Mixing
Determine the matrix A that represents T in the canonical form of R2[x] in the start and in the end
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
Being T: R2[x]->R2[x] defined in T(ax^2+bx+c)=2ax+b
Determine the matrix A that represents T in the canonical form of R2[x] in the start and in the end
My conclusions until now:
T(1)=0
T(x)=1
T(x^2)=2x
I don't know how to use this to make the matrix.
linear-algebra
asked Jul 16 at 21:03
Pedro Quintans
175
add a comment |Â
up vote
0
down vote
favorite
Being T: R2[x]->R2[x] defined in T(ax^2+bx+c)=2ax+b
Determine the matrix A that represents T in the canonical form of R2[x] in the start and in the end
My conclusions until now:
T(1)=0
T(x)=1
T(x^2)=2x
I don't know how to use this to make the matrix.
linear-algebra
asked Jul 16 at 21:03
Pedro Quintans
175
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Being T: R2[x]->R2[x] defined in T(ax^2+bx+c)=2ax+b
Determine the matrix A that represents T in the canonical form of R2[x] in the start and in the end
My conclusions until now:
T(1)=0
T(x)=1
T(x^2)=2x
I don't know how to use this to make the matrix.
linear-algebra
asked Jul 16 at 21:03
Pedro Quintans
175
Being T: R2[x]->R2[x] defined in T(ax^2+bx+c)=2ax+b
Determine the matrix A that represents T in the canonical form of R2[x] in the start and in the end
My conclusions until now:
T(1)=0
T(x)=1
T(x^2)=2x
I don't know how to use this to make the matrix.
linear-algebra
asked Jul 16 at 21:03
Pedro Quintans
175
asked Jul 16 at 21:03
Pedro Quintans
175
asked Jul 16 at 21:03
Pedro Quintans
175
asked Jul 16 at 21:03
Pedro Quintans
175
175
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
To write the matrix of this homomorphism (the method holds for every linear map) $$T: mathbbR_2[x]rightarrow mathbbR_2[x]$$ we first choose a base for for the domain and one for the image: the exercise here requires the canonical basis for either two, which is $mathcalB = 1, x, x^2$. Then to determine the matrix $[T]$ associated to the map you just have to evaluate the application on the basis, in this case: $$T(1) = 0;;;;T(x) = 1;;;; T(x^2) = 2x$$ and write the result as a linear combination of the basis of the image (in this case it remains the same), so: $$mathbfv_1=T(1) = 0*1 + 0*x+0*x^2 \ mathbfv_2=T(x) = 1*1 + 0*x +0*x^2 \ mathbfv_3=T(x^2) = 0*1 + 2*x +0*x^2$$
The coordinate of the three vectors in the base of the image are then $$mathbfv_1 = (0,0,0);;;;mathbfv_2=(1,0,0);;;;mathbfv_3=(0,2,0)$$ Finally the matrix of the map is to be written as
$$left( beginmatrix|&|&|\
mathbfv_1&mathbfv_2&mathbfv_3\
|&|&|endmatrixright) = left( beginmatrix0&1&0\
0&0&2\
0&0&0endmatrixright)$$
You can check if this is right by taking a generic vector $mathbfv=(c,b,a) = c+bx+ax^2$ and seeing if the product of this with the matrix gets you were you wanted to be
$$left( beginmatrix0&1&0\
0&0&2\
0&0&0endmatrixright) left(beginmatrixc\b\aendmatrixright) = left(beginmatrixb\2a\0endmatrixright) = b+2ax$$
as the exercise stated.
Justification
If it wasn't clear, you take the product with the matrix because, if you have the matrix associated with your map (which is unique after choosing the basis) you can write the map as: $$T(mathbfv) = [T]mathbfv^T$$
Just for fun
I really loved the notation used by Lang to write the matrix of a linear map $$L:V_mathcalBrightarrow W_mathcalC$$ where the subscript denotes the basis chosen for each vecor space. He denotes the matrix as $$[L]^mathcalB_mathcalC$$ which I found very useful when dealing with map composition
answered Jul 16 at 21:29
Davide Morgante
1,875220
and write the result as a linear combination of the basis of the image (in this case it remains the same) I couldn't follow the calculus For T(1)=0, following (1,x,x^2) we have 0*1+0*x+0*x^2 i could understand T(x)=1 i got stuck and couldn't understand the 1*1 (i must be dumb)
– Pedro Quintans
Jul 16 at 22:39
That's no shame, it happens to everybody!
