Mixing
Equivalent Definitions of Lines in Projective Space
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I’ve been working with two definitions of lines in $mathbb{P}_mathbb{R}^2$, and tried to show their equivalence.
The first is that, given two points $a=(a_0:a_1:a_2)$ and $b=(b_0:b_1:b_2)$, the line between them is given by ${ua+vb:u,vinmathbb{R}}$.
The second is $${(X_0:X_1:X_2):k_0X_0+k_1X_1+k_2X_2=0}$$ for some $k_iinmathbb{R}$ not all $0$.
Given two distinct points $a$ and $b$, we can use simple linear algebra to find $k_i$ such that our points lie on that line.
However I've been struggling to show the converse. That is, given two distinct points $a=(a_0:a_1:a_2)$ and $b=(b_0:b_1:b_2)$ such that $$k_0a_0+k_1a_1+k_2a_2=0$$ and $$k_0b_0+k_1b_1+k_2b_2=0$$ for some $k_iinmathbb{R}$ not all $0$, then for any point $c=(c_0:c_1:c_2)$ such that $$k_0c_0+k_1c_1+k_2c_2=0$$ we should be able to write $c=ua+vb$ for some $u,vinmathbb{R}$.
Perhaps I'm overlooking something simple, but any help would be much appreciated.
geometry algebraic-geometry projective-geometry projective-space
asked Jan 10 at 17:53
DaveDave
205
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add a comment |
$begingroup$
I’ve been working with two definitions of lines in $mathbb{P}_mathbb{R}^2$, and tried to show their equivalence.
The first is that, given two points $a=(a_0:a_1:a_2)$ and $b=(b_0:b_1:b_2)$, the line between them is given by ${ua+vb:u,vinmathbb{R}}$.
The second is $${(X_0:X_1:X_2):k_0X_0+k_1X_1+k_2X_2=0}$$ for some $k_iinmathbb{R}$ not all $0$.
Given two distinct points $a$ and $b$, we can use simple linear algebra to find $k_i$ such that our points lie on that line.
However I've been struggling to show the converse. That is, given two distinct points $a=(a_0:a_1:a_2)$ and $b=(b_0:b_1:b_2)$ such that $$k_0a_0+k_1a_1+k_2a_2=0$$ and $$k_0b_0+k_1b_1+k_2b_2=0$$ for some $k_iinmathbb{R}$ not all $0$, then for any point $c=(c_0:c_1:c_2)$ such that $$k_0c_0+k_1c_1+k_2c_2=0$$ we should be able to write $c=ua+vb$ for some $u,vinmathbb{R}$.
Perhaps I'm overlooking something simple, but any help would be much appreciated.
geometry algebraic-geometry projective-geometry projective-space
asked Jan 10 at 17:53
DaveDave
205
$endgroup$
$begingroup$
I don't think your definitions are equivalent as stated: take $k_0 = k_1 = k_2 = 0$ and you see that all of $mathbb{P}^2$ lies in your second 'line'. I'm sure that can be easily corrected, however.
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– bounceback
Jan 10 at 18:01
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Edited, thanks.
$endgroup$
– Dave
Jan 10 at 18:07
1
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Hints: if $a$ and $b$ are distinct, then their coordinate tuples are linearly independent. What is the dimension of the orthogonal complement of $(k_0,k_1,k_2)$?
$endgroup$
– amd
Jan 10 at 18:51
add a comment |
$begingroup$
I’ve been working with two definitions of lines in $mathbb{P}_mathbb{R}^2$, and tried to show their equivalence.
The first is that, given two points $a=(a_0:a_1:a_2)$ and $b=(b_0:b_1:b_2)$, the line between them is given by ${ua+vb:u,vinmathbb{R}}$.
The second is $${(X_0:X_1:X_2):k_0X_0+k_1X_1+k_2X_2=0}$$ for some $k_iinmathbb{R}$ not all $0$.
Given two distinct points $a$ and $b$, we can use simple linear algebra to find $k_i$ such that our points lie on that line.
