How to Find A Shear Matrix in 3D?












0












$begingroup$


In particular, I'm struggling with this question:
enter image description here



Give me a rotational matrix, a scaling matrix, or a reflection matrix and I can provide it quite easily. No online resources I've found give me much information on how to find a 3D shear matrix. I even tried asking this question on Chegg and I got a dead wrong answer. Because one thing I do KNOW about shear matrices from 2D is that they preserve area and that can be applied to any dimension so in $R^3$ the volume is preserved. And the "expert" on Chegg said the volume goes to 0, which is obviously not true!



So I'm hoping I'll find better luck asking it here. Thanks in advance!






linear-algebra matrices linear-transformations 3d







share|cite|improve this question









$endgroup$












  • $begingroup$
    Think about what a typical shear matrix (larger than $2times 2$) looks like. An upper triangular matrix with $1$ along the diagonal and very few non-zero entries otherwise.
    $endgroup$
    – Arthur
    Dec 3 '18 at 7:29












  • $begingroup$
    I am aware. How do you algebraically find the non-zero entries?
    $endgroup$
    – Alex
    Dec 3 '18 at 7:31


















0












$begingroup$


In particular, I'm struggling with this question:
enter image description here



Give me a rotational matrix, a scaling matrix, or a reflection matrix and I can provide it quite easily. No online resources I've found give me much information on how to find a 3D shear matrix. I even tried asking this question on Chegg and I got a dead wrong answer. Because one thing I do KNOW about shear matrices from 2D is that they preserve area and that can be applied to any dimension so in $R^3$ the volume is preserved. And the "expert" on Chegg said the volume goes to 0, which is obviously not true!



So I'm hoping I'll find better luck asking it here. Thanks in advance!






linear-algebra matrices linear-transformations 3d







share|cite|improve this question









$endgroup$












  • $begingroup$
    Think about what a typical shear matrix (larger than $2times 2$) looks like. An upper triangular matrix with $1$ along the diagonal and very few non-zero entries otherwise.
    $endgroup$
    – Arthur
    Dec 3 '18 at 7:29












  • $begingroup$
    I am aware. How do you algebraically find the non-zero entries?
    $endgroup$
    – Alex
    Dec 3 '18 at 7:31
















0












0








0





$begingroup$


In particular, I'm struggling with this question:
enter image description here



Give me a rotational matrix, a scaling matrix, or a reflection matrix and I can provide it quite easily. No online resources I've found give me much information on how to find a 3D shear matrix. I even tried asking this question on Chegg and I got a dead wrong answer. Because one thing I do KNOW about shear matrices from 2D is that they preserve area and that can be applied to any dimension so in $R^3$ the volume is preserved. And the "expert" on Chegg said the volume goes to 0, which is obviously not true!



So I'm hoping I'll find better luck asking it here. Thanks in advance!






linear-algebra matrices linear-transformations 3d







share|cite|improve this question









$endgroup$




In particular, I'm struggling with this question:
enter image description here



Give me a rotational matrix, a scaling matrix, or a reflection matrix and I can provide it quite easily. No online resources I've found give me much information on how to find a 3D shear matrix. I even tried asking this question on Chegg and I got a dead wrong answer. Because one thing I do KNOW about shear matrices from 2D is that they preserve area and that can be applied to any dimension so in $R^3$ the volume is preserved. And the "expert" on Chegg said the volume goes to 0, which is obviously not true!



So I'm hoping I'll find better luck asking it here. Thanks in advance!






linear-algebra matrices linear-transformations 3d




linear-algebra matrices linear-transformations 3d






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 3 '18 at 7:24









AlexAlex

525




525












  • $begingroup$
    Think about what a typical shear matrix (larger than $2times 2$) looks like. An upper triangular matrix with $1$ along the diagonal and very few non-zero entries otherwise.
    $endgroup$
    – Arthur
    Dec 3 '18 at 7:29












  • $begingroup$
    I am aware. How do you algebraically find the non-zero entries?
    $endgroup$
    – Alex
    Dec 3 '18 at 7:31




















  • $begingroup$
    Think about what a typical shear matrix (larger than $2times 2$) looks like. An upper triangular matrix with $1$ along the diagonal and very few non-zero entries otherwise.
    $endgroup$
    – Arthur
    Dec 3 '18 at 7:29












  • $begingroup$
    I am aware. How do you algebraically find the non-zero entries?
    $endgroup$
    – Alex
    Dec 3 '18 at 7:31


















