is true but in this case the linear map you're searching is an isomorphism which let's you know that the two spaces on which it operates are isomorphic (so they have the same dimension)
– Davide Morgante
yesterday
What you're saying is true but in this case the linear map you're searching is an isomorphism which let's you know that the two spaces on which it operates are isomorphic (so they have the same dimension)
– Davide Morgante
yesterday
 |Â
show 2 more comments
up vote
1
down vote
Leibnewtz is right for finite dimensional vector spaces. You basically just need to count the number of degrees of freedom in your space to determine the dimension. For example, an element of $R^(3,4)$ contains 12 degrees of freedom, so it has dimension $12$. So it is obvious that $R^12$ is isomorphic to $R^(3,4)$. You could also easily construct the isomorphism between these spaces, for example, by stacking the columns of an element of $R^(3,4)$ into a single long column vector, and check that this map preserves algebraic properties.
answered Aug 3 at 15:49
amarney
978215
Is it false for infinite dimensional spaces?
– leibnewtz
Aug 3 at 23:53
would all of these choices be subspaces: We have 9 <= 12 We have 11 <= 12 We have 6 <=12 -> 6 from 2X3 We have 12<=12
– PERTURBATIONFLOW
Aug 4 at 1:33
add a comment |Â
up vote
1
down vote
Leibnewtz is right for finite dimensional vector spaces. You basically just need to count the number of degrees of freedom in your space to determine the dimension. For example, an element of $R^(3,4)$ contains 12 degrees of freedom, so it has dimension $12$. So it is obvious that $R^12$ is isomorphic to $R^(3,4)$. You could also easily construct the isomorphism between these spaces, for example, by stacking the columns of an element of $R^(3,4)$ into a single long column vector, and check that this map preserves algebraic properties.
answered Aug 3 at 15:49
amarney
978215
Is it false for infinite dimensional spaces?
– leibnewtz
Aug 3 at 23:53
would all of these choices be subspaces: We have 9 <= 12 We have 11 <= 12 We have 6 <=12 -> 6 from 2X3 We have 12<=12
– PERTURBATIONFLOW
Aug 4 at 1:33
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Leibnewtz is right for finite dimensional vector spaces. You basically just need to count the number of degrees of freedom in your space to determine the dimension. For example, an element of $R^(3,4)$ contains 12 degrees of freedom, so it has dimension $12$. So it is obvious that $R^12$ is isomorphic to $R^(3,4)$. You could also easily construct the isomorphism between these spaces, for example, by stacking the columns of an element of $R^(3,4)$ into a single long column vector, and check that this map preserves algebraic properties.
answered Aug 3 at 15:49
amarney
978215
Leibnewtz is right for finite dimensional vector spaces. You basically just need to count the number of degrees of freedom in your space to determine the dimension. For example, an element of $R^(3,4)$ contains 12 degrees of freedom, so it has dimension $12$. So it is obvious that $R^12$ is isomorphic to $R^(3,4)$. You could also easily construct the isomorphism between these spaces, for example, by stacking the columns of an element of $R^(3,4)$ into a single long column vector, and check that this map preserves algebraic properties.
answered Aug 3 at 15:49
amarney
978215
answered Aug 3 at 15:49
amarney
978215
answered Aug 3 at 15:49
amarney
978215
answered Aug 3 at 15:49
amarney
978215
978215
Is it false for infinite dimensional spaces?
– leibnewtz
Aug 3 at 23:53
would all of these choices be subspaces: We have 9 <= 12 We have 11 <= 12 We have 6 <=12 -> 6 from 2X3 We have 12<=12
– PERTURBATIONFLOW
Aug 4 at 1:33
add a comment |Â
Is it false for infinite dimensional spaces?
– leibnewtz
Aug 3 at 23:53
would all of these choices be subspaces: We have 9 <= 12 We have 11 <= 12 We have 6 <=12 -> 6 from 2X3 We have 12<=12
– PERTURBATIONFLOW
Aug 4 at 1:33
Is it false for infinite dimensional spaces?
– leibnewtz
Aug 3 at 23:53
Is it false for infinite dimensional spaces?
– leibnewtz
Aug 3 at 23:53
would all of these choices be subspaces: We have 9 <= 12 We have 11 <= 12 We have 6 <=12 -> 6 from 2X3 We have 12<=12
– PERTURBATIONFLOW
Aug 4 at 1:33
would all of these choices be subspaces: We have 9 <= 12 We have 11 <= 12 We have 6 <=12 -> 6 from 2X3 We have 12<=12
– PERTURBATIONFLOW
Aug 4 at 1:33
add a comment |Â
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It is a little hard to tell which spaces you mean in your enumeration, can you elaborate on that. Also, do you mean with R^3x4 the space $mathbbR^(3,4)$ of $3$ by $4$ matrices?
– zzuussee
Aug 3 at 13:39
Yes, that is the space. P9 refers to characteristic polynomial with degree 9, and p11 refers to characteristic polynomial with degree 11
– PERTURBATIONFLOW
Aug 3 at 14:50
So, I assume that with characteristic polynomial with degree $9$ you mean the matrices which have a characteristic polynomial of degree $9$, i.e. all $9times 9$ matrices?
– zzuussee
Aug 3 at 14:53
1
Vector spaces are classified by their dimension. For vector spaces $V$ and $W$ there exists an injection $V to W$ if and only if $dim(V) leq dim(W)$. Hence for your problem you just need to figure out the dimension of each space
– leibnewtz
Aug 3 at 15:24
@leibnewtz would all of these choices be subspaces:
– PERTURBATIONFLOW
Aug 4 at 0:59