Finding the matrix of $L$ with respect to the following bases:

Clash Royale CLAN TAG#URR8PPP

up vote
0
down vote

favorite

Let $L:mathbbR^3rightarrowmathbbR^4$ be the linear map:
$$L((x_1,x_2,x_3)^T)=(x_1-x_2+x_3,x_1-2x_2,x_2-3x_3,2x_1+x_3)^T$$
Find the matrix of $L$ with respect to the following bases: $(0,0,1)^T,(0,1,1)^T,(1,1,1)^T$ for$mathbbR^3$ and $(0,1,0,0)^T,(1,1,0,0)^T,(0,0,1,0)^T,(0,0,1,1)^T$ for $mathbbR^4$

My try:

Given $beginbmatrix x_1 \ x_2 \ x_3 endbmatrix=beginbmatrix x_1-x_2+x_3 \ x_1-2x_2 \ x_2-3x_3 \ 2x_1+x_3 endbmatrix=x_1beginbmatrix 1 \ 1 \ 0 \ 2 endbmatrix+x_2beginbmatrix -1 \ -2 \ 1 \ 0 endbmatrix+x_3beginbmatrix 1 \ 0 \ -3 \ 1 endbmatrix$

After this I have no idea what to do. Can anyone please explain how to proceed further?

linear-algebra

share|cite|improve this question

asked Jul 31 at 17:12

philip

938

add a comment | 

up vote
0
down vote

favorite

Let $L:mathbbR^3rightarrowmathbbR^4$ be the linear map:
$$L((x_1,x_2,x_3)^T)=(x_1-x_2+x_3,x_1-2x_2,x_2-3x_3,2x_1+x_3)^T$$
Find the matrix of $L$ with respect to the following bases: $(0,0,1)^T,(0,1,1)^T,(1,1,1)^T$ for$mathbbR^3$ and $(0,1,0,0)^T,(1,1,0,0)^T,(0,0,1,0)^T,(0,0,1,1)^T$ for $mathbbR^4$

My try:

Given $beginbmatrix x_1 \ x_2 \ x_3 endbmatrix=beginbmatrix x_1-x_2+x_3 \ x_1-2x_2 \ x_2-3x_3 \ 2x_1+x_3 endbmatrix=x_1beginbmatrix 1 \ 1 \ 0 \ 2 endbmatrix+x_2beginbmatrix -1 \ -2 \ 1 \ 0 endbmatrix+x_3beginbmatrix 1 \ 0 \ -3 \ 1 endbmatrix$

After this I have no idea what to do. Can anyone please explain how to proceed further?

linear-algebra

share|cite|improve this question

asked Jul 31 at 17:12

philip

938

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Let $L:mathbbR^3rightarrowmathbbR^4$ be the linear map:
$$L((x_1,x_2,x_3)^T)=(x_1-x_2+x_3,x_1-2x_2,x_2-3x_3,2x_1+x_3)^T$$
Find the matrix of $L$ with respect to the following bases: $(0,0,1)^T,(0,1,1)^T,(1,1,1)^T$ for$mathbbR^3$ and $(0,1,0,0)^T,(1,1,0,0)^T,(0,0,1,0)^T,(0,0,1,1)^T$ for $mathbbR^4$

My try:

Given $beginbmatrix x_1 \ x_2 \ x_3 endbmatrix=beginbmatrix x_1-x_2+x_3 \ x_1-2x_2 \ x_2-3x_3 \ 2x_1+x_3 endbmatrix=x_1beginbmatrix 1 \ 1 \ 0 \ 2 endbmatrix+x_2beginbmatrix -1 \ -2 \ 1 \ 0 endbmatrix+x_3beginbmatrix 1 \ 0 \ -3 \ 1 endbmatrix$

After this I have no idea what to do. Can anyone please explain how to proceed further?

linear-algebra

share|cite|improve this question

asked Jul 31 at 17:12

philip

938

Let $L:mathbbR^3rightarrowmathbbR^4$ be the linear map:
$$L((x_1,x_2,x_3)^T)=(x_1-x_2+x_3,x_1-2x_2,x_2-3x_3,2x_1+x_3)^T$$
Find the matrix of $L$ with respect to the following bases: $(0,0,1)^T,(0,1,1)^T,(1,1,1)^T$ for$mathbbR^3$ and $(0,1,0,0)^T,(1,1,0,0)^T,(0,0,1,0)^T,(0,0,1,1)^T$ for $mathbbR^4$