$begingroup$
Think about what a typical shear matrix (larger than $2times 2$) looks like. An upper triangular matrix with $1$ along the diagonal and very few non-zero entries otherwise.
$endgroup$
– Arthur
Dec 3 '18 at 7:29






$begingroup$
Think about what a typical shear matrix (larger than $2times 2$) looks like. An upper triangular matrix with $1$ along the diagonal and very few non-zero entries otherwise.
$endgroup$
– Arthur
Dec 3 '18 at 7:29














$begingroup$
I am aware. How do you algebraically find the non-zero entries?
$endgroup$
– Alex
Dec 3 '18 at 7:31






$begingroup$
I am aware. How do you algebraically find the non-zero entries?
$endgroup$
– Alex
Dec 3 '18 at 7:31












1 Answer
1






active

oldest

votes


















1












$begingroup$

A true shear matrix, according to Wikipedia, is an elementary matrix, and with only one non-zero off-diagonal entry, this cannot be done.



However, shear transformations are a bit more flexible. Specifically because they are not bound to shearing along the coordinate axes.



Note that there are infinitely many shear transformations $S$ for which
$$
Sbegin{bmatrix}a\b\cend{bmatrix} = begin{bmatrix}0\0\cend{bmatrix}tag{*}
$$
This is possibly what gives you trouble finding a solution; if you try to solve a problem algebraically and you don't know that there are infinitely many solutions, it's going to look like you're missing an equation somewhere. So, with that said, let's find one such transformation and the matrix representing it.



Firstly, I'm going to assume that $c neq 0$. That's because shears are invertible, and if $c = 0$, then by $(text *)$ we must also have $a = b = 0$, and this becomes quite a boring exercise.



Secondly, we clearly have
$$
Sbegin{bmatrix}a\b\0end{bmatrix} = begin{bmatrix}a\b\0end{bmatrix}
$$
but I will assume that
$$
Sbegin{bmatrix}x\y\0end{bmatrix} = begin{bmatrix}x\y\0end{bmatrix}
$$
for all $x, y$. This is where the multiple solutions come in. If you make a basis $(u, v)$ for the $xy$-plane where $u = [a, b, 0]^T$, then $Su = u$, but we know very little about $Sv$. I assume here that $Sv = v$ as well, but we have no indication in the problem text whether this is so. $Sv$ doesn't even have to be in the $xy$-plane.



So, with this, we are ready to begin making a matrix of $S$:
$$
S = begin{bmatrix}1&0&\0&1&\0&0&1end{bmatrix}
$$
And I believe you can fill in the two missing numbers by using $(text *)$.



Note that expressing this matrix in a basis where $[a, b, 0]^T$ and $[0, 0, 1]^T$ (or constant multiples of these) are two of the basis vectors, this transformation $S$ will have the elementary form required in the Wikipedia article on shear matrices.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    I figured it out! I took your $S = begin{bmatrix}1&0&\0&1&\0&0&1end{bmatrix}$ Then substituted the 2 unknowns with x and y: $S = begin{bmatrix}1&0&x\0&1&y\0&0&1end{bmatrix}$ We have the vector $v =begin{bmatrix}a\b\cend{bmatrix}$ With $Sv=v'$ we get $begin{bmatrix}a+xc\b+yc\cend{bmatrix}$ Of course we are trying to solve for $v' = begin{bmatrix}0\0\cend{bmatrix}$ The $c$ is already done, but we still need to solve for $a+xc=0$ and $b+yc = 0$. Where with a little algebra you get $x=-frac{a}{c}$ and $y=-frac{b}{c}$ And we are done! Is this what you were going for?
    $endgroup$
    – Alex
    Dec 4 '18 at 0:47






  • 1




    $begingroup$
    @Alex Great work!
    $endgroup$
    – Arthur
    Dec 4 '18 at 6:54











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1 Answer
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1 Answer
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1












$begingroup$

A true shear matrix, according to Wikipedia, is an elementary matrix, and with only one non-zero off-diagonal entry, this cannot be done.



However, shear transformations are a bit more flexible. Specifically because they are not bound to shearing along the coordinate axes.



Note that there are infinitely many shear transformations $S$ for which
$$
Sbegin{bmatrix}a\b\cend{bmatrix} = begin{bmatrix}0\0\cend{bmatrix}tag{*}
$$
This is possibly what gives you trouble finding a solution; if you try to solve a problem algebraically and you don't know that there are infinitely many solutions, it's going to look like you're missing an equation somewhere. So, with that said, let's find one such transformation and the matrix representing it.