– Davide Morgante
Jul 16 at 22:43
@PedroQuintans By the way, I think for future reference you should accept the answer so that it will show on MSE as a closed question!
– Davide Morgante
Jul 17 at 17:41
Sorry about that, i'm new to this site, i already closed it, thanks
– Pedro Quintans
Jul 18 at 12:41
add a comment |Â
up vote
0
down vote
I suppose the ‘canonical form’ denotes the canonical basis $1, x, x^2 $ of $:mathbf R_2[X]$.
Hint:
The column vectors of this $3times 3$ matrix are the coordinates in the canonical basis of the images of the vectors of this canonical basis.
answered Jul 16 at 21:11
Bernard
110k635103
Sorry, can you provide an example?
– Pedro Quintans
Jul 16 at 21:20
In your case, for instance $T(x^2)=2x$. So the coordinates of $T(x^2)$ in basis $(1,x,x^2)$ are $(0, 2,0)$ (transposed).
– Bernard
Jul 16 at 21:25
how? i don't understand, you replace the x from the (1,X,x^2) for 2x?
– Pedro Quintans
Jul 16 at 22:44
Because $2x=colorred0cdot 1+colorred2cdot x+colorred0cdot x^2$, that's all!
– Bernard
Jul 16 at 22:52
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
To write the matrix of this homomorphism (the method holds for every linear map) $$T: mathbbR_2[x]rightarrow mathbbR_2[x]$$ we first choose a base for for the domain and one for the image: the exercise here requires the canonical basis for either two, which is $mathcalB = 1, x, x^2$. Then to determine the matrix $[T]$ associated to the map you just have to evaluate the application on the basis, in this case: $$T(1) = 0;;;;T(x) = 1;;;; T(x^2) = 2x$$ and write the result as a linear combination of the basis of the image (in this case it remains the same), so: $$mathbfv_1=T(1) = 0*1 + 0*x+0*x^2 \ mathbfv_2=T(x) = 1*1 + 0*x +0*x^2 \ mathbfv_3=T(x^2) = 0*1 + 2*x +0*x^2$$
The coordinate of the three vectors in the base of the image are then $$mathbfv_1 = (0,0,0);;;;mathbfv_2=(1,0,0);;;;mathbfv_3=(0,2,0)$$ Finally the matrix of the map is to be written as
$$left( beginmatrix|&|&|\
mathbfv_1&mathbfv_2&mathbfv_3\
|&|&|endmatrixright) = left( beginmatrix0&1&0\
0&0&2\
0&0&0endmatrixright)$$
You can check if this is right by taking a generic vector $mathbfv=(c,b,a) = c+bx+ax^2$ and seeing if the product of this with the matrix gets you were you wanted to be
$$left( beginmatrix0&1&0\
0&0&2\
0&0&0endmatrixright) left(beginmatrixc\b\aendmatrixright) = left(beginmatrixb\2a\0endmatrixright) = b+2ax$$
as the exercise stated.
Justification
If it wasn't clear, you take the product with the matrix because, if you have the matrix associated with your map (which is unique after choosing the basis) you can write the map as: $$T(mathbfv) = [T]mathbfv^T$$
Just for fun
I really loved the notation used by Lang to write the matrix of a linear map $$L:V_mathcalBrightarrow W_mathcalC$$ where the subscript denotes the basis chosen for each vecor space. He denotes the matrix as $$[L]^mathcalB_mathcalC$$ which I found very useful when dealing with map composition
answered Jul 16 at 21:29
Davide Morgante
1,875220
and write the result as a linear combination of the basis of the image (in this case it remains the same) I couldn't follow the calculus For T(1)=0, following (1,x,x^2) we have 0*1+0*x+0*x^2 i could understand T(x)=1 i got stuck and couldn't understand the 1*1 (i must be dumb)
– Pedro Quintans
Jul 16 at 22:39
That's no shame, it happens to everybody!
