Determine position of projected points onto a line?
I have a list of points $S$ in the form of $(p, q)$:
$$begin{align}
S = &(43, 58), (44, 60), (40, 60), (41, 61), \
&(46, 60), (40, 57), (53, 62), (50, 61)
end{align}$$
And I wish to center them on the origin $(0, 0)$. I would do this by subtracting from them the midpoints ($(bar{p}, bar{q})$) for each dimension:
$$begin{align}
bar{p} &= frac{p_1 + p_2 + dots + p_n}{n} \
bar{q} &= frac{q_1 + q_2 + dots + q_n}{n}
end{align}$$
I find $bar{p} = 44.625$ and $bar{q} = 59.875$. I find my new $S$ to be:
$$begin{align}
S_{text{new}} = &(-1.625, -1.875), (-0.625, 0.125), (-4.625, 0.125), (-3.625, 1.125), \
&(1.375, 0.125), (-4.625, -2.875), (8.375, 2.125), (5.375, 1.125)
end{align}$$
Using linear regression, I've found the line of best fit for this data set which crosses the origin to be $y = 0.26x + 0$. This is the line in which I want to project points of data onto from right angles.
My question is, how do I find these projected points (marked as red dots)? Taking point $(1.375, 0.125)$, I can make a triangle with vertices at the origin, the point, and the projected point like so:
I know the slope of $c$ ($0.26$), the position of vertex $ba$ ($(1.375, 0.125)$), and position of vertex $ca$ ($(0, 0)$), but how do I find the position of vertex $cb$?
This is for principal component analysis. To find the eigenvalue, I need the sum of squared distances from projected points to the origin. I've already found the eigenvector to be $begin{bmatrix}0.96 \ 0.25end{bmatrix}$.
geometry statistics eigenvalues-eigenvectors vectors
add a comment |
I have a list of points $S$ in the form of $(p, q)$:
$$begin{align}
S = &(43, 58), (44, 60), (40, 60), (41, 61), \
&(46, 60), (40, 57), (53, 62), (50, 61)
end{align}$$
And I wish to center them on the origin $(0, 0)$. I would do this by subtracting from them the midpoints ($(bar{p}, bar{q})$) for each dimension:
$$begin{align}
bar{p} &= frac{p_1 + p_2 + dots + p_n}{n} \
bar{q} &= frac{q_1 + q_2 + dots + q_n}{n}
end{align}$$
I find $bar{p} = 44.625$ and $bar{q} = 59.875$. I find my new $S$ to be:
$$begin{align}
S_{text{new}} = &(-1.625, -1.875), (-0.625, 0.125), (-4.625, 0.125), (-3.625, 1.125), \
&(1.375, 0.125), (-4.625, -2.875), (8.375, 2.125), (5.375, 1.125)
end{align}$$
Using linear regression, I've found the line of best fit for this data set which crosses the origin to be $y = 0.26x + 0$. This is the line in which I want to project points of data onto from right angles.
My question is, how do I find these projected points (marked as red dots)? Taking point $(1.375, 0.125)$, I can make a triangle with vertices at the origin, the point, and the projected point like so:
I know the slope of $c$ ($0.26$), the position of vertex $ba$ ($(1.375, 0.125)$), and position of vertex $ca$ ($(0, 0)$), but how do I find the position of vertex $cb$?
This is for principal component analysis. To find the eigenvalue, I need the sum of squared distances from projected points to the origin. I've already found the eigenvector to be $begin{bmatrix}0.96 \ 0.25end{bmatrix}$.
geometry statistics eigenvalues-eigenvectors vectors
Are you guaranteed that your best-fit line will pass through the origin? In general, it won’t.
– amd
Dec 1 at 0:01
@amd originally, data was in the top-right quadrant and was shifted to center around the origin. For this purpose, it will always pass through $(0,0)$.
