Proving a set of vectors is a basis for the quotient map between two vector spaces











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I want to see if my work is justifiable. I am tasked with the following: enter image description here



I will neglect to prove (a), as the work for this is fairly straight forward. I will center my attention on (b).



$$text{Implication is} mathscr {B}_{mathbb R^4 / ker T} =
left{
begin{bmatrix}
begin{pmatrix}
0 \
1 \
0 \
0 \
end{pmatrix}
end{bmatrix},begin{bmatrix}
begin{pmatrix}
0 \
0 \
1 \
0 \
end{pmatrix}
end{bmatrix}
right} text{, where $mathscr B$ is the basis for $mathbb R^4 / ker T$}$$



$$
implies text{for some $[v] in mathbb R^4 / ker T$,}$$

$$[v] = a_1 begin{bmatrix}
begin{pmatrix}
0 \
1 \
0 \
0 \
end{pmatrix}
end{bmatrix} + a_2 begin{bmatrix}
begin{pmatrix}
0 \
0 \
1 \
0 \
end{pmatrix}
end{bmatrix}$$



$$implies
begin{bmatrix}
begin{pmatrix}
x \
y \
z \
t \
end{pmatrix}
end{bmatrix}
=
a_1 begin{bmatrix}
begin{pmatrix}
0 \
1 \
0 \
0 \
end{pmatrix}
end{bmatrix} + a_2 begin{bmatrix}
begin{pmatrix}
0 \
0 \
1 \
0 \
end{pmatrix}
end{bmatrix}$$



$$implies
begin{bmatrix}
begin{pmatrix}
x \
y \
z \
t \
end{pmatrix}
end{bmatrix}
=
begin{bmatrix}
begin{pmatrix}
0 \
a_1 \
a_2 \
0 \
end{pmatrix}
end{bmatrix} $$



$$implies
begin{pmatrix}
x \
y \
z \
t \
end{pmatrix} - begin{pmatrix}
0 \
a_1 \
a_2 \
0 \
end{pmatrix} in ker T$$

$$
implies
begin{pmatrix}
x \
y-a_1 \
z-a_2 \
t \
end{pmatrix} = b_1 begin{pmatrix}
1 \
1 \
0 \
0 \
end{pmatrix} + b_2 begin{pmatrix}
0 \
0 \
1 \
1 \
end{pmatrix} text{, where $b_1, b_2 in ker T$}$$

After some algebra...



$$implies a_1 = y - x$$
$$implies a_2 = z - t$$



Which I think implies $mathscr B$ is spanning? $x,y,z,t$ are all arbitrary. Now I am left with proving $mathscr B$ is linearly independent. I may skip a step or two.



$$
begin{bmatrix}
begin{pmatrix}
0 \
a_1 \
a_2 \
0 \
end{pmatrix}
end{bmatrix} = [0] iff a_1 = a_2 = 0 $$



$$begin{bmatrix}
begin{pmatrix}
0 \
a_1 \
a_2 \
0 \
end{pmatrix}
end{bmatrix} = [0] implies
begin{pmatrix}
0 \
a_1 \
a_2 \
0 \
end{pmatrix} - begin{pmatrix}
0 \
0 \
0 \
0 \
end{pmatrix} in Ker T$$



$$implies
begin{pmatrix}
0 \
a_1 \
a_2 \
0 \
end{pmatrix} = begin{pmatrix}
b1 \
b1 \
b2 \
b2 \
end{pmatrix} implies a_1 = a_2 = 0$$



Hence, basis. Has what I done made sense?










share|cite|improve this question


























    up vote
    2
    down vote

    favorite












    I want to see if my work is justifiable. I am tasked with the following: enter image description here



    I will neglect to prove (a), as the work for this is fairly straight forward. I will center my attention on (b).