Firstly, I'm going to assume that $c neq 0$. That's because shears are invertible, and if $c = 0$, then by $(text *)$ we must also have $a = b = 0$, and this becomes quite a boring exercise.



Secondly, we clearly have
$$
Sbegin{bmatrix}a\b\0end{bmatrix} = begin{bmatrix}a\b\0end{bmatrix}
$$
but I will assume that
$$
Sbegin{bmatrix}x\y\0end{bmatrix} = begin{bmatrix}x\y\0end{bmatrix}
$$
for all $x, y$. This is where the multiple solutions come in. If you make a basis $(u, v)$ for the $xy$-plane where $u = [a, b, 0]^T$, then $Su = u$, but we know very little about $Sv$. I assume here that $Sv = v$ as well, but we have no indication in the problem text whether this is so. $Sv$ doesn't even have to be in the $xy$-plane.



So, with this, we are ready to begin making a matrix of $S$:
$$
S = begin{bmatrix}1&0&\0&1&\0&0&1end{bmatrix}
$$
And I believe you can fill in the two missing numbers by using $(text *)$.



Note that expressing this matrix in a basis where $[a, b, 0]^T$ and $[0, 0, 1]^T$ (or constant multiples of these) are two of the basis vectors, this transformation $S$ will have the elementary form required in the Wikipedia article on shear matrices.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    I figured it out! I took your $S = begin{bmatrix}1&0&\0&1&\0&0&1end{bmatrix}$ Then substituted the 2 unknowns with x and y: $S = begin{bmatrix}1&0&x\0&1&y\0&0&1end{bmatrix}$ We have the vector $v =begin{bmatrix}a\b\cend{bmatrix}$ With $Sv=v'$ we get $begin{bmatrix}a+xc\b+yc\cend{bmatrix}$ Of course we are trying to solve for $v' = begin{bmatrix}0\0\cend{bmatrix}$ The $c$ is already done, but we still need to solve for $a+xc=0$ and $b+yc = 0$. Where with a little algebra you get $x=-frac{a}{c}$ and $y=-frac{b}{c}$ And we are done! Is this what you were going for?
    $endgroup$
    – Alex
    Dec 4 '18 at 0:47






  • 1




    $begingroup$
    @Alex Great work!
    $endgroup$
    – Arthur
    Dec 4 '18 at 6:54
















1












$begingroup$

A true shear matrix, according to Wikipedia, is an elementary matrix, and with only one non-zero off-diagonal entry, this cannot be done.



However, shear transformations are a bit more flexible. Specifically because they are not bound to shearing along the coordinate axes.



Note that there are infinitely many shear transformations $S$ for which
$$
Sbegin{bmatrix}a\b\cend{bmatrix} = begin{bmatrix}0\0\cend{bmatrix}tag{*}
$$
This is possibly what gives you trouble finding a solution; if you try to solve a problem algebraically and you don't know that there are infinitely many solutions, it's going to look like you're missing an equation somewhere. So, with that said, let's find one such transformation and the matrix representing it.



Firstly, I'm going to assume that $c neq 0$. That's because shears are invertible, and if $c = 0$, then by $(text *)$ we must also have $a = b = 0$, and this becomes quite a boring exercise.



Secondly, we clearly have
$$
Sbegin{bmatrix}a\b\0end{bmatrix} = begin{bmatrix}a\b\0end{bmatrix}
$$
but I will assume that
$$
Sbegin{bmatrix}x\y\0end{bmatrix} = begin{bmatrix}x\y\0end{bmatrix}
$$
for all $x, y$. This is where the multiple solutions come in. If you make a basis $(u, v)$ for the $xy$-plane where $u = [a, b, 0]^T$, then $Su = u$, but we know very little about $Sv$. I assume here that $Sv = v$ as well, but we have no indication in the problem text whether this is so. $Sv$ doesn't even have to be in the $xy$-plane.



So, with this, we are ready to begin making a matrix of $S$:
$$
S = begin{bmatrix}1&0&\0&1&\0&0&1end{bmatrix}
$$
And I believe you can fill in the two missing numbers by using $(text *)$.



