Identifying Isomorphisms [closed]

up vote
-1
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I understand that the conditions for isomorphism is that

close under multiplication/addition

T(-v) = -T(v)

preservation of identity

However, I'm confused about number 2. I know that 1 and 3 are both isomorphic.

But in case 2, how do we determine f(-1) or the matrix indices without the function f itself? Is it necessary to determine the isomorphism.

edited Jul 22 at 10:44

asked Jul 22 at 10:36

wonderkid

closed as unclear what you're asking by JosÃ© Carlos Santos, Xander Henderson, Mostafa Ayaz, John Ma, max_zorn Jul 23 at 0:06

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, itÃ¢Â€Â™s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

Those conditions that you mention are about subspaces. They have nothing to do with isomorphisms.
â€“Â JosÃ© Carlos Santos
Jul 22 at 10:41

@JosÃ©CarlosSantos you are correct. My head went completely somewhere else. Thank you for catching my mistake
â€“Â wonderkid
Jul 22 at 10:43

add a commentÂ |Â

up vote
-1
down vote

favorite

I understand that the conditions for isomorphism is that

close under multiplication/addition

T(-v) = -T(v)

preservation of identity

However, I'm confused about number 2. I know that 1 and 3 are both isomorphic.

But in case 2, how do we determine f(-1) or the matrix indices without the function f itself? Is it necessary to determine the isomorphism.

edited Jul 22 at 10:44

asked Jul 22 at 10:36

wonderkid

closed as unclear what you're asking by JosÃ© Carlos Santos, Xander Henderson, Mostafa Ayaz, John Ma, max_zorn Jul 23 at 0:06

Those conditions that you mention are about subspaces. They have nothing to do with isomorphisms.
â€“Â JosÃ© Carlos Santos
Jul 22 at 10:41

@JosÃ©CarlosSantos you are correct. My head went completely somewhere else. Thank you for catching my mistake
â€“Â wonderkid
Jul 22 at 10:43

add a commentÂ |Â

up vote
-1
down vote

favorite

I understand that the conditions for isomorphism is that

close under multiplication/addition

T(-v) = -T(v)

preservation of identity

However, I'm confused about number 2. I know that 1 and 3 are both isomorphic.

But in case 2, how do we determine f(-1) or the matrix indices without the function f itself? Is it necessary to determine the isomorphism.

edited Jul 22 at 10:44

asked Jul 22 at 10:36

wonderkid

I understand that the conditions for isomorphism is that

close under multiplication/addition

T(-v) = -T(v)

preservation of identity

However, I'm confused about number 2. I know that 1 and 3 are both isomorphic.

But in case 2, how do we determine f(-1) or the matrix indices without the function f itself? Is it necessary to determine the isomorphism.

edited Jul 22 at 10:44

asked Jul 22 at 10:36

wonderkid

edited Jul 22 at 10:44

asked Jul 22 at 10:36

wonderkid

asked Jul 22 at 10:36

wonderkid

asked Jul 22 at 10:36

wonderkid

closed as unclear what you're asking by JosÃ© Carlos Santos, Xander Henderson, Mostafa Ayaz, John Ma, max_zorn Jul 23 at 0:06

Those conditions that you mention are about subspaces. They have nothing to do with isomorphisms.
â€“Â JosÃ© Carlos Santos
Jul 22 at 10:41

@JosÃ©CarlosSantos you are correct. My head went completely somewhere else. Thank you for catching my mistake
â€“Â wonderkid
Jul 22 at 10:43

add a commentÂ |Â

Those conditions that you mention are about subspaces. They have nothing to do with isomorphisms.
â€“Â JosÃ© Carlos Santos
Jul 22 at 10:41

@JosÃ©CarlosSantos you are correct. My head went completely somewhere else. Thank you for catching my mistake
â€“Â wonderkid
Jul 22 at 10:43

Those conditions that you mention are about subspaces. They have nothing to do with isomorphisms.
â€“Â JosÃ© Carlos Santos
Jul 22 at 10:41

@JosÃ©CarlosSantos you are correct. My head went completely somewhere else. Thank you for catching my mistake
â€“Â wonderkid
Jul 22 at 10:43

add a commentÂ |Â

1 Answer
1

active

oldest

votes

up vote
1
down vote

accepted

An isomorphism has to be invertible by definition. Let's see than which of the three you mention it's so:

$T:Prightarrow P$ defined as $T(f)=f'$: this cannot be an isomorphism because by integrating $f'$ you get $f+c$ so a family of functions. Differentiating a polynomial get's rid of the constants and you cannot retrieve them in any way by integrating.