    $$text{Implication is} mathscr {B}_{mathbb R^4 / ker T} =
    left{
    begin{bmatrix}
    begin{pmatrix}
    0 \
    1 \
    0 \
    0 \
    end{pmatrix}
    end{bmatrix},begin{bmatrix}
    begin{pmatrix}
    0 \
    0 \
    1 \
    0 \
    end{pmatrix}
    end{bmatrix}
    right} text{, where $mathscr B$ is the basis for $mathbb R^4 / ker T$}$$



    $$
    implies text{for some $[v] in mathbb R^4 / ker T$,}$$

    $$[v] = a_1 begin{bmatrix}
    begin{pmatrix}
    0 \
    1 \
    0 \
    0 \
    end{pmatrix}
    end{bmatrix} + a_2 begin{bmatrix}
    begin{pmatrix}
    0 \
    0 \
    1 \
    0 \
    end{pmatrix}
    end{bmatrix}$$



    $$implies
    begin{bmatrix}
    begin{pmatrix}
    x \
    y \
    z \
    t \
    end{pmatrix}
    end{bmatrix}
    =
    a_1 begin{bmatrix}
    begin{pmatrix}
    0 \
    1 \
    0 \
    0 \
    end{pmatrix}
    end{bmatrix} + a_2 begin{bmatrix}
    begin{pmatrix}
    0 \
    0 \
    1 \
    0 \
    end{pmatrix}
    end{bmatrix}$$



    $$implies
    begin{bmatrix}
    begin{pmatrix}
    x \
    y \
    z \
    t \
    end{pmatrix}
    end{bmatrix}
    =
    begin{bmatrix}
    begin{pmatrix}
    0 \
    a_1 \
    a_2 \
    0 \
    end{pmatrix}
    end{bmatrix} $$



    $$implies
    begin{pmatrix}
    x \
    y \
    z \
    t \
    end{pmatrix} - begin{pmatrix}
    0 \
    a_1 \
    a_2 \
    0 \
    end{pmatrix} in ker T$$

    $$
    implies
    begin{pmatrix}
    x \
    y-a_1 \
    z-a_2 \
    t \
    end{pmatrix} = b_1 begin{pmatrix}
    1 \
    1 \
    0 \
    0 \
    end{pmatrix} + b_2 begin{pmatrix}
    0 \
    0 \
    1 \
    1 \
    end{pmatrix} text{, where $b_1, b_2 in ker T$}$$

    After some algebra...



    $$implies a_1 = y - x$$
    $$implies a_2 = z - t$$



    Which I think implies $mathscr B$ is spanning? $x,y,z,t$ are all arbitrary. Now I am left with proving $mathscr B$ is linearly independent. I may skip a step or two.



    $$
    begin{bmatrix}
    begin{pmatrix}
    0 \
    a_1 \
    a_2 \
    0 \
    end{pmatrix}
    end{bmatrix} = [0] iff a_1 = a_2 = 0 $$



    $$begin{bmatrix}
    begin{pmatrix}
    0 \
    a_1 \
    a_2 \
    0 \
    end{pmatrix}
    end{bmatrix} = [0] implies
    begin{pmatrix}
    0 \
    a_1 \
    a_2 \
    0 \
    end{pmatrix} - begin{pmatrix}
    0 \
    0 \
    0 \
    0 \
    end{pmatrix} in Ker T$$



    $$implies
    begin{pmatrix}
    0 \
    a_1 \
    a_2 \
    0 \
    end{pmatrix} = begin{pmatrix}
    b1 \
    b1 \
    b2 \
    b2 \
    end{pmatrix} implies a_1 = a_2 = 0$$



    Hence, basis. Has what I done made sense?










    share|cite|improve this question
























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      I want to see if my work is justifiable. I am tasked with the following: enter image description here



      I will neglect to prove (a), as the work for this is fairly straight forward. I will center my attention on (b).



      $$text{Implication is} mathscr {B}_{mathbb R^4 / ker T} =
      left{
      begin{bmatrix}
      begin{pmatrix}
      0 \
      1 \
      0 \
      0 \
      end{pmatrix}
      end{bmatrix},begin{bmatrix}
      begin{pmatrix}
      0 \
      0 \
      1 \
      0 \
      end{pmatrix}
      end{bmatrix}
      right} text{, where $mathscr B$ is the basis for $mathbb R^4 / ker T$}$$