– Davide Morgante
Jul 16 at 22:43
@PedroQuintans By the way, I think for future reference you should accept the answer so that it will show on MSE as a closed question!
– Davide Morgante
Jul 17 at 17:41
Sorry about that, i'm new to this site, i already closed it, thanks
– Pedro Quintans
Jul 18 at 12:41
add a comment |Â
up vote
1
down vote
accepted
To write the matrix of this homomorphism (the method holds for every linear map) $$T: mathbbR_2[x]rightarrow mathbbR_2[x]$$ we first choose a base for for the domain and one for the image: the exercise here requires the canonical basis for either two, which is $mathcalB = 1, x, x^2$. Then to determine the matrix $[T]$ associated to the map you just have to evaluate the application on the basis, in this case: $$T(1) = 0;;;;T(x) = 1;;;; T(x^2) = 2x$$ and write the result as a linear combination of the basis of the image (in this case it remains the same), so: $$mathbfv_1=T(1) = 0*1 + 0*x+0*x^2 \ mathbfv_2=T(x) = 1*1 + 0*x +0*x^2 \ mathbfv_3=T(x^2) = 0*1 + 2*x +0*x^2$$
The coordinate of the three vectors in the base of the image are then $$mathbfv_1 = (0,0,0);;;;mathbfv_2=(1,0,0);;;;mathbfv_3=(0,2,0)$$ Finally the matrix of the map is to be written as
$$left( beginmatrix|&|&|\
mathbfv_1&mathbfv_2&mathbfv_3\
|&|&|endmatrixright) = left( beginmatrix0&1&0\
0&0&2\
0&0&0endmatrixright)$$
You can check if this is right by taking a generic vector $mathbfv=(c,b,a) = c+bx+ax^2$ and seeing if the product of this with the matrix gets you were you wanted to be
$$left( beginmatrix0&1&0\
0&0&2\
0&0&0endmatrixright) left(beginmatrixc\b\aendmatrixright) = left(beginmatrixb\2a\0endmatrixright) = b+2ax$$
as the exercise stated.
Justification
If it wasn't clear, you take the product with the matrix because, if you have the matrix associated with your map (which is unique after choosing the basis) you can write the map as: $$T(mathbfv) = [T]mathbfv^T$$
Just for fun
I really loved the notation used by Lang to write the matrix of a linear map $$L:V_mathcalBrightarrow W_mathcalC$$ where the subscript denotes the basis chosen for each vecor space. He denotes the matrix as $$[L]^mathcalB_mathcalC$$ which I found very useful when dealing with map composition
answered Jul 16 at 21:29
Davide Morgante
1,875220
and write the result as a linear combination of the basis of the image (in this case it remains the same) I couldn't follow the calculus For T(1)=0, following (1,x,x^2) we have 0*1+0*x+0*x^2 i could understand T(x)=1 i got stuck and couldn't understand the 1*1 (i must be dumb)
– Pedro Quintans
Jul 16 at 22:39
That's no shame, it happens to everybody!
– Davide Morgante
Jul 16 at 22:43
@PedroQuintans By the way, I think for future reference you should accept the answer so that it will show on MSE as a closed question!