$T:P_3 rightarrow mathbbR^2times 2$ defined as $T(f) = left[beginmatrixf(-1)&f(0)\f(1)&f(4)endmatrixright]$: for this let's choose a base $mathcalB = 1,x,x^2,x^3$ for $P_3$ and than evaluate the the image of the base by $T$: $$T(1) = left[beginmatrix1&1\1&1endmatrixright]\ T(x) = left[beginmatrix-1&0\1&4endmatrixright] \ T(x^2) = left[beginmatrix1&0\1&16endmatrixright] \ T(x^3) = left[beginmatrix-1&0\1&64endmatrixright]
$$
let's choose the standard base for $mathbbR^2times 2$: $mathcalC = left left[beginmatrix1&0\0&0endmatrixright],left[beginmatrix0&1\0&0endmatrixright] ,left[beginmatrix0&0\1&0endmatrixright] ,left[beginmatrix0&0\0&1endmatrixright] right$ and write down the matrix of the transformation $$A=left[beginmatrix1&-1&1&-1\1&0&0&0\1&1&1&1\1&4&16&64endmatrixright]$$ the determinant of $A$ is positive, so the map is invertible

$T:D_3rightarrow P_2$ defined as $Tleft(left[beginmatrixa&0&0\0&b&0\0&0&cendmatrixright]right) = ax^2+bx+c$: let's use this basis for $D_3$ and $P_2$ $$mathcalB = 1,x,x^2;;;;;mathcalC = leftleft[beginmatrix1&0&0\0&0&0\0&0&0endmatrixright], left[beginmatrix0&0&0\0&1&0\0&0&0endmatrixright], left[beginmatrix0&0&0\0&0&0\0&0&1endmatrixright]right$$ and see where $T$ maps the basis to write down the matrix of this map $$Tleft(left[beginmatrix1&0&0\0&0&0\0&0&0endmatrixright]right) = x^2 \ Tleft(left[beginmatrix0&0&0\0&1&0\0&0&0endmatrixright]right)=x\ Tleft(left[beginmatrix0&0&0\0&0&0\0&0&1endmatrixright]right) = 1$$ then, easily enough, the matrix representation of this map is $$A = left[beginmatrix0&0&1\0&1&0\1&0&0endmatrixright]$$ which has determinant $-1$ so it's invertible.

edited Jul 22 at 11:39

answered Jul 22 at 11:20

Davide Morgante

1,798220

what allows the second case to choose a random basis? Thank you for your response. Likewise, wouldn't a determinant of -1 be non-zero, so that should be invertible?
â€“Â wonderkid
Jul 22 at 11:33

Well they're not random basis. The basis I choose were all standard basis, just to simplify calculations. You could choose any other base because under the choice of a bases $$mathbbM_ntimes n(mathbbR)cong Hom(mathbbR^n, mathbbR^n)$$ which to put it simple, after choosing a base the matrix representation is unique
â€“Â Davide Morgante
Jul 22 at 11:38

Yes, my bad, it's invertible
â€“Â Davide Morgante
Jul 22 at 11:39

I understand now. Regarding the last condition, you attain determinant -1 but say its not invertible. Aren't all non-zero determinant matrices invertible?
â€“Â wonderkid
Jul 22 at 11:39

Yes, they are! I've just corrected it
â€“Â Davide Morgante
Jul 22 at 11:40

Â |Â
show 4 more comments

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

up vote
1
down vote

accepted

An isomorphism has to be invertible by definition. Let's see than which of the three you mention it's so:

$T:Prightarrow P$ defined as $T(f)=f'$: this cannot be an isomorphism because by integrating $f'$ you get $f+c$ so a family of functions. Differentiating a polynomial get's rid of the constants and you cannot retrieve them in any way by integrating.

