How to block diagonalize a real skew-symmetric matrix of 3*3












1












$begingroup$


Suppose $t = [t_1,t_2,t_3]^Tin mathbb R^3,t neq 0$. Then define
$$t^{land} = begin{bmatrix} 0 & -t_3 & t_2 \
t_3 & 0 & -t_1\
-t_2 & t_1 & 0end{bmatrix},
Z= begin{bmatrix} 0 & 1 \
-1 & 0end{bmatrix}.$$
According to exponential map, there exist a corresponding rotation matrix
$$R = exp(t^{land})$$
if $t^{land}$ can be block diagonalized, that is
$$t^{land} = Ubegin{bmatrix} aZ & 0 \
0 & 0end{bmatrix}U^T,a>0$$
then we get $R = exp(t^{land})=Ubegin{bmatrix}cos(|t|_2)&sin(|t|_2) & 0\
-sin(|t|_2)&cos(|t|_2)&0 \
0 & 0 &1end{bmatrix}U^T,a=|t|_2$.



(refer to R.Hartley & A.Zisserman Multiple View Geometry in Computer Vision page 583 to 584.)
So how to do block diagonalization in 3*3 case? And what is $U$ like?






matrices lie-groups lie-algebras numerical-linear-algebra rotations







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Extend $w=t/|t|_2$ to an orthonormal basis ${u,v,w}$. Then take $U=pmatrix{u,v,w}$.
    $endgroup$
    – user1551
    Dec 7 '18 at 9:40










  • $begingroup$
    @user1551 hello, i apply your thread(see following post of mine) while leads to a contradiction, can you help me out?
    $endgroup$
    – Finley
    Dec 7 '18 at 15:37










  • $begingroup$
    I didn't noticed that you need $a>0$. If you want to enforce this constraint and you find that with your chosen $u,v$ the cross product $vtimes u$ is $-w$ rather than $w$, you may simply interchange $u$ and $v$. Alternatively, replace $v$ by $-v$.
    $endgroup$
    – user1551
    Dec 7 '18 at 16:39










  • $begingroup$
    @user1551 But $det(U) = -1$ once I interchange $u text{ and } v$ or negate $v$. Is it reasonable for an orthonormal matrix?
    $endgroup$
    – Finley
    Dec 8 '18 at 2:11












  • $begingroup$
    Why not? Your question doesn't require that $det(U)=1$. Anyway, since it sets $Z$ to $pmatrix{0&1\ -1&0}$ rather than $pmatrix{0&-1\ 1&0}$, the determinant of $U$ has to be $-1$: when $u$ is orthogonal to $t$, the effect of $ttimes u$ in the tangential direction is an anticlockwise rotation of $u$ about the directed axis $t$. However, $Z=pmatrix{0&1\ -1&0}$ represents a clockwise rotation. So, in the change-of-basis matrix $U$, you must incorporate a reflection to preserve the sign, meaning that $det(U)$ has to be $-1$.
    $endgroup$
    – user1551
    Dec 8 '18 at 6:07


















1












$begingroup$


Suppose $t = [t_1,t_2,t_3]^Tin mathbb R^3,t neq 0$. Then define
$$t^{land} = begin{bmatrix} 0 & -t_3 & t_2 \
t_3 & 0 & -t_1\
-t_2 & t_1 & 0end{bmatrix},
Z= begin{bmatrix} 0 & 1 \
-1 & 0end{bmatrix}.$$
According to exponential map, there exist a corresponding rotation matrix
$$R = exp(t^{land})$$
if $t^{land}$ can be block diagonalized, that is
$$t^{land} = Ubegin{bmatrix} aZ & 0 \
0 & 0end{bmatrix}U^T,a>0$$
then we get $R = exp(t^{land})=Ubegin{bmatrix}cos(|t|_2)&sin(|t|_2) & 0\
-sin(|t|_2)&cos(|t|_2)&0 \
0 & 0 &1end{bmatrix}U^T,a=|t|_2$.



