How to simplify equation by change of bases?











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Try to simplify the equation by change of bases (or change of coordinate system) : $2x^2 - 4xy + 5y^2 = 1$



1) Find Transition Matrix P from B' to B:



B' = {(2/$sqrt5$, 1/$sqrt5$),(-1/$sqrt5$, 2/$sqrt5$)



B = {(1,0), (0,1)}



$begin{bmatrix}
2/sqrt5 &
1/sqrt5 &
1 &
0
\
-1/sqrt5 &
2/sqrt5 &
0 & 1
end{bmatrix}$
RREF $begin{bmatrix}
1 &
0 &
2/sqrt5 &
-1/sqrt5
\
0 &
1 &
1/sqrt5 &
2/sqrt5
end{bmatrix}$



P = $begin{bmatrix}
2/sqrt5 &
-1/sqrt5
\
1/sqrt5 &
2/sqrt5
end{bmatrix}$



2) Convert a coordinate matrix $begin{bmatrix}x'\y'end{bmatrix}$ relative to the basis $B'$ into the coordinate matrix $begin{bmatrix}x\yend{bmatrix}$ relative to the basis $B$. Here I choose to make x' and y' = 1.



$begin{bmatrix}x\yend{bmatrix}$ = $begin{bmatrix}
2/sqrt5 &
-1/sqrt5
\
1/sqrt5 &
2/sqrt5
end{bmatrix}$
$begin{bmatrix}x'=1\y'=1end{bmatrix}$ = $begin{bmatrix}
sqrt2
\
sqrt1.8
end{bmatrix}$



3) Plug both $x,y$ found in 2) to the original equation and find a simplified equation in terms of the variables $x', y'$



$2x^2 - 4xy + 5y^2 = 1$ , $x=2$, $y=sqrt1.8$



$2(2) - 4((sqrt2)(sqrt1.8)) + 5(sqrt1.8)^2) = 1$



5.41 != 0




Why didn't my equation equal 1 after I plugged in the x, y values that I got from 2)? I don't know what I did wrong...it is supposed to be 1 = 1 isnt it?











share|cite|improve this question




























    up vote
    -2
    down vote

    favorite












    Try to simplify the equation by change of bases (or change of coordinate system) : $2x^2 - 4xy + 5y^2 = 1$



    1) Find Transition Matrix P from B' to B:



    B' = {(2/$sqrt5$, 1/$sqrt5$),(-1/$sqrt5$, 2/$sqrt5$)



    B = {(1,0), (0,1)}



    $begin{bmatrix}
    2/sqrt5 &
    1/sqrt5 &
    1 &
    0
    \
    -1/sqrt5 &
    2/sqrt5 &
    0 & 1
    end{bmatrix}$
    RREF $begin{bmatrix}
    1 &
    0 &
    2/sqrt5 &
    -1/sqrt5
    \
    0 &
    1 &
    1/sqrt5 &
    2/sqrt5
    end{bmatrix}$



    P = $begin{bmatrix}
    2/sqrt5 &
    -1/sqrt5
    \
    1/sqrt5 &
    2/sqrt5
    end{bmatrix}$



    2) Convert a coordinate matrix $begin{bmatrix}x'\y'end{bmatrix}$ relative to the basis $B'$ into the coordinate matrix $begin{bmatrix}x\yend{bmatrix}$ relative to the basis $B$. Here I choose to make x' and y' = 1.



    $begin{bmatrix}x\yend{bmatrix}$ = $begin{bmatrix}
    2/sqrt5 &
    -1/sqrt5
    \
    1/sqrt5 &
    2/sqrt5
    end{bmatrix}$
    $begin{bmatrix}x'=1\y'=1end{bmatrix}$ = $begin{bmatrix}
    sqrt2
    \
    sqrt1.8
    end{bmatrix}$



    3) Plug both $x,y$ found in 2) to the original equation and find a simplified equation in terms of the variables $x', y'$



    $2x^2 - 4xy + 5y^2 = 1$ , $x=2$, $y=sqrt1.8$



    $2(2) - 4((sqrt2)(sqrt1.8)) + 5(sqrt1.8)^2) = 1$



    5.41 != 0




    Why didn't my equation equal 1 after I plugged in the x, y values that I got from 2)? I don't know what I did wrong...it is supposed to be 1 = 1 isnt it?