However I've been struggling to show the converse. That is, given two distinct points $a=(a_0:a_1:a_2)$ and $b=(b_0:b_1:b_2)$ such that $$k_0a_0+k_1a_1+k_2a_2=0$$ and $$k_0b_0+k_1b_1+k_2b_2=0$$ for some $k_iinmathbb{R}$ not all $0$, then for any point $c=(c_0:c_1:c_2)$ such that $$k_0c_0+k_1c_1+k_2c_2=0$$ we should be able to write $c=ua+vb$ for some $u,vinmathbb{R}$.
Perhaps I'm overlooking something simple, but any help would be much appreciated.
geometry algebraic-geometry projective-geometry projective-space
asked Jan 10 at 17:53
DaveDave
205
$endgroup$
I’ve been working with two definitions of lines in $mathbb{P}_mathbb{R}^2$, and tried to show their equivalence.
The first is that, given two points $a=(a_0:a_1:a_2)$ and $b=(b_0:b_1:b_2)$, the line between them is given by ${ua+vb:u,vinmathbb{R}}$.
The second is $${(X_0:X_1:X_2):k_0X_0+k_1X_1+k_2X_2=0}$$ for some $k_iinmathbb{R}$ not all $0$.
Given two distinct points $a$ and $b$, we can use simple linear algebra to find $k_i$ such that our points lie on that line.
However I've been struggling to show the converse. That is, given two distinct points $a=(a_0:a_1:a_2)$ and $b=(b_0:b_1:b_2)$ such that $$k_0a_0+k_1a_1+k_2a_2=0$$ and $$k_0b_0+k_1b_1+k_2b_2=0$$ for some $k_iinmathbb{R}$ not all $0$, then for any point $c=(c_0:c_1:c_2)$ such that $$k_0c_0+k_1c_1+k_2c_2=0$$ we should be able to write $c=ua+vb$ for some $u,vinmathbb{R}$.
Perhaps I'm overlooking something simple, but any help would be much appreciated.
geometry algebraic-geometry projective-geometry projective-space
geometry algebraic-geometry projective-geometry projective-space
asked Jan 10 at 17:53
DaveDave
205
asked Jan 10 at 17:53
DaveDave
205
edited Jan 10 at 18:43
Dave
asked Jan 10 at 17:53
DaveDave
205
asked Jan 10 at 17:53
DaveDave
205
asked Jan 10 at 17:53
DaveDave
205
205
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I don't think your definitions are equivalent as stated: take $k_0 = k_1 = k_2 = 0$ and you see that all of $mathbb{P}^2$ lies in your second 'line'. I'm sure that can be easily corrected, however.
$endgroup$
– bounceback
Jan 10 at 18:01
$begingroup$
Edited, thanks.
$endgroup$
– Dave
Jan 10 at 18:07
1
$begingroup$
Hints: if $a$ and $b$ are distinct, then their coordinate tuples are linearly independent. What is the dimension of the orthogonal complement of $(k_0,k_1,k_2)$?
$endgroup$
– amd
Jan 10 at 18:51
add a comment |
$begingroup$
I don't think your definitions are equivalent as stated: take $k_0 = k_1 = k_2 = 0$ and you see that all of $mathbb{P}^2$ lies in your second 'line'. I'm sure that can be easily corrected, however.
$endgroup$
– bounceback
Jan 10 at 18:01
$begingroup$
Edited, thanks.
$endgroup$
– Dave
Jan 10 at 18:07
1
$begingroup$
Hints: if $a$ and $b$ are distinct, then their coordinate tuples are linearly independent. What is the dimension of the orthogonal complement of $(k_0,k_1,k_2)$?
$endgroup$
– amd
Jan 10 at 18:51
$begingroup$
I don't think your definitions are equivalent as stated: take $k_0 = k_1 = k_2 = 0$ and you see that all of $mathbb{P}^2$ lies in your second 'line'. I'm sure that can be easily corrected, however.