– gator
Dec 1 at 0:05
add a comment |
I have a list of points $S$ in the form of $(p, q)$:
$$begin{align}
S = &(43, 58), (44, 60), (40, 60), (41, 61), \
&(46, 60), (40, 57), (53, 62), (50, 61)
end{align}$$
And I wish to center them on the origin $(0, 0)$. I would do this by subtracting from them the midpoints ($(bar{p}, bar{q})$) for each dimension:
$$begin{align}
bar{p} &= frac{p_1 + p_2 + dots + p_n}{n} \
bar{q} &= frac{q_1 + q_2 + dots + q_n}{n}
end{align}$$
I find $bar{p} = 44.625$ and $bar{q} = 59.875$. I find my new $S$ to be:
$$begin{align}
S_{text{new}} = &(-1.625, -1.875), (-0.625, 0.125), (-4.625, 0.125), (-3.625, 1.125), \
&(1.375, 0.125), (-4.625, -2.875), (8.375, 2.125), (5.375, 1.125)
end{align}$$
Using linear regression, I've found the line of best fit for this data set which crosses the origin to be $y = 0.26x + 0$. This is the line in which I want to project points of data onto from right angles.
My question is, how do I find these projected points (marked as red dots)? Taking point $(1.375, 0.125)$, I can make a triangle with vertices at the origin, the point, and the projected point like so:
I know the slope of $c$ ($0.26$), the position of vertex $ba$ ($(1.375, 0.125)$), and position of vertex $ca$ ($(0, 0)$), but how do I find the position of vertex $cb$?
This is for principal component analysis. To find the eigenvalue, I need the sum of squared distances from projected points to the origin. I've already found the eigenvector to be $begin{bmatrix}0.96 \ 0.25end{bmatrix}$.
geometry statistics eigenvalues-eigenvectors vectors
I have a list of points $S$ in the form of $(p, q)$:
$$begin{align}
S = &(43, 58), (44, 60), (40, 60), (41, 61), \
&(46, 60), (40, 57), (53, 62), (50, 61)
end{align}$$
And I wish to center them on the origin $(0, 0)$. I would do this by subtracting from them the midpoints ($(bar{p}, bar{q})$) for each dimension:
$$begin{align}
bar{p} &= frac{p_1 + p_2 + dots + p_n}{n} \
bar{q} &= frac{q_1 + q_2 + dots + q_n}{n}
end{align}$$
I find $bar{p} = 44.625$ and $bar{q} = 59.875$. I find my new $S$ to be:
$$begin{align}
S_{text{new}} = &(-1.625, -1.875), (-0.625, 0.125), (-4.625, 0.125), (-3.625, 1.125), \
&(1.375, 0.125), (-4.625, -2.875), (8.375, 2.125), (5.375, 1.125)
end{align}$$
Using linear regression, I've found the line of best fit for this data set which crosses the origin to be $y = 0.26x + 0$. This is the line in which I want to project points of data onto from right angles.
My question is, how do I find these projected points (marked as red dots)? Taking point $(1.375, 0.125)$, I can make a triangle with vertices at the origin, the point, and the projected point like so:
I know the slope of $c$ ($0.26$), the position of vertex $ba$ ($(1.375, 0.125)$), and position of vertex $ca$ ($(0, 0)$), but how do I find the position of vertex $cb$?
This is for principal component analysis. To find the eigenvalue, I need the sum of squared distances from projected points to the origin. I've already found the eigenvector to be $begin{bmatrix}0.96 \ 0.25end{bmatrix}$.
geometry statistics eigenvalues-eigenvectors vectors
geometry statistics eigenvalues-eigenvectors vectors
edited Dec 1 at 5:08
asked Nov 30 at 23:08
gator
65911239
65911239
Are you guaranteed that your best-fit line will pass through the origin? In general, it won’t.
– amd
Dec 1 at 0:01
@amd originally, data was in the top-right quadrant and was shifted to center around the origin. For this purpose, it will always pass through $(0,0)$.
– gator
Dec 1 at 0:05
add a comment |
Are you guaranteed that your best-fit line will pass through the origin? In general, it won’t.
– amd
Dec 1 at 0:01
@amd originally, data was in the top-right quadrant and was shifted to center around the origin. For this purpose, it will always pass through $(0,0)$.