    share|cite|improve this question


























      up vote
      -2
      down vote

      favorite









      up vote
      -2
      down vote

      favorite











      Try to simplify the equation by change of bases (or change of coordinate system) : $2x^2 - 4xy + 5y^2 = 1$



      1) Find Transition Matrix P from B' to B:



      B' = {(2/$sqrt5$, 1/$sqrt5$),(-1/$sqrt5$, 2/$sqrt5$)



      B = {(1,0), (0,1)}



      $begin{bmatrix}
      2/sqrt5 &
      1/sqrt5 &
      1 &
      0
      \
      -1/sqrt5 &
      2/sqrt5 &
      0 & 1
      end{bmatrix}$
      RREF $begin{bmatrix}
      1 &
      0 &
      2/sqrt5 &
      -1/sqrt5
      \
      0 &
      1 &
      1/sqrt5 &
      2/sqrt5
      end{bmatrix}$



      P = $begin{bmatrix}
      2/sqrt5 &
      -1/sqrt5
      \
      1/sqrt5 &
      2/sqrt5
      end{bmatrix}$



      2) Convert a coordinate matrix $begin{bmatrix}x'\y'end{bmatrix}$ relative to the basis $B'$ into the coordinate matrix $begin{bmatrix}x\yend{bmatrix}$ relative to the basis $B$. Here I choose to make x' and y' = 1.



      $begin{bmatrix}x\yend{bmatrix}$ = $begin{bmatrix}
      2/sqrt5 &
      -1/sqrt5
      \
      1/sqrt5 &
      2/sqrt5
      end{bmatrix}$
      $begin{bmatrix}x'=1\y'=1end{bmatrix}$ = $begin{bmatrix}
      sqrt2
      \
      sqrt1.8
      end{bmatrix}$



      3) Plug both $x,y$ found in 2) to the original equation and find a simplified equation in terms of the variables $x', y'$



      $2x^2 - 4xy + 5y^2 = 1$ , $x=2$, $y=sqrt1.8$



      $2(2) - 4((sqrt2)(sqrt1.8)) + 5(sqrt1.8)^2) = 1$



      5.41 != 0




      Why didn't my equation equal 1 after I plugged in the x, y values that I got from 2)? I don't know what I did wrong...it is supposed to be 1 = 1 isnt it?











      share|cite|improve this question















      Try to simplify the equation by change of bases (or change of coordinate system) : $2x^2 - 4xy + 5y^2 = 1$



      1) Find Transition Matrix P from B' to B:



      B' = {(2/$sqrt5$, 1/$sqrt5$),(-1/$sqrt5$, 2/$sqrt5$)



      B = {(1,0), (0,1)}



      $begin{bmatrix}
      2/sqrt5 &
      1/sqrt5 &
      1 &
      0
      \
      -1/sqrt5 &
      2/sqrt5 &
      0 & 1
      end{bmatrix}$
      RREF $begin{bmatrix}
      1 &
      0 &
      2/sqrt5 &
      -1/sqrt5
      \
      0 &
      1 &
      1/sqrt5 &
      2/sqrt5
      end{bmatrix}$



      P = $begin{bmatrix}
      2/sqrt5 &
      -1/sqrt5
      \
      1/sqrt5 &
      2/sqrt5
      end{bmatrix}$



      2) Convert a coordinate matrix $begin{bmatrix}x'\y'end{bmatrix}$ relative to the basis $B'$ into the coordinate matrix $begin{bmatrix}x\yend{bmatrix}$ relative to the basis $B$. Here I choose to make x' and y' = 1.



      $begin{bmatrix}x\yend{bmatrix}$ = $begin{bmatrix}
      2/sqrt5 &
      -1/sqrt5
      \
      1/sqrt5 &
      2/sqrt5
      end{bmatrix}$
      $begin{bmatrix}x'=1\y'=1end{bmatrix}$ = $begin{bmatrix}
      sqrt2
      \
      sqrt1.8
      end{bmatrix}$



      3) Plug both $x,y$ found in 2) to the original equation and find a simplified equation in terms of the variables $x', y'$



      $2x^2 - 4xy + 5y^2 = 1$ , $x=2$, $y=sqrt1.8$



      $2(2) - 4((sqrt2)(sqrt1.8)) + 5(sqrt1.8)^2) = 1$



      5.41 != 0




      Why didn't my equation equal 1 after I plugged in the x, y values that I got from 2)? I don't know what I did wrong...it is supposed to be 1 = 1 isnt it?