My try:

Given $beginbmatrix x_1 \ x_2 \ x_3 endbmatrix=beginbmatrix x_1-x_2+x_3 \ x_1-2x_2 \ x_2-3x_3 \ 2x_1+x_3 endbmatrix=x_1beginbmatrix 1 \ 1 \ 0 \ 2 endbmatrix+x_2beginbmatrix -1 \ -2 \ 1 \ 0 endbmatrix+x_3beginbmatrix 1 \ 0 \ -3 \ 1 endbmatrix$

After this I have no idea what to do. Can anyone please explain how to proceed further?

linear-algebra

share|cite|improve this question

asked Jul 31 at 17:12

philip

938

share|cite|improve this question

share|cite|improve this question

asked Jul 31 at 17:12

philip

938

asked Jul 31 at 17:12

philip

938

asked Jul 31 at 17:12

philip

938

938

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
2
down vote



accepted

Let's give names to the bases-you have a basis $B=(e_1,e_2,e_3)$ in $mathbbR^3$ and a basis $C=(f_1,f_2,f_3,f_4)$ in $mathbbR^4$. By definition, the columns of the transformation matrix with respect to these bases are the coordinates of the vectors $L(e_1),L(e_2),L(e_3)$ with respect to the basis C. So calculate the vectors $L(e_1),L(e_2),L(e_3)$ and then calculate their coordinates in the basis C. That's simply solving systems of linear equations-might be annoying but not hard.

share|cite|improve this answer

answered Jul 31 at 17:22

Mark

5949

• 
  
  
  

  
  

  
  
  
  
  
  

  
  So, I just need to plug in $e_1,e_2,e_3$ in $L(x_1,x_2,x_3)^T$ right?
  – philip
  Jul 31 at 18:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, you need to use the transformation $L$ on each of the vectors $e_1,e_2,e_3$. In the basis you wrote you have $e_1=(0,0,1)$, $e_2=(0,1,1)$, $e_3=(1,1,1)$. And as I wrote, you then need to calculate the coordinates of the vectors you get with respect to the basis in $mathbbR^4$ that is given to you. (and not with respect to the standard basis of $mathbbR^4$)
  – Mark
  Jul 31 at 18:20
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  So, when I plugged in $e_1,e_2,e_3$ in $L$ I got the matrix as $beginbmatrix 1 & 0 & 1 \ 0 & -2 & -1 \ -3 & -2 & -2 \ 1 & 1 & 3 endbmatrix$ is it correct
  – philip
  Jul 31 at 19:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  And how should I plug in $f_1,f_2,f_3,f_4$ because there is no $x_4$ term in the given $L$
  – philip
  Jul 31 at 19:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ok, you calculated the vectors $L(e_1),L(e_2),L(e_3)$. Now you need to find their coordinates with respect to the basis $(f_1,f_2,f_3,f_4)$. Coordinates of $L(e_1)$ with respect to the basis are scalars $a,b,c,d$ such that $L(e_1)=af_1+bf_2+cf_3+df_4$. Find them and then $(a,b,c,d)^T$ will be the first column in your matrix. Then do the same for $L(e_2)$ and $L(e_3)$.
  – Mark
  Jul 31 at 19:53
  

  
  
  
  

 | 
show 4 more comments

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

 

draft saved

draft discarded

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

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2868264%2ffinding-the-matrix-of-l-with-respect-to-the-following-bases%23new-answer', 'question_page');

);

Post as a guest

Name Email

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

Let's give names to the bases-you have a basis $B=(e_1,e_2,e_3)$ in $mathbbR^3$ and a basis $C=(f_1,f_2,f_3,f_4)$ in $mathbbR^4$. By definition, the columns of the transformation matrix with respect to these bases are the coordinates of the vectors $L(e_1),L(e_2),L(e_3)$ with respect to the basis C. So calculate the vectors $L(e_1),L(e_2),L(e_3)$ and then calculate their coordinates in the basis C. That's simply solving systems of linear equations-might be annoying but not hard.