      $$
      implies text{for some $[v] in mathbb R^4 / ker T$,}$$

      $$[v] = a_1 begin{bmatrix}
      begin{pmatrix}
      0 \
      1 \
      0 \
      0 \
      end{pmatrix}
      end{bmatrix} + a_2 begin{bmatrix}
      begin{pmatrix}
      0 \
      0 \
      1 \
      0 \
      end{pmatrix}
      end{bmatrix}$$



      $$implies
      begin{bmatrix}
      begin{pmatrix}
      x \
      y \
      z \
      t \
      end{pmatrix}
      end{bmatrix}
      =
      a_1 begin{bmatrix}
      begin{pmatrix}
      0 \
      1 \
      0 \
      0 \
      end{pmatrix}
      end{bmatrix} + a_2 begin{bmatrix}
      begin{pmatrix}
      0 \
      0 \
      1 \
      0 \
      end{pmatrix}
      end{bmatrix}$$



      $$implies
      begin{bmatrix}
      begin{pmatrix}
      x \
      y \
      z \
      t \
      end{pmatrix}
      end{bmatrix}
      =
      begin{bmatrix}
      begin{pmatrix}
      0 \
      a_1 \
      a_2 \
      0 \
      end{pmatrix}
      end{bmatrix} $$



      $$implies
      begin{pmatrix}
      x \
      y \
      z \
      t \
      end{pmatrix} - begin{pmatrix}
      0 \
      a_1 \
      a_2 \
      0 \
      end{pmatrix} in ker T$$

      $$
      implies
      begin{pmatrix}
      x \
      y-a_1 \
      z-a_2 \
      t \
      end{pmatrix} = b_1 begin{pmatrix}
      1 \
      1 \
      0 \
      0 \
      end{pmatrix} + b_2 begin{pmatrix}
      0 \
      0 \
      1 \
      1 \
      end{pmatrix} text{, where $b_1, b_2 in ker T$}$$

      After some algebra...



      $$implies a_1 = y - x$$
      $$implies a_2 = z - t$$



      Which I think implies $mathscr B$ is spanning? $x,y,z,t$ are all arbitrary. Now I am left with proving $mathscr B$ is linearly independent. I may skip a step or two.



      $$
      begin{bmatrix}
      begin{pmatrix}
      0 \
      a_1 \
      a_2 \
      0 \
      end{pmatrix}
      end{bmatrix} = [0] iff a_1 = a_2 = 0 $$



      $$begin{bmatrix}
      begin{pmatrix}
      0 \
      a_1 \
      a_2 \
      0 \
      end{pmatrix}
      end{bmatrix} = [0] implies
      begin{pmatrix}
      0 \
      a_1 \
      a_2 \
      0 \
      end{pmatrix} - begin{pmatrix}
      0 \
      0 \
      0 \
      0 \
      end{pmatrix} in Ker T$$



      $$implies
      begin{pmatrix}
      0 \
      a_1 \
      a_2 \
      0 \
      end{pmatrix} = begin{pmatrix}
      b1 \
      b1 \
      b2 \
      b2 \
      end{pmatrix} implies a_1 = a_2 = 0$$



      Hence, basis. Has what I done made sense?










      share|cite|improve this question













      I want to see if my work is justifiable. I am tasked with the following: enter image description here



      I will neglect to prove (a), as the work for this is fairly straight forward. I will center my attention on (b).



      $$text{Implication is} mathscr {B}_{mathbb R^4 / ker T} =
      left{
      begin{bmatrix}
      begin{pmatrix}
      0 \
      1 \
      0 \
      0 \
      end{pmatrix}
      end{bmatrix},begin{bmatrix}
      begin{pmatrix}
      0 \
      0 \
      1 \
      0 \
      end{pmatrix}
      end{bmatrix}
      right} text{, where $mathscr B$ is the basis for $mathbb R^4 / ker T$}$$



      $$
      implies text{for some $[v] in mathbb R^4 / ker T$,}$$

      $$[v] = a_1 begin{bmatrix}
      begin{pmatrix}
      0 \
      1 \
      0 \
      0 \
      end{pmatrix}
      end{bmatrix} + a_2 begin{bmatrix}
      begin{pmatrix}
      0 \
      0 \
      1 \
      0 \
      end{pmatrix}
      end{bmatrix}$$