      linear-algebra






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      edited Nov 26 at 23:58

























      asked Nov 26 at 23:26









      Evan Kim

      758




      758






















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          Your matrix $P$ is not the inverse of the matrix$$begin{bmatrix}frac2{sqrt5}&-frac1{sqrt5}\frac1{sqrt5}&frac2{sqrt5}end{bmatrix};$$that would be$$begin{bmatrix}frac2{sqrt5}&frac1{sqrt5}\-frac1{sqrt5}&frac2{sqrt5}end{bmatrix}.$$






          share|cite|improve this answer

















          • 1




            damnit. you're right. I am just starting to learn to use Tex and it is hard to keep track of the details.
            – Evan Kim
            Nov 26 at 23:39










          • I updated my post. Please see the below question
            – Evan Kim
            Nov 26 at 23:58











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          up vote
          0
          down vote













          Your matrix $P$ is not the inverse of the matrix$$begin{bmatrix}frac2{sqrt5}&-frac1{sqrt5}\frac1{sqrt5}&frac2{sqrt5}end{bmatrix};$$that would be$$begin{bmatrix}frac2{sqrt5}&frac1{sqrt5}\-frac1{sqrt5}&frac2{sqrt5}end{bmatrix}.$$






          share|cite|improve this answer

















          • 1




            damnit. you're right. I am just starting to learn to use Tex and it is hard to keep track of the details.
            – Evan Kim
            Nov 26 at 23:39










          • I updated my post. Please see the below question
            – Evan Kim
            Nov 26 at 23:58















          up vote
          0
          down vote













          Your matrix $P$ is not the inverse of the matrix$$begin{bmatrix}frac2{sqrt5}&-frac1{sqrt5}\frac1{sqrt5}&frac2{sqrt5}end{bmatrix};$$that would be$$begin{bmatrix}frac2{sqrt5}&frac1{sqrt5}\-frac1{sqrt5}&frac2{sqrt5}end{bmatrix}.$$






          share|cite|improve this answer

















          • 1




            damnit. you're right. I am just starting to learn to use Tex and it is hard to keep track of the details.
            – Evan Kim
            Nov 26 at 23:39










          • I updated my post. Please see the below question
            – Evan Kim
            Nov 26 at 23:58













          up vote
          0
          down vote










          up vote
          0
          down vote









          Your matrix $P$ is not the inverse of the matrix$$begin{bmatrix}frac2{sqrt5}&-frac1{sqrt5}\frac1{sqrt5}&frac2{sqrt5}end{bmatrix};$$that would be$$begin{bmatrix}frac2{sqrt5}&frac1{sqrt5}\-frac1{sqrt5}&frac2{sqrt5}end{bmatrix}.$$






          share|cite|improve this answer












          Your matrix $P$ is not the inverse of the matrix$$begin{bmatrix}frac2{sqrt5}&-frac1{sqrt5}\frac1{sqrt5}&frac2{sqrt5}end{bmatrix};$$that would be$$begin{bmatrix}frac2{sqrt5}&frac1{sqrt5}\-frac1{sqrt5}&frac2{sqrt5}end{bmatrix}.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 26 at 23:35









          José Carlos Santos

          147k22117217




          147k22117217








          • 1




            damnit. you're right. I am just starting to learn to use Tex and it is hard to keep track of the details.
            – Evan Kim
            Nov 26 at 23:39










          • I updated my post. Please see the below question
            – Evan Kim
            Nov 26 at 23:58














          • 1




            damnit. you're right. I am just starting to learn to use Tex and it is hard to keep track of the details.
            – Evan Kim
            Nov 26 at 23:39










          • I updated my post. Please see the below question
            – Evan Kim
            Nov 26 at 23:58








          1




          1




          damnit. you're right. I am just starting to learn to use Tex and it is hard to keep track of the details.
          – Evan Kim
          Nov 26 at 23:39




          damnit. you're right. I am just starting to learn to use Tex and it is hard to keep track of the details.
          – Evan Kim
          Nov 26 at 23:39












          I updated my post. Please see the below question
          – Evan Kim
          Nov 26 at 23:58




          I updated my post. Please see the below question
          – Evan Kim
          Nov 26 at 23:58


















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