In proving that $f(w)f(v) = vw$ as an orthogonal transformation, why does $v^T = v$?

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I've a proof that says if $f: mathbbR^3 to mathbbR^3, ~ v mapsto Av$, where $A$ is an orthogonal matrix, then for $v,w in mathbbR^3$, we have $f(w)f(v) = (Av)(Aw) = (Av)^T(Aw) = v^TA^TAw = v^Tw = vw$. This is from the proof that the map is an isometry. Could someone explain the last step? It seems to be saying $v^T = v$ but I'm not sure how that's true. Thanks.




linear-algebra




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  • Is $vw$ supposed to be dot product? It is not true in a technical sense that $v=v^T$; one will be a row vector and the other a column vector
    – ThomasGrubb
    Aug 6 at 20:47






  • 1




    It is just a change in notation :) $v^T w$ is seen as the matrix product of the row $v^T$ with the column $w$, but this is the same thing as the dot product of the two vectors, which is often just written as $vw$.
    – Lukas Miristwhisky
    Aug 6 at 20:47










  • Oh, I see. I actually missed the dot in the last step, thanks guys!
    – Nebulae
    Aug 6 at 20:49














up vote
2
down vote

favorite












I've a proof that says if $f: mathbbR^3 to mathbbR^3, ~ v mapsto Av$, where $A$ is an orthogonal matrix, then for $v,w in mathbbR^3$, we have $f(w)f(v) = (Av)(Aw) = (Av)^T(Aw) = v^TA^TAw = v^Tw = vw$. This is from the proof that the map is an isometry. Could someone explain the last step? It seems to be saying $v^T = v$ but I'm not sure how that's true. Thanks.




linear-algebra




share|cite|improve this question



















  • Is $vw$ supposed to be dot product? It is not true in a technical sense that $v=v^T$; one will be a row vector and the other a column vector
    – ThomasGrubb
    Aug 6 at 20:47






  • 1




    It is just a change in notation :) $v^T w$ is seen as the matrix product of the row $v^T$ with the column $w$, but this is the same thing as the dot product of the two vectors, which is often just written as $vw$.
    – Lukas Miristwhisky
    Aug 6 at 20:47










  • Oh, I see. I actually missed the dot in the last step, thanks guys!
    – Nebulae
    Aug 6 at 20:49












up vote
2
down vote

favorite









up vote
2
down vote

favorite











I've a proof that says if $f: mathbbR^3 to mathbbR^3, ~ v mapsto Av$, where $A$ is an orthogonal matrix, then for $v,w in mathbbR^3$, we have $f(w)f(v) = (Av)(Aw) = (Av)^T(Aw) = v^TA^TAw = v^Tw = vw$. This is from the proof that the map is an isometry. Could someone explain the last step? It seems to be saying $v^T = v$ but I'm not sure how that's true. Thanks.




linear-algebra




share|cite|improve this question











I've a proof that says if $f: mathbbR^3 to mathbbR^3, ~ v mapsto Av$, where $A$ is an orthogonal matrix, then for $v,w in mathbbR^3$, we have $f(w)f(v) = (Av)(Aw) = (Av)^T(Aw) = v^TA^TAw = v^Tw = vw$. This is from the proof that the map is an isometry. Could someone explain the last step? It seems to be saying $v^T = v$ but I'm not sure how that's true. Thanks.





linear-algebra





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share|cite|improve this question




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asked Aug 6 at 20:41









Nebulae

525




525











  • Is $vw$ supposed to be dot product? It is not true in a technical sense that $v=v^T$; one will be a row vector and the other a column vector
    – ThomasGrubb
    Aug 6 at 20:47






  • 1




    It is just a change in notation :) $v^T w$ is seen as the matrix product of the row $v^T$ with the column $w$, but this is the same thing as the dot product of the two vectors, which is often just written as $vw$.
    – Lukas Miristwhisky
    Aug 6 at 20:47










  • Oh, I see. I actually missed the dot in the last step, thanks guys!
    – Nebulae
    Aug 6 at 20:49
















  • Is $vw$ supposed to be dot product? It is not true in a technical sense that $v=v^T$; one will be a row vector and the other a column vector
    – ThomasGrubb
    Aug 6 at 20:47






  • 1




    It is just a change in notation :) $v^T w$ is seen as the matrix product of the row $v^T$ with the column $w$, but this is the same thing as the dot product of the two vectors, which is often just written as $vw$.
    – Lukas Miristwhisky
    Aug 6 at 20:47










  • Oh, I see. I actually missed the dot in the last step, thanks guys!
    – Nebulae
    Aug 6 at 20:49















Is $vw$ supposed to be dot product? It is not true in a technical sense that $v=v^T$; one will be a row vector and the other a column vector
– ThomasGrubb
Aug 6 at 20:47




Is $vw$ supposed to be dot product? It is not true in a technical sense that $v=v^T$; one will be a row vector and the other a column vector
– ThomasGrubb
Aug 6 at 20:47