– gator
Dec 1 at 0:05
Are you guaranteed that your best-fit line will pass through the origin? In general, it won’t.
– amd
Dec 1 at 0:01
Are you guaranteed that your best-fit line will pass through the origin? In general, it won’t.
– amd
Dec 1 at 0:01
@amd originally, data was in the top-right quadrant and was shifted to center around the origin. For this purpose, it will always pass through $(0,0)$.
– gator
Dec 1 at 0:05
@amd originally, data was in the top-right quadrant and was shifted to center around the origin. For this purpose, it will always pass through $(0,0)$.
– gator
Dec 1 at 0:05
add a comment |
2 Answers
2
active
oldest
votes
One way to do this is by calculating the euclidean vector of the blue line, in this case it is $begin{bmatrix} 1 \ 0.26 end{bmatrix}$, you want its norm to be $1$ so you divide it by its norm to get: $v = begin{bmatrix} 0.97 \ 0.25 end{bmatrix}$.
Then see every point as vector and to get the coordinates of point A one the blue line you just have to calculate $(A cdot v) cdot v$.
For example, for the point $(-1.625, −1.875)$, you would find:
$$left( begin{bmatrix} -1.625 \ -1.875 end{bmatrix} cdot begin{bmatrix} 0.97 \ 0.25end{bmatrix}right) cdot begin{bmatrix} 0.97 \ 0.25 end{bmatrix} = begin{bmatrix} -1.98 \ -0.51 end{bmatrix}$$
This works for this particular case, but in general the best-fit line might not pass through the origin.
– amd
Dec 1 at 0:02
This is true, but he can still remove the $y-intercept$ of the line to all the points, calculate all the new points and then add the $y-intercept$ again.
– Euler Pythagoras
Dec 1 at 0:14
add a comment |
The other answer to totally correct, but I would like to add a bit more explanation. What you would like to do is to transform the original data to a new coordinate frame. Let the new coordinate frame be represented by the vectors $textbf{v}$ and $textbf{w}$, as shown in the image below. Because $textbf{v}$ and $textbf{w}$ are perpendicular, their dot product is zero: $textbf{v} cdot textbf{w} = 0$. Furthermore, I assume that $textbf{v}$ is a unit vector (which is not the case in the image due to my bad drawing skills...), so $textbf{v} cdot textbf{v}=1$.
Any datapoint $textbf{x}$ can be represented in the new coordinate frame:
$$
textbf{x} = a textbf{v} + b textbf{w}.
$$
As you are interested in the position on the line, you are only interested in $a$. To get $a$, we multiply the equation on both sides with $textbf{v}$:
$$
textbf{x} cdot textbf{v} = a textbf{v} cdot textbf{v} + b textbf{w} cdot textbf{v} = a,
$$
where I used $textbf{v} cdot textbf{w} = 0$ and $textbf{v} cdot textbf{v}=1$.
The only thing you now need to do is to get the position on the line in the original frame by simply multiplying $a$ with $textbf{v}$, i.e., $(textbf{x} cdot textbf{v}) textbf{v}$.
In your case, $textbf{v} = frac{1}{sqrt{1+0.26^2}} begin{bmatrix} 1 \ 0.26end{bmatrix} approx begin{bmatrix} 0.96 \ 0.25 end{bmatrix}$, and this results in:
-1.9787 -0.5145
-0.5550 -0.1443
-4.3017 -1.1184
-3.1215 -0.8116
1.3184 0.3428
-5.0323 -1.3084
8.3622 2.1742
5.3086 1.3802
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
One way to do this is by calculating the euclidean vector of the blue line, in this case it is $begin{bmatrix} 1 \ 0.26 end{bmatrix}$, you want its norm to be $1$ so you divide it by its norm to get: $v = begin{bmatrix} 0.97 \ 0.25 end{bmatrix}$.
Then see every point as vector and to get the coordinates of point A one the blue line you just have to calculate $(A cdot v) cdot v$.
