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Are all collinearity preserving bijections from $mathbb R^3$ to itself of the form $Ax+b$ for $A in M_3times3(mathbb R)$ and $b inmathbb R^3$?
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Are all collinearity preserving bijections from $newcommandRRmathbb RRR^3$ to $mathbb R^3$ of the form $Ax+b$ for $A$ in $M_3times3(RR)$ and $b in mathbb R^3$?
In particular, let bijective $f:mathbb R^3tomathbb R^3$ such that if $a,b,c$ in $mathbb R^3$ are collinear then $f(a),f(b),f(c)$ are collinear. Prove or disprove that there exists $A$ in $M_3times3(mathbb R)$ and $b$ in $mathbb R^3$ such that $f(x)=Ax+b$.
My work so far:
Because $f$ is bijective, $f^-1$ exists, thus we can find
$a = f^-1((0,0,0))$, $b = f^-1((1,0,0))$, $c = f^-1((0,1,0))$, $d = f^-1((0,0,1))$
and construct the unique linear transform T such that
$a = T((0,0,0))$, $b = T((1,0,0))$, $c = T((0,1,0))$, $d = T((0,0,1))$
notice $g$ fixes $(0,0,0), (1,0,0), (0,1,0), (0,0,1)$
Then let $g(x) = f(T(x))$, if we can show that $g$ is of the form $Ax+b$ then because $f(x) = g(T^-1(x))$ and $g$ and $T$ are both of the form $Ax+b$, then we will have show $f$ is of the form $Ax+b$, thus it is sufficient to consider a collinearity preserving bijection that fixes the origin and 3 standard basis vectors.
I'm not sure where to go from here, any help is appreciated, thanks.
linear-algebra affine-geometry
edited Jul 21 at 18:49
jgon
8,54611435
asked Jul 20 at 18:57
mathew
356114
add a comment |Â
up vote
2
down vote
favorite
Are all collinearity preserving bijections from $newcommandRRmathbb RRR^3$ to $mathbb R^3$ of the form $Ax+b$ for $A$ in $M_3times3(RR)$ and $b in mathbb R^3$?
In particular, let bijective $f:mathbb R^3tomathbb R^3$ such that if $a,b,c$ in $mathbb R^3$ are collinear then $f(a),f(b),f(c)$ are collinear. Prove or disprove that there exists $A$ in $M_3times3(mathbb R)$ and $b$ in $mathbb R^3$ such that $f(x)=Ax+b$.
My work so far:
Because $f$ is bijective, $f^-1$ exists, thus we can find
$a = f^-1((0,0,0))$, $b = f^-1((1,0,0))$, $c = f^-1((0,1,0))$, $d = f^-1((0,0,1))$
and construct the unique linear transform T such that
$a = T((0,0,0))$, $b = T((1,0,0))$, $c = T((0,1,0))$, $d = T((0,0,1))$
notice $g$ fixes $(0,0,0), (1,0,0), (0,1,0), (0,0,1)$
Then let $g(x) = f(T(x))$, if we can show that $g$ is of the form $Ax+b$ then because $f(x) = g(T^-1(x))$ and $g$ and $T$ are both of the form $Ax+b$, then we will have show $f$ is of the form $Ax+b$, thus it is sufficient to consider a collinearity preserving bijection that fixes the origin and 3 standard basis vectors.
I'm not sure where to go from here, any help is appreciated, thanks.
linear-algebra affine-geometry
edited Jul 21 at 18:49
jgon
8,54611435
asked Jul 20 at 18:57
mathew
356114
Here's a MathJax tutorial :)
– Shaun
Jul 20 at 18:59
Note the typographical difference between $3x3$ and $3times3.$
– Michael Hardy
Jul 20 at 19:08
thanks a lot Micheal, I'm not very good with mathJax yet, I did do a bit though, I used the $ a few times so its not as bad as it could have been (or at least its not as bad as some of my questions have been in the past). Also thanks Shaun, I'm reading the link you gave me right now
– mathew
Jul 20 at 19:09
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Are all collinearity preserving bijections from $newcommandRRmathbb RRR^3$ to $mathbb R^3$ of the form $Ax+b$ for $A$ in $M_3times3(RR)$ and $b in mathbb R^3$?
In particular, let bijective $f:mathbb R^3tomathbb R^3$ such that if $a,b,c$ in $mathbb R^3$ are collinear then $f(a),f(b),f(c)$ are collinear. Prove or disprove that there exists $A$ in $M_3times3(mathbb R)$ and $b$ in $mathbb R^3$ such that $f(x)=Ax+b$.
My work so far:
Because $f$ is bijective, $f^-1$ exists, thus we can find
$a = f^-1((0,0,0))$, $b = f^-1((1,0,0))$, $c = f^-1((0,1,0))$, $d = f^-1((0,0,1))$
and construct the unique linear transform T such that
$a = T((0,0,0))$, $b = T((1,0,0))$, $c = T((0,1,0))$, $d = T((0,0,1))$
notice $g$ fixes $(0,0,0), (1,0,0), (0,1,0), (0,0,1)$
Then let $g(x) = f(T(x))$, if we can show that $g$ is of the form $Ax+b$ then because $f(x) = g(T^-1(x))$ and $g$ and $T$ are both of the form $Ax+b$, then we will have show $f$ is of the form $Ax+b$, thus it is sufficient to consider a collinearity preserving bijection that fixes the origin and 3 standard basis vectors.
