4x4 Matrix Visualized as a Cartesian Coordinate System
$begingroup$
I'm reading this text on computer graphics, and came across the following section.
The 4x4 homogenous matrix I'm accustomed to looks like this
$begin{bmatrix}
r_{11} & r_{12} & r_{13} & t_{14} \
r_{21} & r_{22} & r_{23} & t_{24} \
r_{31} & r_{32} & r_{33} & t_{34} \
0 & 0 & 0 & 1
end{bmatrix}$
Where the $r_{11}$ through $r_{33}$ define a 3x3 rotation matrix, and $t_{14}$, $t_{24}$, and $t_{34}$ define a 3x1 translation vector.
I'm also familiar with the "change of coordinates" matrix. Suppose we have vector spaces A and B in $mathbb{R}^3$. I can use the following matrix to take a vector in B, to a vector in A.
$P_{A leftarrow B} = begin{bmatrix} [b_1]_A & [b_2]_A & [b_3]_Aend{bmatrix}$
where $[b_i]_A$ is the i-th basis vector of B in A-coordinates.
Given this information, I'm trying to interpret how the rows of a 4x4 matrix represent the x-axis, y-axis, z-axis, and translation.
linear-algebra geometry euclidean-geometry
asked Dec 27 '18 at 22:30
CarpetfizzCarpetfizz
486313
$endgroup$
add a comment |
$begingroup$
I'm reading this text on computer graphics, and came across the following section.
The 4x4 homogenous matrix I'm accustomed to looks like this
$begin{bmatrix}
r_{11} & r_{12} & r_{13} & t_{14} \
r_{21} & r_{22} & r_{23} & t_{24} \
r_{31} & r_{32} & r_{33} & t_{34} \
0 & 0 & 0 & 1
end{bmatrix}$
Where the $r_{11}$ through $r_{33}$ define a 3x3 rotation matrix, and $t_{14}$, $t_{24}$, and $t_{34}$ define a 3x1 translation vector.
I'm also familiar with the "change of coordinates" matrix. Suppose we have vector spaces A and B in $mathbb{R}^3$. I can use the following matrix to take a vector in B, to a vector in A.
$P_{A leftarrow B} = begin{bmatrix} [b_1]_A & [b_2]_A & [b_3]_Aend{bmatrix}$
where $[b_i]_A$ is the i-th basis vector of B in A-coordinates.
Given this information, I'm trying to interpret how the rows of a 4x4 matrix represent the x-axis, y-axis, z-axis, and translation.
linear-algebra geometry euclidean-geometry
asked Dec 27 '18 at 22:30
CarpetfizzCarpetfizz
486313
$endgroup$
1
$begingroup$
For one thing, that text represents points by row vectors, so transformations involve post-multiplying by a matrix, which is thus the transpose of what you’re used to.
$endgroup$
– amd
Dec 28 '18 at 1:50
$begingroup$
Got it, can’t believe I missed that detail. Thank you so much!
$endgroup$
– Carpetfizz
Dec 28 '18 at 1:51
add a comment |
$begingroup$
I'm reading this text on computer graphics, and came across the following section.
The 4x4 homogenous matrix I'm accustomed to looks like this
$begin{bmatrix}
r_{11} & r_{12} & r_{13} & t_{14} \
r_{21} & r_{22} & r_{23} & t_{24} \
r_{31} & r_{32} & r_{33} & t_{34} \
0 & 0 & 0 & 1
end{bmatrix}$
Where the $r_{11}$ through $r_{33}$ define a 3x3 rotation matrix, and $t_{14}$, $t_{24}$, and $t_{34}$ define a 3x1 translation vector.
I'm also familiar with the "change of coordinates" matrix. Suppose we have vector spaces A and B in $mathbb{R}^3$. I can use the following matrix to take a vector in B, to a vector in A.
$P_{A leftarrow B} = begin{bmatrix} [b_1]_A & [b_2]_A & [b_3]_Aend{bmatrix}$
where $[b_i]_A$ is the i-th basis vector of B in A-coordinates.