For example, for the point $(-1.625, −1.875)$, you would find:
$$left( begin{bmatrix} -1.625 \ -1.875 end{bmatrix} cdot begin{bmatrix} 0.97 \ 0.25end{bmatrix}right) cdot begin{bmatrix} 0.97 \ 0.25 end{bmatrix} = begin{bmatrix} -1.98 \ -0.51 end{bmatrix}$$
This works for this particular case, but in general the best-fit line might not pass through the origin.
– amd
Dec 1 at 0:02
This is true, but he can still remove the $y-intercept$ of the line to all the points, calculate all the new points and then add the $y-intercept$ again.
– Euler Pythagoras
Dec 1 at 0:14
add a comment |
One way to do this is by calculating the euclidean vector of the blue line, in this case it is $begin{bmatrix} 1 \ 0.26 end{bmatrix}$, you want its norm to be $1$ so you divide it by its norm to get: $v = begin{bmatrix} 0.97 \ 0.25 end{bmatrix}$.
Then see every point as vector and to get the coordinates of point A one the blue line you just have to calculate $(A cdot v) cdot v$.
For example, for the point $(-1.625, −1.875)$, you would find:
$$left( begin{bmatrix} -1.625 \ -1.875 end{bmatrix} cdot begin{bmatrix} 0.97 \ 0.25end{bmatrix}right) cdot begin{bmatrix} 0.97 \ 0.25 end{bmatrix} = begin{bmatrix} -1.98 \ -0.51 end{bmatrix}$$
This works for this particular case, but in general the best-fit line might not pass through the origin.
– amd
Dec 1 at 0:02
This is true, but he can still remove the $y-intercept$ of the line to all the points, calculate all the new points and then add the $y-intercept$ again.
– Euler Pythagoras
Dec 1 at 0:14
add a comment |
One way to do this is by calculating the euclidean vector of the blue line, in this case it is $begin{bmatrix} 1 \ 0.26 end{bmatrix}$, you want its norm to be $1$ so you divide it by its norm to get: $v = begin{bmatrix} 0.97 \ 0.25 end{bmatrix}$.
Then see every point as vector and to get the coordinates of point A one the blue line you just have to calculate $(A cdot v) cdot v$.
For example, for the point $(-1.625, −1.875)$, you would find:
$$left( begin{bmatrix} -1.625 \ -1.875 end{bmatrix} cdot begin{bmatrix} 0.97 \ 0.25end{bmatrix}right) cdot begin{bmatrix} 0.97 \ 0.25 end{bmatrix} = begin{bmatrix} -1.98 \ -0.51 end{bmatrix}$$
One way to do this is by calculating the euclidean vector of the blue line, in this case it is $begin{bmatrix} 1 \ 0.26 end{bmatrix}$, you want its norm to be $1$ so you divide it by its norm to get: $v = begin{bmatrix} 0.97 \ 0.25 end{bmatrix}$.
Then see every point as vector and to get the coordinates of point A one the blue line you just have to calculate $(A cdot v) cdot v$.
For example, for the point $(-1.625, −1.875)$, you would find:
$$left( begin{bmatrix} -1.625 \ -1.875 end{bmatrix} cdot begin{bmatrix} 0.97 \ 0.25end{bmatrix}right) cdot begin{bmatrix} 0.97 \ 0.25 end{bmatrix} = begin{bmatrix} -1.98 \ -0.51 end{bmatrix}$$
edited Dec 1 at 13:37
answered Nov 30 at 23:34
Euler Pythagoras
4949
4949
This works for this particular case, but in general the best-fit line might not pass through the origin.
– amd
Dec 1 at 0:02
This is true, but he can still remove the $y-intercept$ of the line to all the points, calculate all the new points and then add the $y-intercept$ again.
– Euler Pythagoras
Dec 1 at 0:14
add a comment |
This works for this particular case, but in general the best-fit line might not pass through the origin.
– amd
Dec 1 at 0:02
This is true, but he can still remove the $y-intercept$ of the line to all the points, calculate all the new points and then add the $y-intercept$ again.
