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?
linear-algebra
add a comment |
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?
linear-algebra
add a comment |
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?
linear-algebra
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
linear-algebra
edited Nov 26 at 23:58
asked Nov 26 at 23:26
Evan Kim
758
758
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add a comment |
1 Answer
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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$$begin{bmatrix}frac2{sqrt5}&frac1{sqrt5}\-frac1{sqrt5}&frac2{sqrt5}end{bmatrix}.$$
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
add a comment |
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1 Answer
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up vote
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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$$begin{bmatrix}frac2{sqrt5}&frac1{sqrt5}\-frac1{sqrt5}&frac2{sqrt5}end{bmatrix}.$$
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
add a comment |
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$$begin{bmatrix}frac2{sqrt5}&frac1{sqrt5}\-frac1{sqrt5}&frac2{sqrt5}end{bmatrix}.$$
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
add a comment |
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$$begin{bmatrix}frac2{sqrt5}&frac1{sqrt5}\-frac1{sqrt5}&frac2{sqrt5}end{bmatrix}.$$
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$$begin{bmatrix}frac2{sqrt5}&frac1{sqrt5}\-frac1{sqrt5}&frac2{sqrt5}end{bmatrix}.$$
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
add a comment |
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
add a comment |
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