share|cite|improve this answer

answered Jul 31 at 17:22

Mark

5949

• 
  
  
  

  
  

  
  
  
  
  
  

  
  So, I just need to plug in $e_1,e_2,e_3$ in $L(x_1,x_2,x_3)^T$ right?
  – philip
  Jul 31 at 18:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, you need to use the transformation $L$ on each of the vectors $e_1,e_2,e_3$. In the basis you wrote you have $e_1=(0,0,1)$, $e_2=(0,1,1)$, $e_3=(1,1,1)$. And as I wrote, you then need to calculate the coordinates of the vectors you get with respect to the basis in $mathbbR^4$ that is given to you. (and not with respect to the standard basis of $mathbbR^4$)
  – Mark
  Jul 31 at 18:20
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  So, when I plugged in $e_1,e_2,e_3$ in $L$ I got the matrix as $beginbmatrix 1 & 0 & 1 \ 0 & -2 & -1 \ -3 & -2 & -2 \ 1 & 1 & 3 endbmatrix$ is it correct
  – philip
  Jul 31 at 19:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  And how should I plug in $f_1,f_2,f_3,f_4$ because there is no $x_4$ term in the given $L$
  – philip
  Jul 31 at 19:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ok, you calculated the vectors $L(e_1),L(e_2),L(e_3)$. Now you need to find their coordinates with respect to the basis $(f_1,f_2,f_3,f_4)$. Coordinates of $L(e_1)$ with respect to the basis are scalars $a,b,c,d$ such that $L(e_1)=af_1+bf_2+cf_3+df_4$. Find them and then $(a,b,c,d)^T$ will be the first column in your matrix. Then do the same for $L(e_2)$ and $L(e_3)$.
  – Mark
  Jul 31 at 19:53
  

  
  
  
  

 | 
show 4 more comments

up vote
2
down vote



accepted

Let's give names to the bases-you have a basis $B=(e_1,e_2,e_3)$ in $mathbbR^3$ and a basis $C=(f_1,f_2,f_3,f_4)$ in $mathbbR^4$. By definition, the columns of the transformation matrix with respect to these bases are the coordinates of the vectors $L(e_1),L(e_2),L(e_3)$ with respect to the basis C. So calculate the vectors $L(e_1),L(e_2),L(e_3)$ and then calculate their coordinates in the basis C. That's simply solving systems of linear equations-might be annoying but not hard.

share|cite|improve this answer

answered Jul 31 at 17:22

Mark

5949

• 
  
  
  

  
  

  
  
  
  
  
  

  
  So, I just need to plug in $e_1,e_2,e_3$ in $L(x_1,x_2,x_3)^T$ right?
  – philip
  Jul 31 at 18:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, you need to use the transformation $L$ on each of the vectors $e_1,e_2,e_3$. In the basis you wrote you have $e_1=(0,0,1)$, $e_2=(0,1,1)$, $e_3=(1,1,1)$. And as I wrote, you then need to calculate the coordinates of the vectors you get with respect to the basis in $mathbbR^4$ that is given to you. (and not with respect to the standard basis of $mathbbR^4$)
  – Mark
  Jul 31 at 18:20
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  So, when I plugged in $e_1,e_2,e_3$ in $L$ I got the matrix as $beginbmatrix 1 & 0 & 1 \ 0 & -2 & -1 \ -3 & -2 & -2 \ 1 & 1 & 3 endbmatrix$ is it correct
  – philip
  Jul 31 at 19:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  And how should I plug in $f_1,f_2,f_3,f_4$ because there is no $x_4$ term in the given $L$
  – philip
  Jul 31 at 19:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ok, you calculated the vectors $L(e_1),L(e_2),L(e_3)$. Now you need to find their coordinates with respect to the basis $(f_1,f_2,f_3,f_4)$. Coordinates of $L(e_1)$ with respect to the basis are scalars $a,b,c,d$ such that $L(e_1)=af_1+bf_2+cf_3+df_4$. Find them and then $(a,b,c,d)^T$ will be the first column in your matrix. Then do the same for $L(e_2)$ and $L(e_3)$.
  – Mark
  Jul 31 at 19:53
  

  
  