(refer to R.Hartley & A.Zisserman Multiple View Geometry in Computer Vision page 583 to 584.)
So how to do block diagonalization in 3*3 case? And what is $U$ like?






matrices lie-groups lie-algebras numerical-linear-algebra rotations







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Extend $w=t/|t|_2$ to an orthonormal basis ${u,v,w}$. Then take $U=pmatrix{u,v,w}$.
    $endgroup$
    – user1551
    Dec 7 '18 at 9:40










  • $begingroup$
    @user1551 hello, i apply your thread(see following post of mine) while leads to a contradiction, can you help me out?
    $endgroup$
    – Finley
    Dec 7 '18 at 15:37










  • $begingroup$
    I didn't noticed that you need $a>0$. If you want to enforce this constraint and you find that with your chosen $u,v$ the cross product $vtimes u$ is $-w$ rather than $w$, you may simply interchange $u$ and $v$. Alternatively, replace $v$ by $-v$.
    $endgroup$
    – user1551
    Dec 7 '18 at 16:39










  • $begingroup$
    @user1551 But $det(U) = -1$ once I interchange $u text{ and } v$ or negate $v$. Is it reasonable for an orthonormal matrix?
    $endgroup$
    – Finley
    Dec 8 '18 at 2:11












  • $begingroup$
    Why not? Your question doesn't require that $det(U)=1$. Anyway, since it sets $Z$ to $pmatrix{0&1\ -1&0}$ rather than $pmatrix{0&-1\ 1&0}$, the determinant of $U$ has to be $-1$: when $u$ is orthogonal to $t$, the effect of $ttimes u$ in the tangential direction is an anticlockwise rotation of $u$ about the directed axis $t$. However, $Z=pmatrix{0&1\ -1&0}$ represents a clockwise rotation. So, in the change-of-basis matrix $U$, you must incorporate a reflection to preserve the sign, meaning that $det(U)$ has to be $-1$.
    $endgroup$
    – user1551
    Dec 8 '18 at 6:07
















1












1








1


0



$begingroup$


Suppose $t = [t_1,t_2,t_3]^Tin mathbb R^3,t neq 0$. Then define
$$t^{land} = begin{bmatrix} 0 & -t_3 & t_2 \
t_3 & 0 & -t_1\
-t_2 & t_1 & 0end{bmatrix},
Z= begin{bmatrix} 0 & 1 \
-1 & 0end{bmatrix}.$$
According to exponential map, there exist a corresponding rotation matrix
$$R = exp(t^{land})$$
if $t^{land}$ can be block diagonalized, that is
$$t^{land} = Ubegin{bmatrix} aZ & 0 \
0 & 0end{bmatrix}U^T,a>0$$
then we get $R = exp(t^{land})=Ubegin{bmatrix}cos(|t|_2)&sin(|t|_2) & 0\
-sin(|t|_2)&cos(|t|_2)&0 \
0 & 0 &1end{bmatrix}U^T,a=|t|_2$.



(refer to R.Hartley & A.Zisserman Multiple View Geometry in Computer Vision page 583 to 584.)
So how to do block diagonalization in 3*3 case? And what is $U$ like?






matrices lie-groups lie-algebras numerical-linear-algebra rotations







share|cite|improve this question









$endgroup$




Suppose $t = [t_1,t_2,t_3]^Tin mathbb R^3,t neq 0$. Then define
$$t^{land} = begin{bmatrix} 0 & -t_3 & t_2 \
t_3 & 0 & -t_1\
-t_2 & t_1 & 0end{bmatrix},
Z= begin{bmatrix} 0 & 1 \
-1 & 0end{bmatrix}.$$
According to exponential map, there exist a corresponding rotation matrix
$$R = exp(t^{land})$$
if $t^{land}$ can be block diagonalized, that is
$$t^{land} = Ubegin{bmatrix} aZ & 0 \
0 & 0end{bmatrix}U^T,a>0$$
then we get $R = exp(t^{land})=Ubegin{bmatrix}cos(|t|_2)&sin(|t|_2) & 0\
-sin(|t|_2)&cos(|t|_2)&0 \
0 & 0 &1end{bmatrix}U^T,a=|t|_2$.