      $$implies
      begin{bmatrix}
      begin{pmatrix}
      x \
      y \
      z \
      t \
      end{pmatrix}
      end{bmatrix}
      =
      a_1 begin{bmatrix}
      begin{pmatrix}
      0 \
      1 \
      0 \
      0 \
      end{pmatrix}
      end{bmatrix} + a_2 begin{bmatrix}
      begin{pmatrix}
      0 \
      0 \
      1 \
      0 \
      end{pmatrix}
      end{bmatrix}$$



      $$implies
      begin{bmatrix}
      begin{pmatrix}
      x \
      y \
      z \
      t \
      end{pmatrix}
      end{bmatrix}
      =
      begin{bmatrix}
      begin{pmatrix}
      0 \
      a_1 \
      a_2 \
      0 \
      end{pmatrix}
      end{bmatrix} $$



      $$implies
      begin{pmatrix}
      x \
      y \
      z \
      t \
      end{pmatrix} - begin{pmatrix}
      0 \
      a_1 \
      a_2 \
      0 \
      end{pmatrix} in ker T$$

      $$
      implies
      begin{pmatrix}
      x \
      y-a_1 \
      z-a_2 \
      t \
      end{pmatrix} = b_1 begin{pmatrix}
      1 \
      1 \
      0 \
      0 \
      end{pmatrix} + b_2 begin{pmatrix}
      0 \
      0 \
      1 \
      1 \
      end{pmatrix} text{, where $b_1, b_2 in ker T$}$$

      After some algebra...



      $$implies a_1 = y - x$$
      $$implies a_2 = z - t$$



      Which I think implies $mathscr B$ is spanning? $x,y,z,t$ are all arbitrary. Now I am left with proving $mathscr B$ is linearly independent. I may skip a step or two.



      $$
      begin{bmatrix}
      begin{pmatrix}
      0 \
      a_1 \
      a_2 \
      0 \
      end{pmatrix}
      end{bmatrix} = [0] iff a_1 = a_2 = 0 $$



      $$begin{bmatrix}
      begin{pmatrix}
      0 \
      a_1 \
      a_2 \
      0 \
      end{pmatrix}
      end{bmatrix} = [0] implies
      begin{pmatrix}
      0 \
      a_1 \
      a_2 \
      0 \
      end{pmatrix} - begin{pmatrix}
      0 \
      0 \
      0 \
      0 \
      end{pmatrix} in Ker T$$



      $$implies
      begin{pmatrix}
      0 \
      a_1 \
      a_2 \
      0 \
      end{pmatrix} = begin{pmatrix}
      b1 \
      b1 \
      b2 \
      b2 \
      end{pmatrix} implies a_1 = a_2 = 0$$



      Hence, basis. Has what I done made sense?







      linear-algebra proof-verification quotient-spaces






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 17 at 18:12









      sangstar

      838214




      838214






















          1 Answer
          1






          active

          oldest

          votes

















          up vote
          1
          down vote



          accepted










          It's all correct.



          Note that the 4 given vectors form a basis of $Bbb R^4$, two of them spans the kernel, so the (images of) the other two will be a basis of the quotient space.






          share|cite|improve this answer





















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            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            1
            down vote



            accepted










            It's all correct.



            Note that the 4 given vectors form a basis of $Bbb R^4$, two of them spans the kernel, so the (images of) the other two will be a basis of the quotient space.






            share|cite|improve this answer

























              up vote
              1
              down vote



              accepted










              It's all correct.



              Note that the 4 given vectors form a basis of $Bbb R^4$, two of them spans the kernel, so the (images of) the other two will be a basis of the quotient space.






              share|cite|improve this answer























                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                It's all correct.



                Note that the 4 given vectors form a basis of $Bbb R^4$, two of them spans the kernel, so the (images of) the other two will be a basis of the quotient space.






                share|cite|improve this answer












                It's all correct.



                Note that the 4 given vectors form a basis of $Bbb R^4$, two of them spans the kernel, so the (images of) the other two will be a basis of the quotient space.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 17 at 22:02









                Berci

                59.1k23671




                59.1k23671






























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