– Euler Pythagoras
Dec 1 at 0:14
This works for this particular case, but in general the best-fit line might not pass through the origin.
– amd
Dec 1 at 0:02
This works for this particular case, but in general the best-fit line might not pass through the origin.
– amd
Dec 1 at 0:02
This is true, but he can still remove the $y-intercept$ of the line to all the points, calculate all the new points and then add the $y-intercept$ again.
– Euler Pythagoras
Dec 1 at 0:14
This is true, but he can still remove the $y-intercept$ of the line to all the points, calculate all the new points and then add the $y-intercept$ again.
– Euler Pythagoras
Dec 1 at 0:14
add a comment |
The other answer to totally correct, but I would like to add a bit more explanation. What you would like to do is to transform the original data to a new coordinate frame. Let the new coordinate frame be represented by the vectors $textbf{v}$ and $textbf{w}$, as shown in the image below. Because $textbf{v}$ and $textbf{w}$ are perpendicular, their dot product is zero: $textbf{v} cdot textbf{w} = 0$. Furthermore, I assume that $textbf{v}$ is a unit vector (which is not the case in the image due to my bad drawing skills...), so $textbf{v} cdot textbf{v}=1$.
Any datapoint $textbf{x}$ can be represented in the new coordinate frame:
$$
textbf{x} = a textbf{v} + b textbf{w}.
$$
As you are interested in the position on the line, you are only interested in $a$. To get $a$, we multiply the equation on both sides with $textbf{v}$:
$$
textbf{x} cdot textbf{v} = a textbf{v} cdot textbf{v} + b textbf{w} cdot textbf{v} = a,
$$
where I used $textbf{v} cdot textbf{w} = 0$ and $textbf{v} cdot textbf{v}=1$.
The only thing you now need to do is to get the position on the line in the original frame by simply multiplying $a$ with $textbf{v}$, i.e., $(textbf{x} cdot textbf{v}) textbf{v}$.
In your case, $textbf{v} = frac{1}{sqrt{1+0.26^2}} begin{bmatrix} 1 \ 0.26end{bmatrix} approx begin{bmatrix} 0.96 \ 0.25 end{bmatrix}$, and this results in:
-1.9787 -0.5145
-0.5550 -0.1443
-4.3017 -1.1184
-3.1215 -0.8116
1.3184 0.3428
-5.0323 -1.3084
8.3622 2.1742
5.3086 1.3802
add a comment |
The other answer to totally correct, but I would like to add a bit more explanation. What you would like to do is to transform the original data to a new coordinate frame. Let the new coordinate frame be represented by the vectors $textbf{v}$ and $textbf{w}$, as shown in the image below. Because $textbf{v}$ and $textbf{w}$ are perpendicular, their dot product is zero: $textbf{v} cdot textbf{w} = 0$. Furthermore, I assume that $textbf{v}$ is a unit vector (which is not the case in the image due to my bad drawing skills...), so $textbf{v} cdot textbf{v}=1$.
Any datapoint $textbf{x}$ can be represented in the new coordinate frame:
$$
textbf{x} = a textbf{v} + b textbf{w}.
$$
As you are interested in the position on the line, you are only interested in $a$. To get $a$, we multiply the equation on both sides with $textbf{v}$:
$$
textbf{x} cdot textbf{v} = a textbf{v} cdot textbf{v} + b textbf{w} cdot textbf{v} = a,
$$
where I used $textbf{v} cdot textbf{w} = 0$ and $textbf{v} cdot textbf{v}=1$.
The only thing you now need to do is to get the position on the line in the original frame by simply multiplying $a$ with $textbf{v}$, i.e., $(textbf{x} cdot textbf{v}) textbf{v}$.
