solving a basis with two free variables

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I've been trying to solve a basis for this question but nothing seems to be working. I've written the linear system in a matrix which consists of the two equations, put it into reduced row echelon form, found that $z$ and $w$ are free variables. what's next?






linear-algebra







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  • $w=-x-y$ then $x-z-2x-2y=0$ so $z=-x-2y$. Now what can you say for the vectors $(1, 0, -1, -1), (0, 1, -2, -1)$ ?
    – dmtri
    Sep 1 at 5:22










  • they form the basis? how did you arrive at $1,0,-1,-1 and 0,1,-2,-2$ though?
    – lohboys
    Sep 1 at 5:47










  • You can solve the system of your two equations and you get a solution with 2 free variables. Yes they are independent.
    – dmtri
    Sep 1 at 6:46










  • How to read a kernel basis from the RREF.
    – amd
    Sep 1 at 8:53














up vote
1
down vote

favorite












enter image description here



I've been trying to solve a basis for this question but nothing seems to be working. I've written the linear system in a matrix which consists of the two equations, put it into reduced row echelon form, found that $z$ and $w$ are free variables. what's next?






linear-algebra







share|cite|improve this question





















  • $w=-x-y$ then $x-z-2x-2y=0$ so $z=-x-2y$. Now what can you say for the vectors $(1, 0, -1, -1), (0, 1, -2, -1)$ ?
    – dmtri
    Sep 1 at 5:22










  • they form the basis? how did you arrive at $1,0,-1,-1 and 0,1,-2,-2$ though?
    – lohboys
    Sep 1 at 5:47










  • You can solve the system of your two equations and you get a solution with 2 free variables. Yes they are independent.
    – dmtri
    Sep 1 at 6:46










  • How to read a kernel basis from the RREF.
    – amd
    Sep 1 at 8:53












up vote
1
down vote

favorite









up vote
1
down vote

favorite











enter image description here



I've been trying to solve a basis for this question but nothing seems to be working. I've written the linear system in a matrix which consists of the two equations, put it into reduced row echelon form, found that $z$ and $w$ are free variables. what's next?






linear-algebra







share|cite|improve this question













enter image description here



I've been trying to solve a basis for this question but nothing seems to be working. I've written the linear system in a matrix which consists of the two equations, put it into reduced row echelon form, found that $z$ and $w$ are free variables. what's next?






linear-algebra




linear-algebra






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share|cite|improve this question











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share|cite|improve this question










asked Sep 1 at 5:11









lohboys

617




617











  • $w=-x-y$ then $x-z-2x-2y=0$ so $z=-x-2y$. Now what can you say for the vectors $(1, 0, -1, -1), (0, 1, -2, -1)$ ?
    – dmtri
    Sep 1 at 5:22










  • they form the basis? how did you arrive at $1,0,-1,-1 and 0,1,-2,-2$ though?
    – lohboys
    Sep 1 at 5:47










  • You can solve the system of your two equations and you get a solution with 2 free variables. Yes they are independent.
    – dmtri
    Sep 1 at 6:46










  • How to read a kernel basis from the RREF.
    – amd
    Sep 1 at 8:53
















  • $w=-x-y$ then $x-z-2x-2y=0$ so $z=-x-2y$. Now what can you say for the vectors $(1, 0, -1, -1), (0, 1, -2, -1)$ ?
    – dmtri
    Sep 1 at 5:22










  • they form the basis? how did you arrive at $1,0,-1,-1 and 0,1,-2,-2$ though?
    – lohboys
    Sep 1 at 5:47










  • You can solve the system of your two equations and you get a solution with 2 free variables. Yes they are independent.
    – dmtri
    Sep 1 at 6:46










  • How to read a kernel basis from the RREF.
    – amd
    Sep 1 at 8:53















$w=-x-y$ then $x-z-2x-2y=0$ so $z=-x-2y$. Now what can you say for the vectors $(1, 0, -1, -1), (0, 1, -2, -1)$ ?
– dmtri
Sep 1 at 5:22