(refer to R.Hartley & A.Zisserman Multiple View Geometry in Computer Vision page 583 to 584.)
So how to do block diagonalization in 3*3 case? And what is $U$ like?






matrices lie-groups lie-algebras numerical-linear-algebra rotations




matrices lie-groups lie-algebras numerical-linear-algebra rotations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 7 '18 at 9:37









FinleyFinley

375213




375213








  • 1




    $begingroup$
    Extend $w=t/|t|_2$ to an orthonormal basis ${u,v,w}$. Then take $U=pmatrix{u,v,w}$.
    $endgroup$
    – user1551
    Dec 7 '18 at 9:40










  • $begingroup$
    @user1551 hello, i apply your thread(see following post of mine) while leads to a contradiction, can you help me out?
    $endgroup$
    – Finley
    Dec 7 '18 at 15:37










  • $begingroup$
    I didn't noticed that you need $a>0$. If you want to enforce this constraint and you find that with your chosen $u,v$ the cross product $vtimes u$ is $-w$ rather than $w$, you may simply interchange $u$ and $v$. Alternatively, replace $v$ by $-v$.
    $endgroup$
    – user1551
    Dec 7 '18 at 16:39










  • $begingroup$
    @user1551 But $det(U) = -1$ once I interchange $u text{ and } v$ or negate $v$. Is it reasonable for an orthonormal matrix?
    $endgroup$
    – Finley
    Dec 8 '18 at 2:11












  • $begingroup$
    Why not? Your question doesn't require that $det(U)=1$. Anyway, since it sets $Z$ to $pmatrix{0&1\ -1&0}$ rather than $pmatrix{0&-1\ 1&0}$, the determinant of $U$ has to be $-1$: when $u$ is orthogonal to $t$, the effect of $ttimes u$ in the tangential direction is an anticlockwise rotation of $u$ about the directed axis $t$. However, $Z=pmatrix{0&1\ -1&0}$ represents a clockwise rotation. So, in the change-of-basis matrix $U$, you must incorporate a reflection to preserve the sign, meaning that $det(U)$ has to be $-1$.
    $endgroup$
    – user1551
    Dec 8 '18 at 6:07
















  • 1




    $begingroup$
    Extend $w=t/|t|_2$ to an orthonormal basis ${u,v,w}$. Then take $U=pmatrix{u,v,w}$.
    $endgroup$
    – user1551
    Dec 7 '18 at 9:40










  • $begingroup$
    @user1551 hello, i apply your thread(see following post of mine) while leads to a contradiction, can you help me out?
    $endgroup$
    – Finley
    Dec 7 '18 at 15:37










  • $begingroup$
    I didn't noticed that you need $a>0$. If you want to enforce this constraint and you find that with your chosen $u,v$ the cross product $vtimes u$ is $-w$ rather than $w$, you may simply interchange $u$ and $v$. Alternatively, replace $v$ by $-v$.
    $endgroup$
    – user1551
    Dec 7 '18 at 16:39










  • $begingroup$
    @user1551 But $det(U) = -1$ once I interchange $u text{ and } v$ or negate $v$. Is it reasonable for an orthonormal matrix?
    $endgroup$
    – Finley
    Dec 8 '18 at 2:11












  • $begingroup$
    Why not? Your question doesn't require that $det(U)=1$. Anyway, since it sets $Z$ to $pmatrix{0&1\ -1&0}$ rather than $pmatrix{0&-1\ 1&0}$, the determinant of $U$ has to be $-1$: when $u$ is orthogonal to $t$, the effect of $ttimes u$ in the tangential direction is an anticlockwise rotation of $u$ about the directed axis $t$. However, $Z=pmatrix{0&1\ -1&0}$ represents a clockwise rotation. So, in the change-of-basis matrix $U$, you must incorporate a reflection to preserve the sign, meaning that $det(U)$ has to be $-1$.
    $endgroup$
    – user1551
    Dec 8 '18 at 6:07










1




1




$begingroup$
Extend $w=t/|t|_2$ to an orthonormal basis ${u,v,w}$. Then take $U=pmatrix{u,v,w}$.
$endgroup$
– user1551
Dec 7 '18 at 9:40




$begingroup$
Extend $w=t/|t|_2$ to an orthonormal basis ${u,v,w}$. Then take $U=pmatrix{u,v,w}$.
$endgroup$
– user1551
Dec 7 '18 at 9:40












$begingroup$
@user1551 hello, i apply your thread(see following post of mine) while leads to a contradiction, can you help me out?
$endgroup$
– Finley
Dec 7 '18 at 15:37