$T:P_3 rightarrow mathbbR^2times 2$ defined as $T(f) = left[beginmatrixf(-1)&f(0)\f(1)&f(4)endmatrixright]$: for this let's choose a base $mathcalB = 1,x,x^2,x^3$ for $P_3$ and than evaluate the the image of the base by $T$: $$T(1) = left[beginmatrix1&1\1&1endmatrixright]\ T(x) = left[beginmatrix-1&0\1&4endmatrixright] \ T(x^2) = left[beginmatrix1&0\1&16endmatrixright] \ T(x^3) = left[beginmatrix-1&0\1&64endmatrixright]
$$
let's choose the standard base for $mathbbR^2times 2$: $mathcalC = left left[beginmatrix1&0\0&0endmatrixright],left[beginmatrix0&1\0&0endmatrixright] ,left[beginmatrix0&0\1&0endmatrixright] ,left[beginmatrix0&0\0&1endmatrixright] right$ and write down the matrix of the transformation $$A=left[beginmatrix1&-1&1&-1\1&0&0&0\1&1&1&1\1&4&16&64endmatrixright]$$ the determinant of $A$ is positive, so the map is invertible

$T:D_3rightarrow P_2$ defined as $Tleft(left[beginmatrixa&0&0\0&b&0\0&0&cendmatrixright]right) = ax^2+bx+c$: let's use this basis for $D_3$ and $P_2$ $$mathcalB = 1,x,x^2;;;;;mathcalC = leftleft[beginmatrix1&0&0\0&0&0\0&0&0endmatrixright], left[beginmatrix0&0&0\0&1&0\0&0&0endmatrixright], left[beginmatrix0&0&0\0&0&0\0&0&1endmatrixright]right$$ and see where $T$ maps the basis to write down the matrix of this map $$Tleft(left[beginmatrix1&0&0\0&0&0\0&0&0endmatrixright]right) = x^2 \ Tleft(left[beginmatrix0&0&0\0&1&0\0&0&0endmatrixright]right)=x\ Tleft(left[beginmatrix0&0&0\0&0&0\0&0&1endmatrixright]right) = 1$$ then, easily enough, the matrix representation of this map is $$A = left[beginmatrix0&0&1\0&1&0\1&0&0endmatrixright]$$ which has determinant $-1$ so it's invertible.

edited Jul 22 at 11:39

answered Jul 22 at 11:20

Davide Morgante

1,798220

what allows the second case to choose a random basis? Thank you for your response. Likewise, wouldn't a determinant of -1 be non-zero, so that should be invertible?
â€“Â wonderkid
Jul 22 at 11:33

Well they're not random basis. The basis I choose were all standard basis, just to simplify calculations. You could choose any other base because under the choice of a bases $$mathbbM_ntimes n(mathbbR)cong Hom(mathbbR^n, mathbbR^n)$$ which to put it simple, after choosing a base the matrix representation is unique
â€“Â Davide Morgante
Jul 22 at 11:38

Yes, my bad, it's invertible
â€“Â Davide Morgante
Jul 22 at 11:39

I understand now. Regarding the last condition, you attain determinant -1 but say its not invertible. Aren't all non-zero determinant matrices invertible?
â€“Â wonderkid
Jul 22 at 11:39

Yes, they are! I've just corrected it
â€“Â Davide Morgante
Jul 22 at 11:40

Â |Â
show 4 more comments

up vote
1
down vote

accepted

An isomorphism has to be invertible by definition. Let's see than which of the three you mention it's so:

$T:Prightarrow P$ defined as $T(f)=f'$: this cannot be an isomorphism because by integrating $f'$ you get $f+c$ so a family of functions. Differentiating a polynomial get's rid of the constants and you cannot retrieve them in any way by integrating.