$w=-x-y$ then $x-z-2x-2y=0$ so $z=-x-2y$. Now what can you say for the vectors $(1, 0, -1, -1), (0, 1, -2, -1)$ ?
– dmtri
Sep 1 at 5:22












they form the basis? how did you arrive at $1,0,-1,-1 and 0,1,-2,-2$ though?
– lohboys
Sep 1 at 5:47




they form the basis? how did you arrive at $1,0,-1,-1 and 0,1,-2,-2$ though?
– lohboys
Sep 1 at 5:47












You can solve the system of your two equations and you get a solution with 2 free variables. Yes they are independent.
– dmtri
Sep 1 at 6:46




You can solve the system of your two equations and you get a solution with 2 free variables. Yes they are independent.
– dmtri
Sep 1 at 6:46












How to read a kernel basis from the RREF.
– amd
Sep 1 at 8:53




How to read a kernel basis from the RREF.
– amd
Sep 1 at 8:53










1 Answer
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$x+y+w=0$



$x-z+2w=0$



The first equation leads to $x = -y-w$ and the second leads to $x = z-2w$. Then we have $-y-w = z-2w$, so $y = -z+w$



Then $x = z-2w$ and $y = -z+w$. We have two variables $z,w$ that are free, as expected.



So you have a vector $(x,y,z,w)$ and want to write it as a combination of basis vectors. As the equations showed: $(x,y,z,w) = (z-2w,-z+w,z,w) = (1,-1,1,0)z + (-2,1,0,1)w$, hence a basis would be $(1,-1,1,0),(-2,1,0,1)$






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    1 Answer
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    active

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    1 Answer
    1






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    active

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    up vote
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    $x+y+w=0$



    $x-z+2w=0$



    The first equation leads to $x = -y-w$ and the second leads to $x = z-2w$. Then we have $-y-w = z-2w$, so $y = -z+w$



    Then $x = z-2w$ and $y = -z+w$. We have two variables $z,w$ that are free, as expected.



    So you have a vector $(x,y,z,w)$ and want to write it as a combination of basis vectors. As the equations showed: $(x,y,z,w) = (z-2w,-z+w,z,w) = (1,-1,1,0)z + (-2,1,0,1)w$, hence a basis would be $(1,-1,1,0),(-2,1,0,1)$






    share|cite|improve this answer
























      up vote
      2
      down vote













      $x+y+w=0$



      $x-z+2w=0$



      The first equation leads to $x = -y-w$ and the second leads to $x = z-2w$. Then we have $-y-w = z-2w$, so $y = -z+w$



      Then $x = z-2w$ and $y = -z+w$. We have two variables $z,w$ that are free, as expected.



      So you have a vector $(x,y,z,w)$ and want to write it as a combination of basis vectors. As the equations showed: $(x,y,z,w) = (z-2w,-z+w,z,w) = (1,-1,1,0)z + (-2,1,0,1)w$, hence a basis would be $(1,-1,1,0),(-2,1,0,1)$






      share|cite|improve this answer






















        up vote
        2
        down vote










        up vote
        2
        down vote









        $x+y+w=0$



        $x-z+2w=0$



        The first equation leads to $x = -y-w$ and the second leads to $x = z-2w$. Then we have $-y-w = z-2w$, so $y = -z+w$



        Then $x = z-2w$ and $y = -z+w$. We have two variables $z,w$ that are free, as expected.



        So you have a vector $(x,y,z,w)$ and want to write it as a combination of basis vectors. As the equations showed: $(x,y,z,w) = (z-2w,-z+w,z,w) = (1,-1,1,0)z + (-2,1,0,1)w$, hence a basis would be $(1,-1,1,0),(-2,1,0,1)$






        share|cite|improve this answer












        $x+y+w=0$



        $x-z+2w=0$



        The first equation leads to $x = -y-w$ and the second leads to $x = z-2w$. Then we have $-y-w = z-2w$, so $y = -z+w$



        Then $x = z-2w$ and $y = -z+w$. We have two variables $z,w$ that are free, as expected.



        So you have a vector $(x,y,z,w)$ and want to write it as a combination of basis vectors. As the equations showed: $(x,y,z,w) = (z-2w,-z+w,z,w) = (1,-1,1,0)z + (-2,1,0,1)w$, hence a basis would be $(1,-1,1,0),(-2,1,0,1)$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Sep 1 at 6:16









        dude3221

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