$endgroup$
– bounceback
Jan 10 at 18:01
$begingroup$
I don't think your definitions are equivalent as stated: take $k_0 = k_1 = k_2 = 0$ and you see that all of $mathbb{P}^2$ lies in your second 'line'. I'm sure that can be easily corrected, however.
$endgroup$
– bounceback
Jan 10 at 18:01
$begingroup$
Edited, thanks.
$endgroup$
– Dave
Jan 10 at 18:07
$begingroup$
Edited, thanks.
$endgroup$
– Dave
Jan 10 at 18:07
1
1
$begingroup$
Hints: if $a$ and $b$ are distinct, then their coordinate tuples are linearly independent. What is the dimension of the orthogonal complement of $(k_0,k_1,k_2)$?
$endgroup$
– amd
Jan 10 at 18:51
$begingroup$
Hints: if $a$ and $b$ are distinct, then their coordinate tuples are linearly independent. What is the dimension of the orthogonal complement of $(k_0,k_1,k_2)$?
$endgroup$
– amd
Jan 10 at 18:51
add a comment |
1 Answer
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Since $(k_0,k_1,k_2$) isn't in its own orthogonal complement it can't be of dimension $3$. Then because both $a$ and $b$ are, and are linearly independent, it must be of dimension $2$ with $a$ and $b$ as a basis. Then since $c$ is in the orthogonal complement it must be expressible in terms of $a$ and $b$, and we can use simple linear algebra to find the coefficients.
Thanks to amd for their comment.
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add a comment |
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$begingroup$
Since $(k_0,k_1,k_2$) isn't in its own orthogonal complement it can't be of dimension $3$. Then because both $a$ and $b$ are, and are linearly independent, it must be of dimension $2$ with $a$ and $b$ as a basis. Then since $c$ is in the orthogonal complement it must be expressible in terms of $a$ and $b$, and we can use simple linear algebra to find the coefficients.
Thanks to amd for their comment.
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add a comment |
$begingroup$
Since $(k_0,k_1,k_2$) isn't in its own orthogonal complement it can't be of dimension $3$. Then because both $a$ and $b$ are, and are linearly independent, it must be of dimension $2$ with $a$ and $b$ as a basis. Then since $c$ is in the orthogonal complement it must be expressible in terms of $a$ and $b$, and we can use simple linear algebra to find the coefficients.
Thanks to amd for their comment.
$endgroup$
add a comment |
$begingroup$
Since $(k_0,k_1,k_2$) isn't in its own orthogonal complement it can't be of dimension $3$. Then because both $a$ and $b$ are, and are linearly independent, it must be of dimension $2$ with $a$ and $b$ as a basis. Then since $c$ is in the orthogonal complement it must be expressible in terms of $a$ and $b$, and we can use simple linear algebra to find the coefficients.
Thanks to amd for their comment.
$endgroup$
Since $(k_0,k_1,k_2$) isn't in its own orthogonal complement it can't be of dimension $3$. Then because both $a$ and $b$ are, and are linearly independent, it must be of dimension $2$ with $a$ and $b$ as a basis. Then since $c$ is in the orthogonal complement it must be expressible in terms of $a$ and $b$, and we can use simple linear algebra to find the coefficients.
Thanks to amd for their comment.
answered Jan 13 at 18:03
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Dave
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$begingroup$
I don't think your definitions are equivalent as stated: take $k_0 = k_1 = k_2 = 0$ and you see that all of $mathbb{P}^2$ lies in your second 'line'. I'm sure that can be easily corrected, however.
$endgroup$
– bounceback
Jan 10 at 18:01
$begingroup$
Edited, thanks.
$endgroup$
– Dave
Jan 10 at 18:07
1
$begingroup$
Hints: if $a$ and $b$ are distinct, then their coordinate tuples are linearly independent. What is the dimension of the orthogonal complement of $(k_0,k_1,k_2)$?
$endgroup$
– amd
Jan 10 at 18:51