$T:P_3 rightarrow mathbbR^2times 2$ defined as $T(f) = left[beginmatrixf(-1)&f(0)\f(1)&f(4)endmatrixright]$: for this let's choose a base $mathcalB = 1,x,x^2,x^3$ for $P_3$ and than evaluate the the image of the base by $T$: $$T(1) = left[beginmatrix1&1\1&1endmatrixright]\ T(x) = left[beginmatrix-1&0\1&4endmatrixright] \ T(x^2) = left[beginmatrix1&0\1&16endmatrixright] \ T(x^3) = left[beginmatrix-1&0\1&64endmatrixright]
$$
let's choose the standard base for $mathbbR^2times 2$: $mathcalC = left left[beginmatrix1&0\0&0endmatrixright],left[beginmatrix0&1\0&0endmatrixright] ,left[beginmatrix0&0\1&0endmatrixright] ,left[beginmatrix0&0\0&1endmatrixright] right$ and write down the matrix of the transformation $$A=left[beginmatrix1&-1&1&-1\1&0&0&0\1&1&1&1\1&4&16&64endmatrixright]$$ the determinant of $A$ is positive, so the map is invertible

$T:D_3rightarrow P_2$ defined as $Tleft(left[beginmatrixa&0&0\0&b&0\0&0&cendmatrixright]right) = ax^2+bx+c$: let's use this basis for $D_3$ and $P_2$ $$mathcalB = 1,x,x^2;;;;;mathcalC = leftleft[beginmatrix1&0&0\0&0&0\0&0&0endmatrixright], left[beginmatrix0&0&0\0&1&0\0&0&0endmatrixright], left[beginmatrix0&0&0\0&0&0\0&0&1endmatrixright]right$$ and see where $T$ maps the basis to write down the matrix of this map $$Tleft(left[beginmatrix1&0&0\0&0&0\0&0&0endmatrixright]right) = x^2 \ Tleft(left[beginmatrix0&0&0\0&1&0\0&0&0endmatrixright]right)=x\ Tleft(left[beginmatrix0&0&0\0&0&0\0&0&1endmatrixright]right) = 1$$ then, easily enough, the matrix representation of this map is $$A = left[beginmatrix0&0&1\0&1&0\1&0&0endmatrixright]$$ which has determinant $-1$ so it's invertible.

edited Jul 22 at 11:39

answered Jul 22 at 11:20

Davide Morgante

1,798220

what allows the second case to choose a random basis? Thank you for your response. Likewise, wouldn't a determinant of -1 be non-zero, so that should be invertible?
â€“Â wonderkid
Jul 22 at 11:33

Well they're not random basis. The basis I choose were all standard basis, just to simplify calculations. You could choose any other base because under the choice of a bases $$mathbbM_ntimes n(mathbbR)cong Hom(mathbbR^n, mathbbR^n)$$ which to put it simple, after choosing a base the matrix representation is unique
â€“Â Davide Morgante
Jul 22 at 11:38

Yes, my bad, it's invertible
â€“Â Davide Morgante
Jul 22 at 11:39

I understand now. Regarding the last condition, you attain determinant -1 but say its not invertible. Aren't all non-zero determinant matrices invertible?
â€“Â wonderkid
Jul 22 at 11:39

Yes, they are! I've just corrected it
â€“Â Davide Morgante
Jul 22 at 11:40

Â |Â
show 4 more comments

up vote
1
down vote

accepted

An isomorphism has to be invertible by definition. Let's see than which of the three you mention it's so:

$T:Prightarrow P$ defined as $T(f)=f'$: this cannot be an isomorphism because by integrating $f'$ you get $f+c$ so a family of functions. Differentiating a polynomial get's rid of the constants and you cannot retrieve them in any way by integrating.