I'm not sure where to go from here, any help is appreciated, thanks.
linear-algebra affine-geometry
edited Jul 21 at 18:49
jgon
8,54611435
asked Jul 20 at 18:57
mathew
356114
Are all collinearity preserving bijections from $newcommandRRmathbb RRR^3$ to $mathbb R^3$ of the form $Ax+b$ for $A$ in $M_3times3(RR)$ and $b in mathbb R^3$?
In particular, let bijective $f:mathbb R^3tomathbb R^3$ such that if $a,b,c$ in $mathbb R^3$ are collinear then $f(a),f(b),f(c)$ are collinear. Prove or disprove that there exists $A$ in $M_3times3(mathbb R)$ and $b$ in $mathbb R^3$ such that $f(x)=Ax+b$.
My work so far:
Because $f$ is bijective, $f^-1$ exists, thus we can find
$a = f^-1((0,0,0))$, $b = f^-1((1,0,0))$, $c = f^-1((0,1,0))$, $d = f^-1((0,0,1))$
and construct the unique linear transform T such that
$a = T((0,0,0))$, $b = T((1,0,0))$, $c = T((0,1,0))$, $d = T((0,0,1))$
notice $g$ fixes $(0,0,0), (1,0,0), (0,1,0), (0,0,1)$
Then let $g(x) = f(T(x))$, if we can show that $g$ is of the form $Ax+b$ then because $f(x) = g(T^-1(x))$ and $g$ and $T$ are both of the form $Ax+b$, then we will have show $f$ is of the form $Ax+b$, thus it is sufficient to consider a collinearity preserving bijection that fixes the origin and 3 standard basis vectors.
I'm not sure where to go from here, any help is appreciated, thanks.
linear-algebra affine-geometry
edited Jul 21 at 18:49
jgon
8,54611435
asked Jul 20 at 18:57
mathew
356114
edited Jul 21 at 18:49
jgon
8,54611435
edited Jul 21 at 18:49
jgon
8,54611435
edited Jul 21 at 18:49
jgon
8,54611435
8,54611435
asked Jul 20 at 18:57
mathew
356114
asked Jul 20 at 18:57
mathew
356114
asked Jul 20 at 18:57
mathew
356114
356114
Here's a MathJax tutorial :)
– Shaun
Jul 20 at 18:59
Note the typographical difference between $3x3$ and $3times3.$
– Michael Hardy
Jul 20 at 19:08
thanks a lot Micheal, I'm not very good with mathJax yet, I did do a bit though, I used the $ a few times so its not as bad as it could have been (or at least its not as bad as some of my questions have been in the past). Also thanks Shaun, I'm reading the link you gave me right now
– mathew
Jul 20 at 19:09
add a comment |Â
Here's a MathJax tutorial :)
– Shaun
Jul 20 at 18:59
Note the typographical difference between $3x3$ and $3times3.$
– Michael Hardy
Jul 20 at 19:08
thanks a lot Micheal, I'm not very good with mathJax yet, I did do a bit though, I used the $ a few times so its not as bad as it could have been (or at least its not as bad as some of my questions have been in the past). Also thanks Shaun, I'm reading the link you gave me right now
– mathew
Jul 20 at 19:09
Here's a MathJax tutorial :)
– Shaun
Jul 20 at 18:59
Here's a MathJax tutorial :)
– Shaun
Jul 20 at 18:59
Note the typographical difference between $3x3$ and $3times3.$
– Michael Hardy
Jul 20 at 19:08
Note the typographical difference between $3x3$ and $3times3.$
– Michael Hardy
Jul 20 at 19:08
thanks a lot Micheal, I'm not very good with mathJax yet, I did do a bit though, I used the $ a few times so its not as bad as it could have been (or at least its not as bad as some of my questions have been in the past). Also thanks Shaun, I'm reading the link you gave me right now
– mathew
Jul 20 at 19:09
thanks a lot Micheal, I'm not very good with mathJax yet, I did do a bit though, I used the $ a few times so its not as bad as it could have been (or at least its not as bad as some of my questions have been in the past). Also thanks Shaun, I'm reading the link you gave me right now
– mathew
Jul 20 at 19:09
add a comment |Â
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Here's a MathJax tutorial :)
– Shaun
Jul 20 at 18:59
Note the typographical difference between $3x3$ and $3times3.$
– Michael Hardy
Jul 20 at 19:08
thanks a lot Micheal, I'm not very good with mathJax yet, I did do a bit though, I used the $ a few times so its not as bad as it could have been (or at least its not as bad as some of my questions have been in the past). Also thanks Shaun, I'm reading the link you gave me right now
– mathew
Jul 20 at 19:09