1




1




It is just a change in notation :) $v^T w$ is seen as the matrix product of the row $v^T$ with the column $w$, but this is the same thing as the dot product of the two vectors, which is often just written as $vw$.
– Lukas Miristwhisky
Aug 6 at 20:47




It is just a change in notation :) $v^T w$ is seen as the matrix product of the row $v^T$ with the column $w$, but this is the same thing as the dot product of the two vectors, which is often just written as $vw$.
– Lukas Miristwhisky
Aug 6 at 20:47












Oh, I see. I actually missed the dot in the last step, thanks guys!
– Nebulae
Aug 6 at 20:49




Oh, I see. I actually missed the dot in the last step, thanks guys!
– Nebulae
Aug 6 at 20:49










1 Answer
1






active

oldest

votes

















up vote
2
down vote



accepted










There could be some confusion in general going on. E.g. $f(w)$ and $f(v)$ are column vectors in $mathbbR^3$, so that $f(w)f(v)$ does not make sense in a matrix/vector multiplication way, as we'd try to calculate the matrix product



$$beginpmatrixx\y\zendpmatrixbeginpmatrixx'\y'\z'endpmatrix$$



which is ill-defined.



There is however one way I could read the statement, namely involving the classical dot product over real vectors, where you'd have the relationship $vcdot w=v^top w$. Then, to write this out with the standard notation for the dot product, $langlecdot,cdotrangle$, you'd have:



$$
langle f(w),f(v)rangle=langle Aw,Avrangle=(Aw)^top(Av)=w^top A^top Av=w^top v=langle w,vrangle
$$



EDIT: Note, that there is a slight twist in the way you wrote it, if I'm not mistaken. This makes also sense as note that $f$ is a so called orthogonal transformation, i.e. it is a linear transformation that additionally preserves a corresponding scalar product, i.e. it is an isometry of the standard euclidean vector space $(mathbbR^3,langlecdot,cdotrangle)$ and thus preserves lengths, angles, etc.



Preservation of the scalar product means exactly that $langle w,vrangle=langle f(w),f(v)rangle$ for all $w,vinmathbbR^3$ and you with this verified the statement that an orthogonal matrix induces an orthogonal map. There is actually(up to orthonormal bases), a 1-to-1 correspondence between orthogonal maps and orthogonal matrices.






share|cite|improve this answer























  • Makes perfect sense. Thank you!
    – Nebulae
    Aug 6 at 20:49










  • Thanks for the addendum. Appreciate it.
    – Nebulae
    Aug 6 at 21:24










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










There could be some confusion in general going on. E.g. $f(w)$ and $f(v)$ are column vectors in $mathbbR^3$, so that $f(w)f(v)$ does not make sense in a matrix/vector multiplication way, as we'd try to calculate the matrix product



$$beginpmatrixx\y\zendpmatrixbeginpmatrixx'\y'\z'endpmatrix$$



which is ill-defined.



There is however one way I could read the statement, namely involving the classical dot product over real vectors, where you'd have the relationship $vcdot w=v^top w$. Then, to write this out with the standard notation for the dot product, $langlecdot,cdotrangle$, you'd have:



$$
langle f(w),f(v)rangle=langle Aw,Avrangle=(Aw)^top(Av)=w^top A^top Av=w^top v=langle w,vrangle
$$



EDIT: Note, that there is a slight twist in the way you wrote it, if I'm not mistaken. This makes also sense as note that $f$ is a so called orthogonal transformation, i.e. it is a linear transformation that additionally preserves a corresponding scalar product, i.e. it is an isometry of the standard euclidean vector space $(mathbbR^3,langlecdot,cdotrangle)$ and thus preserves lengths, angles, etc.



Preservation of the scalar product means exactly that $langle w,vrangle=langle f(w),f(v)rangle$ for all $w,vinmathbbR^3$ and you with this verified the statement that an orthogonal matrix induces an orthogonal map. There is actually(up to orthonormal bases), a 1-to-1 correspondence between orthogonal maps and orthogonal matrices.






share|cite|improve this answer























  • Makes perfect sense. Thank you!
    – Nebulae
    Aug 6 at 20:49










  • Thanks for the addendum. Appreciate it.
    – Nebulae
    Aug 6 at 21:24














up vote
2
down vote



accepted










There could be some confusion in general going on. E.g. $f(w)$ and $f(v)$ are column vectors in $mathbbR^3$, so that $f(w)f(v)$ does not make sense in a matrix/vector multiplication way, as we'd try to calculate the matrix product



$$beginpmatrixx\y\zendpmatrixbeginpmatrixx'\y'\z'endpmatrix$$



which is ill-defined.