– Davide Morgante
Jul 17 at 17:41
Sorry about that, i'm new to this site, i already closed it, thanks
– Pedro Quintans
Jul 18 at 12:41
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
To write the matrix of this homomorphism (the method holds for every linear map) $$T: mathbbR_2[x]rightarrow mathbbR_2[x]$$ we first choose a base for for the domain and one for the image: the exercise here requires the canonical basis for either two, which is $mathcalB = 1, x, x^2$. Then to determine the matrix $[T]$ associated to the map you just have to evaluate the application on the basis, in this case: $$T(1) = 0;;;;T(x) = 1;;;; T(x^2) = 2x$$ and write the result as a linear combination of the basis of the image (in this case it remains the same), so: $$mathbfv_1=T(1) = 0*1 + 0*x+0*x^2 \ mathbfv_2=T(x) = 1*1 + 0*x +0*x^2 \ mathbfv_3=T(x^2) = 0*1 + 2*x +0*x^2$$
The coordinate of the three vectors in the base of the image are then $$mathbfv_1 = (0,0,0);;;;mathbfv_2=(1,0,0);;;;mathbfv_3=(0,2,0)$$ Finally the matrix of the map is to be written as
$$left( beginmatrix|&|&|\
mathbfv_1&mathbfv_2&mathbfv_3\
|&|&|endmatrixright) = left( beginmatrix0&1&0\
0&0&2\
0&0&0endmatrixright)$$
You can check if this is right by taking a generic vector $mathbfv=(c,b,a) = c+bx+ax^2$ and seeing if the product of this with the matrix gets you were you wanted to be
$$left( beginmatrix0&1&0\
0&0&2\
0&0&0endmatrixright) left(beginmatrixc\b\aendmatrixright) = left(beginmatrixb\2a\0endmatrixright) = b+2ax$$
as the exercise stated.
Justification
If it wasn't clear, you take the product with the matrix because, if you have the matrix associated with your map (which is unique after choosing the basis) you can write the map as: $$T(mathbfv) = [T]mathbfv^T$$
Just for fun
I really loved the notation used by Lang to write the matrix of a linear map $$L:V_mathcalBrightarrow W_mathcalC$$ where the subscript denotes the basis chosen for each vecor space. He denotes the matrix as $$[L]^mathcalB_mathcalC$$ which I found very useful when dealing with map composition
answered Jul 16 at 21:29
Davide Morgante
1,875220
To write the matrix of this homomorphism (the method holds for every linear map) $$T: mathbbR_2[x]rightarrow mathbbR_2[x]$$ we first choose a base for for the domain and one for the image: the exercise here requires the canonical basis for either two, which is $mathcalB = 1, x, x^2$. Then to determine the matrix $[T]$ associated to the map you just have to evaluate the application on the basis, in this case: $$T(1) = 0;;;;T(x) = 1;;;; T(x^2) = 2x$$ and write the result as a linear combination of the basis of the image (in this case it remains the same), so: $$mathbfv_1=T(1) = 0*1 + 0*x+0*x^2 \ mathbfv_2=T(x) = 1*1 + 0*x +0*x^2 \ mathbfv_3=T(x^2) = 0*1 + 2*x +0*x^2$$
The coordinate of the three vectors in the base of the image are then $$mathbfv_1 = (0,0,0);;;;mathbfv_2=(1,0,0);;;;mathbfv_3=(0,2,0)$$ Finally the matrix of the map is to be written as
$$left( beginmatrix|&|&|\
mathbfv_1&mathbfv_2&mathbfv_3\
|&|&|endmatrixright) = left( beginmatrix0&1&0\
0&0&2\
0&0&0endmatrixright)$$
You can check if this is right by taking a generic vector $mathbfv=(c,b,a) = c+bx+ax^2$ and seeing if the product of this with the matrix gets you were you wanted to be
$$left( beginmatrix0&1&0\
0&0&2\
0&0&0endmatrixright) left(beginmatrixc\b\aendmatrixright) = left(beginmatrixb\2a\0endmatrixright) = b+2ax$$
as the exercise stated.
Justification
If it wasn't clear, you take the product with the matrix because, if you have the matrix associated with your map (which is unique after choosing the basis) you can write the map as: $$T(mathbfv) = [T]mathbfv^T$$
Just for fun
I really loved the notation used by Lang to write the matrix of a linear map $$L:V_mathcalBrightarrow W_mathcalC$$ where the subscript denotes the basis chosen for each vecor space. He denotes the matrix as $$[L]^mathcalB_mathcalC$$ which I found very useful when dealing with map composition
answered Jul 16 at 21:29
Davide Morgante
1,875220
edited Jul 16 at 21:44
answered Jul 16 at 21:29
Davide Morgante
1,875220
answered Jul 16 at 21:29
Davide Morgante
1,875220
answered Jul 16 at 21:29
Davide Morgante
1,875220
1,875220
and write the result as a linear combination of the basis of the image (in this case it remains the same) I couldn't follow the calculus For T(1)=0, following (1,x,x^2) we have 0*1+0*x+0*x^2 i could understand T(x)=1 i got stuck and couldn't understand the 1*1 (i must be dumb)
– Pedro Quintans
Jul 16 at 22:39
That's no shame, it happens to everybody!