  
  

 | 
show 4 more comments

up vote
2
down vote



accepted

up vote
2
down vote



accepted

Let's give names to the bases-you have a basis $B=(e_1,e_2,e_3)$ in $mathbbR^3$ and a basis $C=(f_1,f_2,f_3,f_4)$ in $mathbbR^4$. By definition, the columns of the transformation matrix with respect to these bases are the coordinates of the vectors $L(e_1),L(e_2),L(e_3)$ with respect to the basis C. So calculate the vectors $L(e_1),L(e_2),L(e_3)$ and then calculate their coordinates in the basis C. That's simply solving systems of linear equations-might be annoying but not hard.

share|cite|improve this answer

answered Jul 31 at 17:22

Mark

5949

Let's give names to the bases-you have a basis $B=(e_1,e_2,e_3)$ in $mathbbR^3$ and a basis $C=(f_1,f_2,f_3,f_4)$ in $mathbbR^4$. By definition, the columns of the transformation matrix with respect to these bases are the coordinates of the vectors $L(e_1),L(e_2),L(e_3)$ with respect to the basis C. So calculate the vectors $L(e_1),L(e_2),L(e_3)$ and then calculate their coordinates in the basis C. That's simply solving systems of linear equations-might be annoying but not hard.

share|cite|improve this answer

answered Jul 31 at 17:22

Mark

5949

share|cite|improve this answer

share|cite|improve this answer

answered Jul 31 at 17:22

Mark

5949

answered Jul 31 at 17:22

Mark

5949

answered Jul 31 at 17:22

Mark

5949

5949

• 
  
  
  

  
  

  
  
  
  
  
  

  
  So, I just need to plug in $e_1,e_2,e_3$ in $L(x_1,x_2,x_3)^T$ right?
  – philip
  Jul 31 at 18:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, you need to use the transformation $L$ on each of the vectors $e_1,e_2,e_3$. In the basis you wrote you have $e_1=(0,0,1)$, $e_2=(0,1,1)$, $e_3=(1,1,1)$. And as I wrote, you then need to calculate the coordinates of the vectors you get with respect to the basis in $mathbbR^4$ that is given to you. (and not with respect to the standard basis of $mathbbR^4$)
  – Mark
  Jul 31 at 18:20
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  So, when I plugged in $e_1,e_2,e_3$ in $L$ I got the matrix as $beginbmatrix 1 & 0 & 1 \ 0 & -2 & -1 \ -3 & -2 & -2 \ 1 & 1 & 3 endbmatrix$ is it correct
  – philip
  Jul 31 at 19:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  And how should I plug in $f_1,f_2,f_3,f_4$ because there is no $x_4$ term in the given $L$
  – philip
  Jul 31 at 19:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ok, you calculated the vectors $L(e_1),L(e_2),L(e_3)$. Now you need to find their coordinates with respect to the basis $(f_1,f_2,f_3,f_4)$. Coordinates of $L(e_1)$ with respect to the basis are scalars $a,b,c,d$ such that $L(e_1)=af_1+bf_2+cf_3+df_4$. Find them and then $(a,b,c,d)^T$ will be the first column in your matrix. Then do the same for $L(e_2)$ and $L(e_3)$.
  – Mark
  Jul 31 at 19:53
  

  
  
  
  

 | 
show 4 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  So, I just need to plug in $e_1,e_2,e_3$ in $L(x_1,x_2,x_3)^T$ right?
  – philip
  Jul 31 at 18:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, you need to use the transformation $L$ on each of the vectors $e_1,e_2,e_3$. In the basis you wrote you have $e_1=(0,0,1)$, $e_2=(0,1,1)$, $e_3=(1,1,1)$. And as I wrote, you then need to calculate the coordinates of the vectors you get with respect to the basis in $mathbbR^4$ that is given to you. (and not with respect to the standard basis of $mathbbR^4$)
  – Mark
  Jul 31 at 18:20
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  So, when I plugged in $e_1,e_2,e_3$ in $L$ I got the matrix as $beginbmatrix 1 & 0 & 1 \ 0 & -2 & -1 \ -3 & -2 & -2 \ 1 & 1 & 3 endbmatrix$ is it correct
  – philip
  Jul 31 at 19:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  And how should I plug in $f_1,f_2,f_3,f_4$ because there is no $x_4$ term in the given $L$
  – philip
  Jul 31 at 19:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ok, you calculated the vectors $L(e_1),L(e_2),L(e_3)$. Now you need to find their coordinates with respect to the basis $(f_1,f_2,f_3,f_4)$. Coordinates of $L(e_1)$ with respect to the basis are scalars $a,b,c,d$ such that $L(e_1)=af_1+bf_2+cf_3+df_4$. Find them and then $(a,b,c,d)^T$ will be the first column in your matrix. Then do the same for $L(e_2)$ and $L(e_3)$.
  – Mark
  Jul 31 at 19:53
  