In your case, $textbf{v} = frac{1}{sqrt{1+0.26^2}} begin{bmatrix} 1 \ 0.26end{bmatrix} approx begin{bmatrix} 0.96 \ 0.25 end{bmatrix}$, and this results in:
-1.9787 -0.5145
-0.5550 -0.1443
-4.3017 -1.1184
-3.1215 -0.8116
1.3184 0.3428
-5.0323 -1.3084
8.3622 2.1742
5.3086 1.3802
add a comment |
The other answer to totally correct, but I would like to add a bit more explanation. What you would like to do is to transform the original data to a new coordinate frame. Let the new coordinate frame be represented by the vectors $textbf{v}$ and $textbf{w}$, as shown in the image below. Because $textbf{v}$ and $textbf{w}$ are perpendicular, their dot product is zero: $textbf{v} cdot textbf{w} = 0$. Furthermore, I assume that $textbf{v}$ is a unit vector (which is not the case in the image due to my bad drawing skills...), so $textbf{v} cdot textbf{v}=1$.
Any datapoint $textbf{x}$ can be represented in the new coordinate frame:
$$
textbf{x} = a textbf{v} + b textbf{w}.
$$
As you are interested in the position on the line, you are only interested in $a$. To get $a$, we multiply the equation on both sides with $textbf{v}$:
$$
textbf{x} cdot textbf{v} = a textbf{v} cdot textbf{v} + b textbf{w} cdot textbf{v} = a,
$$
where I used $textbf{v} cdot textbf{w} = 0$ and $textbf{v} cdot textbf{v}=1$.
The only thing you now need to do is to get the position on the line in the original frame by simply multiplying $a$ with $textbf{v}$, i.e., $(textbf{x} cdot textbf{v}) textbf{v}$.
In your case, $textbf{v} = frac{1}{sqrt{1+0.26^2}} begin{bmatrix} 1 \ 0.26end{bmatrix} approx begin{bmatrix} 0.96 \ 0.25 end{bmatrix}$, and this results in:
-1.9787 -0.5145
-0.5550 -0.1443
-4.3017 -1.1184
-3.1215 -0.8116
1.3184 0.3428
-5.0323 -1.3084
8.3622 2.1742
5.3086 1.3802
The other answer to totally correct, but I would like to add a bit more explanation. What you would like to do is to transform the original data to a new coordinate frame. Let the new coordinate frame be represented by the vectors $textbf{v}$ and $textbf{w}$, as shown in the image below. Because $textbf{v}$ and $textbf{w}$ are perpendicular, their dot product is zero: $textbf{v} cdot textbf{w} = 0$. Furthermore, I assume that $textbf{v}$ is a unit vector (which is not the case in the image due to my bad drawing skills...), so $textbf{v} cdot textbf{v}=1$.
Any datapoint $textbf{x}$ can be represented in the new coordinate frame:
$$
textbf{x} = a textbf{v} + b textbf{w}.
$$
As you are interested in the position on the line, you are only interested in $a$. To get $a$, we multiply the equation on both sides with $textbf{v}$:
$$
textbf{x} cdot textbf{v} = a textbf{v} cdot textbf{v} + b textbf{w} cdot textbf{v} = a,
$$
where I used $textbf{v} cdot textbf{w} = 0$ and $textbf{v} cdot textbf{v}=1$.
The only thing you now need to do is to get the position on the line in the original frame by simply multiplying $a$ with $textbf{v}$, i.e., $(textbf{x} cdot textbf{v}) textbf{v}$.
In your case, $textbf{v} = frac{1}{sqrt{1+0.26^2}} begin{bmatrix} 1 \ 0.26end{bmatrix} approx begin{bmatrix} 0.96 \ 0.25 end{bmatrix}$, and this results in:
-1.9787 -0.5145
-0.5550 -0.1443
-4.3017 -1.1184
-3.1215 -0.8116
1.3184 0.3428
-5.0323 -1.3084
8.3622 2.1742
5.3086 1.3802
edited Dec 1 at 7:44
answered Dec 1 at 2:19
EdG
1,193314
1,193314
add a comment |
add a comment |
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Are you guaranteed that your best-fit line will pass through the origin? In general, it won’t.
– amd
Dec 1 at 0:01
@amd originally, data was in the top-right quadrant and was shifted to center around the origin. For this purpose, it will always pass through $(0,0)$.
– gator
Dec 1 at 0:05