– Davide Morgante
Jul 16 at 22:43
@PedroQuintans By the way, I think for future reference you should accept the answer so that it will show on MSE as a closed question!
– Davide Morgante
Jul 17 at 17:41
Sorry about that, i'm new to this site, i already closed it, thanks
– Pedro Quintans
Jul 18 at 12:41
add a comment |Â
and write the result as a linear combination of the basis of the image (in this case it remains the same) I couldn't follow the calculus For T(1)=0, following (1,x,x^2) we have 0*1+0*x+0*x^2 i could understand T(x)=1 i got stuck and couldn't understand the 1*1 (i must be dumb)
– Pedro Quintans
Jul 16 at 22:39
That's no shame, it happens to everybody!
– Davide Morgante
Jul 16 at 22:43
@PedroQuintans By the way, I think for future reference you should accept the answer so that it will show on MSE as a closed question!
– Davide Morgante
Jul 17 at 17:41
Sorry about that, i'm new to this site, i already closed it, thanks
– Pedro Quintans
Jul 18 at 12:41
and write the result as a linear combination of the basis of the image (in this case it remains the same) I couldn't follow the calculus For T(1)=0, following (1,x,x^2) we have 0*1+0*x+0*x^2 i could understand T(x)=1 i got stuck and couldn't understand the 1*1 (i must be dumb)
– Pedro Quintans
Jul 16 at 22:39
and write the result as a linear combination of the basis of the image (in this case it remains the same) I couldn't follow the calculus For T(1)=0, following (1,x,x^2) we have 0*1+0*x+0*x^2 i could understand T(x)=1 i got stuck and couldn't understand the 1*1 (i must be dumb)
– Pedro Quintans
Jul 16 at 22:39
That's no shame, it happens to everybody!
– Davide Morgante
Jul 16 at 22:43
That's no shame, it happens to everybody!
– Davide Morgante
Jul 16 at 22:43
@PedroQuintans By the way, I think for future reference you should accept the answer so that it will show on MSE as a closed question!
– Davide Morgante
Jul 17 at 17:41
@PedroQuintans By the way, I think for future reference you should accept the answer so that it will show on MSE as a closed question!
– Davide Morgante
Jul 17 at 17:41
Sorry about that, i'm new to this site, i already closed it, thanks
– Pedro Quintans
Jul 18 at 12:41
Sorry about that, i'm new to this site, i already closed it, thanks
– Pedro Quintans
Jul 18 at 12:41
add a comment |Â
up vote
0
down vote
I suppose the ‘canonical form’ denotes the canonical basis $1, x, x^2 $ of $:mathbf R_2[X]$.
Hint:
The column vectors of this $3times 3$ matrix are the coordinates in the canonical basis of the images of the vectors of this canonical basis.
answered Jul 16 at 21:11
Bernard
110k635103
Sorry, can you provide an example?
– Pedro Quintans
Jul 16 at 21:20
In your case, for instance $T(x^2)=2x$. So the coordinates of $T(x^2)$ in basis $(1,x,x^2)$ are $(0, 2,0)$ (transposed).
– Bernard
Jul 16 at 21:25
how? i don't understand, you replace the x from the (1,X,x^2) for 2x?
– Pedro Quintans
Jul 16 at 22:44
Because $2x=colorred0cdot 1+colorred2cdot x+colorred0cdot x^2$, that's all!
– Bernard
Jul 16 at 22:52
add a comment |Â
up vote
0
down vote
I suppose the ‘canonical form’ denotes the canonical basis $1, x, x^2 $ of $:mathbf R_2[X]$.