Given this information, I'm trying to interpret how the rows of a 4x4 matrix represent the x-axis, y-axis, z-axis, and translation.
linear-algebra geometry euclidean-geometry
asked Dec 27 '18 at 22:30
CarpetfizzCarpetfizz
486313
$endgroup$
I'm reading this text on computer graphics, and came across the following section.
The 4x4 homogenous matrix I'm accustomed to looks like this
$begin{bmatrix}
r_{11} & r_{12} & r_{13} & t_{14} \
r_{21} & r_{22} & r_{23} & t_{24} \
r_{31} & r_{32} & r_{33} & t_{34} \
0 & 0 & 0 & 1
end{bmatrix}$
Where the $r_{11}$ through $r_{33}$ define a 3x3 rotation matrix, and $t_{14}$, $t_{24}$, and $t_{34}$ define a 3x1 translation vector.
I'm also familiar with the "change of coordinates" matrix. Suppose we have vector spaces A and B in $mathbb{R}^3$. I can use the following matrix to take a vector in B, to a vector in A.
$P_{A leftarrow B} = begin{bmatrix} [b_1]_A & [b_2]_A & [b_3]_Aend{bmatrix}$
where $[b_i]_A$ is the i-th basis vector of B in A-coordinates.
Given this information, I'm trying to interpret how the rows of a 4x4 matrix represent the x-axis, y-axis, z-axis, and translation.
linear-algebra geometry euclidean-geometry
linear-algebra geometry euclidean-geometry
asked Dec 27 '18 at 22:30
CarpetfizzCarpetfizz
486313
asked Dec 27 '18 at 22:30
CarpetfizzCarpetfizz
486313
asked Dec 27 '18 at 22:30
CarpetfizzCarpetfizz
486313
asked Dec 27 '18 at 22:30
CarpetfizzCarpetfizz
486313
asked Dec 27 '18 at 22:30
CarpetfizzCarpetfizz
486313
486313
1
$begingroup$
For one thing, that text represents points by row vectors, so transformations involve post-multiplying by a matrix, which is thus the transpose of what you’re used to.
$endgroup$
– amd
Dec 28 '18 at 1:50
$begingroup$
Got it, can’t believe I missed that detail. Thank you so much!
$endgroup$
– Carpetfizz
Dec 28 '18 at 1:51
add a comment |
1
$begingroup$
For one thing, that text represents points by row vectors, so transformations involve post-multiplying by a matrix, which is thus the transpose of what you’re used to.
$endgroup$
– amd
Dec 28 '18 at 1:50
$begingroup$
Got it, can’t believe I missed that detail. Thank you so much!
$endgroup$
– Carpetfizz
Dec 28 '18 at 1:51
1
1
$begingroup$
For one thing, that text represents points by row vectors, so transformations involve post-multiplying by a matrix, which is thus the transpose of what you’re used to.
$endgroup$
– amd
Dec 28 '18 at 1:50
$begingroup$
For one thing, that text represents points by row vectors, so transformations involve post-multiplying by a matrix, which is thus the transpose of what you’re used to.
$endgroup$
– amd
Dec 28 '18 at 1:50
$begingroup$
Got it, can’t believe I missed that detail. Thank you so much!
$endgroup$
– Carpetfizz
Dec 28 '18 at 1:51
$begingroup$
Got it, can’t believe I missed that detail. Thank you so much!
$endgroup$
– Carpetfizz
Dec 28 '18 at 1:51
add a comment |
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$begingroup$
For one thing, that text represents points by row vectors, so transformations involve post-multiplying by a matrix, which is thus the transpose of what you’re used to.
$endgroup$
– amd
Dec 28 '18 at 1:50
$begingroup$
Got it, can’t believe I missed that detail. Thank you so much!
$endgroup$
– Carpetfizz
Dec 28 '18 at 1:51