$T:P_3 rightarrow mathbbR^2times 2$ defined as $T(f) = left[beginmatrixf(-1)&f(0)\f(1)&f(4)endmatrixright]$: for this let's choose a base $mathcalB = 1,x,x^2,x^3$ for $P_3$ and than evaluate the the image of the base by $T$: $$T(1) = left[beginmatrix1&1\1&1endmatrixright]\ T(x) = left[beginmatrix-1&0\1&4endmatrixright] \ T(x^2) = left[beginmatrix1&0\1&16endmatrixright] \ T(x^3) = left[beginmatrix-1&0\1&64endmatrixright]
$$
let's choose the standard base for $mathbbR^2times 2$: $mathcalC = left left[beginmatrix1&0\0&0endmatrixright],left[beginmatrix0&1\0&0endmatrixright] ,left[beginmatrix0&0\1&0endmatrixright] ,left[beginmatrix0&0\0&1endmatrixright] right$ and write down the matrix of the transformation $$A=left[beginmatrix1&-1&1&-1\1&0&0&0\1&1&1&1\1&4&16&64endmatrixright]$$ the determinant of $A$ is positive, so the map is invertible

$T:D_3rightarrow P_2$ defined as $Tleft(left[beginmatrixa&0&0\0&b&0\0&0&cendmatrixright]right) = ax^2+bx+c$: let's use this basis for $D_3$ and $P_2$ $$mathcalB = 1,x,x^2;;;;;mathcalC = leftleft[beginmatrix1&0&0\0&0&0\0&0&0endmatrixright], left[beginmatrix0&0&0\0&1&0\0&0&0endmatrixright], left[beginmatrix0&0&0\0&0&0\0&0&1endmatrixright]right$$ and see where $T$ maps the basis to write down the matrix of this map $$Tleft(left[beginmatrix1&0&0\0&0&0\0&0&0endmatrixright]right) = x^2 \ Tleft(left[beginmatrix0&0&0\0&1&0\0&0&0endmatrixright]right)=x\ Tleft(left[beginmatrix0&0&0\0&0&0\0&0&1endmatrixright]right) = 1$$ then, easily enough, the matrix representation of this map is $$A = left[beginmatrix0&0&1\0&1&0\1&0&0endmatrixright]$$ which has determinant $-1$ so it's invertible.

edited Jul 22 at 11:39

answered Jul 22 at 11:20

Davide Morgante

1,798220

An isomorphism has to be invertible by definition. Let's see than which of the three you mention it's so:

$T:Prightarrow P$ defined as $T(f)=f'$: this cannot be an isomorphism because by integrating $f'$ you get $f+c$ so a family of functions. Differentiating a polynomial get's rid of the constants and you cannot retrieve them in any way by integrating.

$T:P_3 rightarrow mathbbR^2times 2$ defined as $T(f) = left[beginmatrixf(-1)&f(0)\f(1)&f(4)endmatrixright]$: for this let's choose a base $mathcalB = 1,x,x^2,x^3$ for $P_3$ and than evaluate the the image of the base by $T$: $$T(1) = left[beginmatrix1&1\1&1endmatrixright]\ T(x) = left[beginmatrix-1&0\1&4endmatrixright] \ T(x^2) = left[beginmatrix1&0\1&16endmatrixright] \ T(x^3) = left[beginmatrix-1&0\1&64endmatrixright]
$$
let's choose the standard base for $mathbbR^2times 2$: $mathcalC = left left[beginmatrix1&0\0&0endmatrixright],left[beginmatrix0&1\0&0endmatrixright] ,left[beginmatrix0&0\1&0endmatrixright] ,left[beginmatrix0&0\0&1endmatrixright] right$ and write down the matrix of the transformation $$A=left[beginmatrix1&-1&1&-1\1&0&0&0\1&1&1&1\1&4&16&64endmatrixright]$$ the determinant of $A$ is positive, so the map is invertible