Hint:
The column vectors of this $3times 3$ matrix are the coordinates in the canonical basis of the images of the vectors of this canonical basis.
answered Jul 16 at 21:11
Bernard
110k635103
Sorry, can you provide an example?
– Pedro Quintans
Jul 16 at 21:20
In your case, for instance $T(x^2)=2x$. So the coordinates of $T(x^2)$ in basis $(1,x,x^2)$ are $(0, 2,0)$ (transposed).
– Bernard
Jul 16 at 21:25
how? i don't understand, you replace the x from the (1,X,x^2) for 2x?
– Pedro Quintans
Jul 16 at 22:44
Because $2x=colorred0cdot 1+colorred2cdot x+colorred0cdot x^2$, that's all!
– Bernard
Jul 16 at 22:52
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I suppose the ‘canonical form’ denotes the canonical basis $1, x, x^2 $ of $:mathbf R_2[X]$.
Hint:
The column vectors of this $3times 3$ matrix are the coordinates in the canonical basis of the images of the vectors of this canonical basis.
answered Jul 16 at 21:11
Bernard
110k635103
I suppose the ‘canonical form’ denotes the canonical basis $1, x, x^2 $ of $:mathbf R_2[X]$.
Hint:
The column vectors of this $3times 3$ matrix are the coordinates in the canonical basis of the images of the vectors of this canonical basis.
answered Jul 16 at 21:11
Bernard
110k635103
answered Jul 16 at 21:11
Bernard
110k635103
answered Jul 16 at 21:11
Bernard
110k635103
answered Jul 16 at 21:11
Bernard
110k635103
110k635103
Sorry, can you provide an example?
– Pedro Quintans
Jul 16 at 21:20
In your case, for instance $T(x^2)=2x$. So the coordinates of $T(x^2)$ in basis $(1,x,x^2)$ are $(0, 2,0)$ (transposed).
– Bernard
Jul 16 at 21:25
how? i don't understand, you replace the x from the (1,X,x^2) for 2x?
– Pedro Quintans
Jul 16 at 22:44
Because $2x=colorred0cdot 1+colorred2cdot x+colorred0cdot x^2$, that's all!
– Bernard
Jul 16 at 22:52
add a comment |Â
Sorry, can you provide an example?
– Pedro Quintans
Jul 16 at 21:20
In your case, for instance $T(x^2)=2x$. So the coordinates of $T(x^2)$ in basis $(1,x,x^2)$ are $(0, 2,0)$ (transposed).
– Bernard
Jul 16 at 21:25
how? i don't understand, you replace the x from the (1,X,x^2) for 2x?
– Pedro Quintans
Jul 16 at 22:44
Because $2x=colorred0cdot 1+colorred2cdot x+colorred0cdot x^2$, that's all!
– Bernard
Jul 16 at 22:52
Sorry, can you provide an example?
– Pedro Quintans
Jul 16 at 21:20
Sorry, can you provide an example?
– Pedro Quintans
Jul 16 at 21:20
In your case, for instance $T(x^2)=2x$. So the coordinates of $T(x^2)$ in basis $(1,x,x^2)$ are $(0, 2,0)$ (transposed).
– Bernard
Jul 16 at 21:25
In your case, for instance $T(x^2)=2x$. So the coordinates of $T(x^2)$ in basis $(1,x,x^2)$ are $(0, 2,0)$ (transposed).
– Bernard
Jul 16 at 21:25
how? i don't understand, you replace the x from the (1,X,x^2) for 2x?
– Pedro Quintans
Jul 16 at 22:44
how? i don't understand, you replace the x from the (1,X,x^2) for 2x?
– Pedro Quintans
Jul 16 at 22:44
Because $2x=colorred0cdot 1+colorred2cdot x+colorred0cdot x^2$, that's all!
– Bernard
Jul 16 at 22:52
Because $2x=colorred0cdot 1+colorred2cdot x+colorred0cdot x^2$, that's all!
– Bernard
Jul 16 at 22:52
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2853854%2fdetermine-the-matrix-a-that-represents-t-in-the-canonical-form-of-r2x-in-the-s%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password