$begingroup$
@user1551 hello, i apply your thread(see following post of mine) while leads to a contradiction, can you help me out?
$endgroup$
– Finley
Dec 7 '18 at 15:37












$begingroup$
I didn't noticed that you need $a>0$. If you want to enforce this constraint and you find that with your chosen $u,v$ the cross product $vtimes u$ is $-w$ rather than $w$, you may simply interchange $u$ and $v$. Alternatively, replace $v$ by $-v$.
$endgroup$
– user1551
Dec 7 '18 at 16:39




$begingroup$
I didn't noticed that you need $a>0$. If you want to enforce this constraint and you find that with your chosen $u,v$ the cross product $vtimes u$ is $-w$ rather than $w$, you may simply interchange $u$ and $v$. Alternatively, replace $v$ by $-v$.
$endgroup$
– user1551
Dec 7 '18 at 16:39












$begingroup$
@user1551 But $det(U) = -1$ once I interchange $u text{ and } v$ or negate $v$. Is it reasonable for an orthonormal matrix?
$endgroup$
– Finley
Dec 8 '18 at 2:11






$begingroup$
@user1551 But $det(U) = -1$ once I interchange $u text{ and } v$ or negate $v$. Is it reasonable for an orthonormal matrix?
$endgroup$
– Finley
Dec 8 '18 at 2:11














$begingroup$
Why not? Your question doesn't require that $det(U)=1$. Anyway, since it sets $Z$ to $pmatrix{0&1\ -1&0}$ rather than $pmatrix{0&-1\ 1&0}$, the determinant of $U$ has to be $-1$: when $u$ is orthogonal to $t$, the effect of $ttimes u$ in the tangential direction is an anticlockwise rotation of $u$ about the directed axis $t$. However, $Z=pmatrix{0&1\ -1&0}$ represents a clockwise rotation. So, in the change-of-basis matrix $U$, you must incorporate a reflection to preserve the sign, meaning that $det(U)$ has to be $-1$.
$endgroup$
– user1551
Dec 8 '18 at 6:07






$begingroup$
Why not? Your question doesn't require that $det(U)=1$. Anyway, since it sets $Z$ to $pmatrix{0&1\ -1&0}$ rather than $pmatrix{0&-1\ 1&0}$, the determinant of $U$ has to be $-1$: when $u$ is orthogonal to $t$, the effect of $ttimes u$ in the tangential direction is an anticlockwise rotation of $u$ about the directed axis $t$. However, $Z=pmatrix{0&1\ -1&0}$ represents a clockwise rotation. So, in the change-of-basis matrix $U$, you must incorporate a reflection to preserve the sign, meaning that $det(U)$ has to be $-1$.
$endgroup$
– user1551
Dec 8 '18 at 6:07












1 Answer
1






active

oldest

votes


















0












$begingroup$

Update



enter image description here



let $w=t/|t|$, then we take three orthonormal basis which form a orthonormal matrix $U = [u,v,w]$.Therefore
$$begin{align} Udiag(aZ,0)U^T&=[u,v,w]begin{bmatrix}
0 & a & 0 \
-a & 0 & 0 \
0 & 0 & 0end{bmatrix}begin{bmatrix} u^T \ v^T \ w^T end{bmatrix} \
&=a(uv^T-vu^T) \
&=abegin{bmatrix}
0 & u_1v_2-u_2v_1 & u_1v_3-u_3v_1 \
u_2v_1-u_1v_2 & 0 & u_2v_3-u_3v_2 \
u_3v_1-u_1v_3 & u_3v_2-u_2v_3 & 0end{bmatrix} \
&=a(v times u)^{land}end{align} $$
So above derivation leads to
$$t^{land} = a(v times u)^{land}$$that is
$$t = a*(-w) = |t|(-w)$$
but it contradicts with $t = |t|w$, who can help?