  
  
  
  

So, I just need to plug in $e_1,e_2,e_3$ in $L(x_1,x_2,x_3)^T$ right?
– philip
Jul 31 at 18:08

So, I just need to plug in $e_1,e_2,e_3$ in $L(x_1,x_2,x_3)^T$ right?
– philip
Jul 31 at 18:08

Yes, you need to use the transformation $L$ on each of the vectors $e_1,e_2,e_3$. In the basis you wrote you have $e_1=(0,0,1)$, $e_2=(0,1,1)$, $e_3=(1,1,1)$. And as I wrote, you then need to calculate the coordinates of the vectors you get with respect to the basis in $mathbbR^4$ that is given to you. (and not with respect to the standard basis of $mathbbR^4$)
– Mark
Jul 31 at 18:20

Yes, you need to use the transformation $L$ on each of the vectors $e_1,e_2,e_3$. In the basis you wrote you have $e_1=(0,0,1)$, $e_2=(0,1,1)$, $e_3=(1,1,1)$. And as I wrote, you then need to calculate the coordinates of the vectors you get with respect to the basis in $mathbbR^4$ that is given to you. (and not with respect to the standard basis of $mathbbR^4$)
– Mark
Jul 31 at 18:20

So, when I plugged in $e_1,e_2,e_3$ in $L$ I got the matrix as $beginbmatrix 1 & 0 & 1 \ 0 & -2 & -1 \ -3 & -2 & -2 \ 1 & 1 & 3 endbmatrix$ is it correct
– philip
Jul 31 at 19:15

So, when I plugged in $e_1,e_2,e_3$ in $L$ I got the matrix as $beginbmatrix 1 & 0 & 1 \ 0 & -2 & -1 \ -3 & -2 & -2 \ 1 & 1 & 3 endbmatrix$ is it correct
– philip
Jul 31 at 19:15

And how should I plug in $f_1,f_2,f_3,f_4$ because there is no $x_4$ term in the given $L$
– philip
Jul 31 at 19:16

And how should I plug in $f_1,f_2,f_3,f_4$ because there is no $x_4$ term in the given $L$
– philip
Jul 31 at 19:16

Ok, you calculated the vectors $L(e_1),L(e_2),L(e_3)$. Now you need to find their coordinates with respect to the basis $(f_1,f_2,f_3,f_4)$. Coordinates of $L(e_1)$ with respect to the basis are scalars $a,b,c,d$ such that $L(e_1)=af_1+bf_2+cf_3+df_4$. Find them and then $(a,b,c,d)^T$ will be the first column in your matrix. Then do the same for $L(e_2)$ and $L(e_3)$.
– Mark
Jul 31 at 19:53

Ok, you calculated the vectors $L(e_1),L(e_2),L(e_3)$. Now you need to find their coordinates with respect to the basis $(f_1,f_2,f_3,f_4)$. Coordinates of $L(e_1)$ with respect to the basis are scalars $a,b,c,d$ such that $L(e_1)=af_1+bf_2+cf_3+df_4$. Find them and then $(a,b,c,d)^T$ will be the first column in your matrix. Then do the same for $L(e_2)$ and $L(e_3)$.
– Mark
Jul 31 at 19:53

 | 
show 4 more comments

 

draft saved

draft discarded

 

draft saved

draft discarded

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

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2868264%2ffinding-the-matrix-of-l-with-respect-to-the-following-bases%23new-answer', 'question_page');

);

Post as a guest

Name Email

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

Name Email

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

Name Email

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

Post as a guest

Name Email

Name Email

Name

Email

Name Email

Name

Email