Note that expressing this matrix in a basis where $[a, b, 0]^T$ and $[0, 0, 1]^T$ (or constant multiples of these) are two of the basis vectors, this transformation $S$ will have the elementary form required in the Wikipedia article on shear matrices.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    I figured it out! I took your $S = begin{bmatrix}1&0&\0&1&\0&0&1end{bmatrix}$ Then substituted the 2 unknowns with x and y: $S = begin{bmatrix}1&0&x\0&1&y\0&0&1end{bmatrix}$ We have the vector $v =begin{bmatrix}a\b\cend{bmatrix}$ With $Sv=v'$ we get $begin{bmatrix}a+xc\b+yc\cend{bmatrix}$ Of course we are trying to solve for $v' = begin{bmatrix}0\0\cend{bmatrix}$ The $c$ is already done, but we still need to solve for $a+xc=0$ and $b+yc = 0$. Where with a little algebra you get $x=-frac{a}{c}$ and $y=-frac{b}{c}$ And we are done! Is this what you were going for?
    $endgroup$
    – Alex
    Dec 4 '18 at 0:47






  • 1




    $begingroup$
    @Alex Great work!
    $endgroup$
    – Arthur
    Dec 4 '18 at 6:54














1












1








1





$begingroup$

A true shear matrix, according to Wikipedia, is an elementary matrix, and with only one non-zero off-diagonal entry, this cannot be done.



However, shear transformations are a bit more flexible. Specifically because they are not bound to shearing along the coordinate axes.



Note that there are infinitely many shear transformations $S$ for which
$$
Sbegin{bmatrix}a\b\cend{bmatrix} = begin{bmatrix}0\0\cend{bmatrix}tag{*}
$$
This is possibly what gives you trouble finding a solution; if you try to solve a problem algebraically and you don't know that there are infinitely many solutions, it's going to look like you're missing an equation somewhere. So, with that said, let's find one such transformation and the matrix representing it.



Firstly, I'm going to assume that $c neq 0$. That's because shears are invertible, and if $c = 0$, then by $(text *)$ we must also have $a = b = 0$, and this becomes quite a boring exercise.



Secondly, we clearly have
$$
Sbegin{bmatrix}a\b\0end{bmatrix} = begin{bmatrix}a\b\0end{bmatrix}
$$
but I will assume that
$$
Sbegin{bmatrix}x\y\0end{bmatrix} = begin{bmatrix}x\y\0end{bmatrix}
$$
for all $x, y$. This is where the multiple solutions come in. If you make a basis $(u, v)$ for the $xy$-plane where $u = [a, b, 0]^T$, then $Su = u$, but we know very little about $Sv$. I assume here that $Sv = v$ as well, but we have no indication in the problem text whether this is so. $Sv$ doesn't even have to be in the $xy$-plane.



So, with this, we are ready to begin making a matrix of $S$:
$$
S = begin{bmatrix}1&0&\0&1&\0&0&1end{bmatrix}
$$
And I believe you can fill in the two missing numbers by using $(text *)$.



Note that expressing this matrix in a basis where $[a, b, 0]^T$ and $[0, 0, 1]^T$ (or constant multiples of these) are two of the basis vectors, this transformation $S$ will have the elementary form required in the Wikipedia article on shear matrices.






share|cite|improve this answer









$endgroup$



A true shear matrix, according to Wikipedia, is an elementary matrix, and with only one non-zero off-diagonal entry, this cannot be done.



However, shear transformations are a bit more flexible. Specifically because they are not bound to shearing along the coordinate axes.



Note that there are infinitely many shear transformations $S$ for which
$$
Sbegin{bmatrix}a\b\cend{bmatrix} = begin{bmatrix}0\0\cend{bmatrix}tag{*}
$$
This is possibly what gives you trouble finding a solution; if you try to solve a problem algebraically and you don't know that there are infinitely many solutions, it's going to look like you're missing an equation somewhere. So, with that said, let's find one such transformation and the matrix representing it.



Firstly, I'm going to assume that $c neq 0$. That's because shears are invertible, and if $c = 0$, then by $(text *)$ we must also have $a = b = 0$, and this becomes quite a boring exercise.



Secondly, we clearly have
$$
Sbegin{bmatrix}a\b\0end{bmatrix} = begin{bmatrix}a\b\0end{bmatrix}
$$
but I will assume that
$$
Sbegin{bmatrix}x\y\0end{bmatrix} = begin{bmatrix}x\y\0end{bmatrix}
$$
for all $x, y$. This is where the multiple solutions come in. If you make a basis $(u, v)$ for the $xy$-plane where $u = [a, b, 0]^T$, then $Su = u$, but we know very little about $Sv$. I assume here that $Sv = v$ as well, but we have no indication in the problem text whether this is so. $Sv$ doesn't even have to be in the $xy$-plane.