share|cite|improve this answer









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    1






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    0












    $begingroup$

    Update



    enter image description here



    let $w=t/|t|$, then we take three orthonormal basis which form a orthonormal matrix $U = [u,v,w]$.Therefore
    $$begin{align} Udiag(aZ,0)U^T&=[u,v,w]begin{bmatrix}
    0 & a & 0 \
    -a & 0 & 0 \
    0 & 0 & 0end{bmatrix}begin{bmatrix} u^T \ v^T \ w^T end{bmatrix} \
    &=a(uv^T-vu^T) \
    &=abegin{bmatrix}
    0 & u_1v_2-u_2v_1 & u_1v_3-u_3v_1 \
    u_2v_1-u_1v_2 & 0 & u_2v_3-u_3v_2 \
    u_3v_1-u_1v_3 & u_3v_2-u_2v_3 & 0end{bmatrix} \
    &=a(v times u)^{land}end{align} $$
    So above derivation leads to
    $$t^{land} = a(v times u)^{land}$$that is
    $$t = a*(-w) = |t|(-w)$$
    but it contradicts with $t = |t|w$, who can help?






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Update



      enter image description here



      let $w=t/|t|$, then we take three orthonormal basis which form a orthonormal matrix $U = [u,v,w]$.Therefore
      $$begin{align} Udiag(aZ,0)U^T&=[u,v,w]begin{bmatrix}
      0 & a & 0 \
      -a & 0 & 0 \
      0 & 0 & 0end{bmatrix}begin{bmatrix} u^T \ v^T \ w^T end{bmatrix} \
      &=a(uv^T-vu^T) \
      &=abegin{bmatrix}
      0 & u_1v_2-u_2v_1 & u_1v_3-u_3v_1 \
      u_2v_1-u_1v_2 & 0 & u_2v_3-u_3v_2 \
      u_3v_1-u_1v_3 & u_3v_2-u_2v_3 & 0end{bmatrix} \
      &=a(v times u)^{land}end{align} $$
      So above derivation leads to
      $$t^{land} = a(v times u)^{land}$$that is
      $$t = a*(-w) = |t|(-w)$$
      but it contradicts with $t = |t|w$, who can help?






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Update



        enter image description here



        let $w=t/|t|$, then we take three orthonormal basis which form a orthonormal matrix $U = [u,v,w]$.Therefore
        $$begin{align} Udiag(aZ,0)U^T&=[u,v,w]begin{bmatrix}
        0 & a & 0 \
        -a & 0 & 0 \
        0 & 0 & 0end{bmatrix}begin{bmatrix} u^T \ v^T \ w^T end{bmatrix} \
        &=a(uv^T-vu^T) \
        &=abegin{bmatrix}
        0 & u_1v_2-u_2v_1 & u_1v_3-u_3v_1 \
        u_2v_1-u_1v_2 & 0 & u_2v_3-u_3v_2 \
        u_3v_1-u_1v_3 & u_3v_2-u_2v_3 & 0end{bmatrix} \
        &=a(v times u)^{land}end{align} $$
        So above derivation leads to
        $$t^{land} = a(v times u)^{land}$$that is
        $$t = a*(-w) = |t|(-w)$$
        but it contradicts with $t = |t|w$, who can help?






        share|cite|improve this answer









        $endgroup$



        Update



        enter image description here



        let $w=t/|t|$, then we take three orthonormal basis which form a orthonormal matrix $U = [u,v,w]$.Therefore
        $$begin{align} Udiag(aZ,0)U^T&=[u,v,w]begin{bmatrix}
        0 & a & 0 \
        -a & 0 & 0 \
        0 & 0 & 0end{bmatrix}begin{bmatrix} u^T \ v^T \ w^T end{bmatrix} \
        &=a(uv^T-vu^T) \
        &=abegin{bmatrix}
        0 & u_1v_2-u_2v_1 & u_1v_3-u_3v_1 \
        u_2v_1-u_1v_2 & 0 & u_2v_3-u_3v_2 \
        u_3v_1-u_1v_3 & u_3v_2-u_2v_3 & 0end{bmatrix} \
        &=a(v times u)^{land}end{align} $$
        So above derivation leads to
        $$t^{land} = a(v times u)^{land}$$that is
        $$t = a*(-w) = |t|(-w)$$
        but it contradicts with $t = |t|w$, who can help?







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 7 '18 at 15:35









        FinleyFinley

        375213




        375213






























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            Puebla de Zaragoza

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            Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
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            Musa

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            Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
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