$T:D_3rightarrow P_2$ defined as $Tleft(left[beginmatrixa&0&0\0&b&0\0&0&cendmatrixright]right) = ax^2+bx+c$: let's use this basis for $D_3$ and $P_2$ $$mathcalB = 1,x,x^2;;;;;mathcalC = leftleft[beginmatrix1&0&0\0&0&0\0&0&0endmatrixright], left[beginmatrix0&0&0\0&1&0\0&0&0endmatrixright], left[beginmatrix0&0&0\0&0&0\0&0&1endmatrixright]right$$ and see where $T$ maps the basis to write down the matrix of this map $$Tleft(left[beginmatrix1&0&0\0&0&0\0&0&0endmatrixright]right) = x^2 \ Tleft(left[beginmatrix0&0&0\0&1&0\0&0&0endmatrixright]right)=x\ Tleft(left[beginmatrix0&0&0\0&0&0\0&0&1endmatrixright]right) = 1$$ then, easily enough, the matrix representation of this map is $$A = left[beginmatrix0&0&1\0&1&0\1&0&0endmatrixright]$$ which has determinant $-1$ so it's invertible.

edited Jul 22 at 11:39

answered Jul 22 at 11:20

Davide Morgante

1,798220

edited Jul 22 at 11:39

answered Jul 22 at 11:20

Davide Morgante

1,798220

answered Jul 22 at 11:20

Davide Morgante

1,798220

answered Jul 22 at 11:20

Davide Morgante

1,798220

what allows the second case to choose a random basis? Thank you for your response. Likewise, wouldn't a determinant of -1 be non-zero, so that should be invertible?
â€“Â wonderkid
Jul 22 at 11:33

Well they're not random basis. The basis I choose were all standard basis, just to simplify calculations. You could choose any other base because under the choice of a bases $$mathbbM_ntimes n(mathbbR)cong Hom(mathbbR^n, mathbbR^n)$$ which to put it simple, after choosing a base the matrix representation is unique
â€“Â Davide Morgante
Jul 22 at 11:38

Yes, my bad, it's invertible
â€“Â Davide Morgante
Jul 22 at 11:39

I understand now. Regarding the last condition, you attain determinant -1 but say its not invertible. Aren't all non-zero determinant matrices invertible?
â€“Â wonderkid
Jul 22 at 11:39

Yes, they are! I've just corrected it
â€“Â Davide Morgante
Jul 22 at 11:40

Â |Â
show 4 more comments

what allows the second case to choose a random basis? Thank you for your response. Likewise, wouldn't a determinant of -1 be non-zero, so that should be invertible?
â€“Â wonderkid
Jul 22 at 11:33

Well they're not random basis. The basis I choose were all standard basis, just to simplify calculations. You could choose any other base because under the choice of a bases $$mathbbM_ntimes n(mathbbR)cong Hom(mathbbR^n, mathbbR^n)$$ which to put it simple, after choosing a base the matrix representation is unique
â€“Â Davide Morgante
Jul 22 at 11:38

Yes, my bad, it's invertible
â€“Â Davide Morgante
Jul 22 at 11:39

I understand now. Regarding the last condition, you attain determinant -1 but say its not invertible. Aren't all non-zero determinant matrices invertible?
â€“Â wonderkid
Jul 22 at 11:39

Yes, they are! I've just corrected it
â€“Â Davide Morgante
Jul 22 at 11:40

what allows the second case to choose a random basis? Thank you for your response. Likewise, wouldn't a determinant of -1 be non-zero, so that should be invertible?
â€“Â wonderkid
Jul 22 at 11:33

Well they're not random basis. The basis I choose were all standard basis, just to simplify calculations. You could choose any other base because under the choice of a bases $$mathbbM_ntimes n(mathbbR)cong Hom(mathbbR^n, mathbbR^n)$$ which to put it simple, after choosing a base the matrix representation is unique
â€“Â Davide Morgante
Jul 22 at 11:38

Yes, my bad, it's invertible
â€“Â Davide Morgante
Jul 22 at 11:39

I understand now. Regarding the last condition, you attain determinant -1 but say its not invertible. Aren't all non-zero determinant matrices invertible?
â€“Â wonderkid
Jul 22 at 11:39

Yes, they are! I've just corrected it
â€“Â Davide Morgante
Jul 22 at 11:40

Â |Â
show 4 more comments

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