So, with this, we are ready to begin making a matrix of $S$:
$$
S = begin{bmatrix}1&0&\0&1&\0&0&1end{bmatrix}
$$
And I believe you can fill in the two missing numbers by using $(text *)$.



Note that expressing this matrix in a basis where $[a, b, 0]^T$ and $[0, 0, 1]^T$ (or constant multiples of these) are two of the basis vectors, this transformation $S$ will have the elementary form required in the Wikipedia article on shear matrices.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 3 '18 at 11:25









ArthurArthur

113k7110193




113k7110193








  • 1




    $begingroup$
    I figured it out! I took your $S = begin{bmatrix}1&0&\0&1&\0&0&1end{bmatrix}$ Then substituted the 2 unknowns with x and y: $S = begin{bmatrix}1&0&x\0&1&y\0&0&1end{bmatrix}$ We have the vector $v =begin{bmatrix}a\b\cend{bmatrix}$ With $Sv=v'$ we get $begin{bmatrix}a+xc\b+yc\cend{bmatrix}$ Of course we are trying to solve for $v' = begin{bmatrix}0\0\cend{bmatrix}$ The $c$ is already done, but we still need to solve for $a+xc=0$ and $b+yc = 0$. Where with a little algebra you get $x=-frac{a}{c}$ and $y=-frac{b}{c}$ And we are done! Is this what you were going for?
    $endgroup$
    – Alex
    Dec 4 '18 at 0:47






  • 1




    $begingroup$
    @Alex Great work!
    $endgroup$
    – Arthur
    Dec 4 '18 at 6:54














  • 1




    $begingroup$
    I figured it out! I took your $S = begin{bmatrix}1&0&\0&1&\0&0&1end{bmatrix}$ Then substituted the 2 unknowns with x and y: $S = begin{bmatrix}1&0&x\0&1&y\0&0&1end{bmatrix}$ We have the vector $v =begin{bmatrix}a\b\cend{bmatrix}$ With $Sv=v'$ we get $begin{bmatrix}a+xc\b+yc\cend{bmatrix}$ Of course we are trying to solve for $v' = begin{bmatrix}0\0\cend{bmatrix}$ The $c$ is already done, but we still need to solve for $a+xc=0$ and $b+yc = 0$. Where with a little algebra you get $x=-frac{a}{c}$ and $y=-frac{b}{c}$ And we are done! Is this what you were going for?
    $endgroup$
    – Alex
    Dec 4 '18 at 0:47






  • 1




    $begingroup$
    @Alex Great work!
    $endgroup$
    – Arthur
    Dec 4 '18 at 6:54








1




1




$begingroup$
I figured it out! I took your $S = begin{bmatrix}1&0&\0&1&\0&0&1end{bmatrix}$ Then substituted the 2 unknowns with x and y: $S = begin{bmatrix}1&0&x\0&1&y\0&0&1end{bmatrix}$ We have the vector $v =begin{bmatrix}a\b\cend{bmatrix}$ With $Sv=v'$ we get $begin{bmatrix}a+xc\b+yc\cend{bmatrix}$ Of course we are trying to solve for $v' = begin{bmatrix}0\0\cend{bmatrix}$ The $c$ is already done, but we still need to solve for $a+xc=0$ and $b+yc = 0$. Where with a little algebra you get $x=-frac{a}{c}$ and $y=-frac{b}{c}$ And we are done! Is this what you were going for?
$endgroup$
– Alex
Dec 4 '18 at 0:47




$begingroup$
I figured it out! I took your $S = begin{bmatrix}1&0&\0&1&\0&0&1end{bmatrix}$ Then substituted the 2 unknowns with x and y: $S = begin{bmatrix}1&0&x\0&1&y\0&0&1end{bmatrix}$ We have the vector $v =begin{bmatrix}a\b\cend{bmatrix}$ With $Sv=v'$ we get $begin{bmatrix}a+xc\b+yc\cend{bmatrix}$ Of course we are trying to solve for $v' = begin{bmatrix}0\0\cend{bmatrix}$ The $c$ is already done, but we still need to solve for $a+xc=0$ and $b+yc = 0$. Where with a little algebra you get $x=-frac{a}{c}$ and $y=-frac{b}{c}$ And we are done! Is this what you were going for?
$endgroup$
– Alex
Dec 4 '18 at 0:47




1




1




$begingroup$
@Alex Great work!
$endgroup$
– Arthur
Dec 4 '18 at 6:54




$begingroup$
@Alex Great work!
$endgroup$
– Arthur
Dec 4 '18 at 6:54


















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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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