There is however one way I could read the statement, namely involving the classical dot product over real vectors, where you'd have the relationship $vcdot w=v^top w$. Then, to write this out with the standard notation for the dot product, $langlecdot,cdotrangle$, you'd have:



$$
langle f(w),f(v)rangle=langle Aw,Avrangle=(Aw)^top(Av)=w^top A^top Av=w^top v=langle w,vrangle
$$



EDIT: Note, that there is a slight twist in the way you wrote it, if I'm not mistaken. This makes also sense as note that $f$ is a so called orthogonal transformation, i.e. it is a linear transformation that additionally preserves a corresponding scalar product, i.e. it is an isometry of the standard euclidean vector space $(mathbbR^3,langlecdot,cdotrangle)$ and thus preserves lengths, angles, etc.



Preservation of the scalar product means exactly that $langle w,vrangle=langle f(w),f(v)rangle$ for all $w,vinmathbbR^3$ and you with this verified the statement that an orthogonal matrix induces an orthogonal map. There is actually(up to orthonormal bases), a 1-to-1 correspondence between orthogonal maps and orthogonal matrices.






share|cite|improve this answer























  • Makes perfect sense. Thank you!
    – Nebulae
    Aug 6 at 20:49










  • Thanks for the addendum. Appreciate it.
    – Nebulae
    Aug 6 at 21:24












up vote
2
down vote



accepted







up vote
2
down vote



accepted






There could be some confusion in general going on. E.g. $f(w)$ and $f(v)$ are column vectors in $mathbbR^3$, so that $f(w)f(v)$ does not make sense in a matrix/vector multiplication way, as we'd try to calculate the matrix product



$$beginpmatrixx\y\zendpmatrixbeginpmatrixx'\y'\z'endpmatrix$$



which is ill-defined.



There is however one way I could read the statement, namely involving the classical dot product over real vectors, where you'd have the relationship $vcdot w=v^top w$. Then, to write this out with the standard notation for the dot product, $langlecdot,cdotrangle$, you'd have:



$$
langle f(w),f(v)rangle=langle Aw,Avrangle=(Aw)^top(Av)=w^top A^top Av=w^top v=langle w,vrangle
$$



EDIT: Note, that there is a slight twist in the way you wrote it, if I'm not mistaken. This makes also sense as note that $f$ is a so called orthogonal transformation, i.e. it is a linear transformation that additionally preserves a corresponding scalar product, i.e. it is an isometry of the standard euclidean vector space $(mathbbR^3,langlecdot,cdotrangle)$ and thus preserves lengths, angles, etc.



Preservation of the scalar product means exactly that $langle w,vrangle=langle f(w),f(v)rangle$ for all $w,vinmathbbR^3$ and you with this verified the statement that an orthogonal matrix induces an orthogonal map. There is actually(up to orthonormal bases), a 1-to-1 correspondence between orthogonal maps and orthogonal matrices.






share|cite|improve this answer















There could be some confusion in general going on. E.g. $f(w)$ and $f(v)$ are column vectors in $mathbbR^3$, so that $f(w)f(v)$ does not make sense in a matrix/vector multiplication way, as we'd try to calculate the matrix product



$$beginpmatrixx\y\zendpmatrixbeginpmatrixx'\y'\z'endpmatrix$$



which is ill-defined.



There is however one way I could read the statement, namely involving the classical dot product over real vectors, where you'd have the relationship $vcdot w=v^top w$. Then, to write this out with the standard notation for the dot product, $langlecdot,cdotrangle$, you'd have:



$$
langle f(w),f(v)rangle=langle Aw,Avrangle=(Aw)^top(Av)=w^top A^top Av=w^top v=langle w,vrangle
$$



EDIT: Note, that there is a slight twist in the way you wrote it, if I'm not mistaken. This makes also sense as note that $f$ is a so called orthogonal transformation, i.e. it is a linear transformation that additionally preserves a corresponding scalar product, i.e. it is an isometry of the standard euclidean vector space $(mathbbR^3,langlecdot,cdotrangle)$ and thus preserves lengths, angles, etc.



Preservation of the scalar product means exactly that $langle w,vrangle=langle f(w),f(v)rangle$ for all $w,vinmathbbR^3$ and you with this verified the statement that an orthogonal matrix induces an orthogonal map. There is actually(up to orthonormal bases), a 1-to-1 correspondence between orthogonal maps and orthogonal matrices.







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited Aug 6 at 21:10


























answered Aug 6 at 20:48









zzuussee

1,537419




1,537419











  • Makes perfect sense. Thank you!
    – Nebulae
    Aug 6 at 20:49










  • Thanks for the addendum. Appreciate it.
    – Nebulae
    Aug 6 at 21:24
















  • Makes perfect sense. Thank you!
    – Nebulae
    Aug 6 at 20:49










  • Thanks for the addendum. Appreciate it.
    – Nebulae
    Aug 6 at 21:24















Makes perfect sense. Thank you!
– Nebulae
Aug 6 at 20:49




Makes perfect sense. Thank you!
– Nebulae
Aug 6 at 20:49












Thanks for the addendum. Appreciate it.
– Nebulae
Aug 6 at 21:24




Thanks for the addendum. Appreciate it.
– Nebulae
Aug 6 at 21:24












 

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