Change of coordinate transformation matrix
$begingroup$
I want to convert the following matrix:
$$
left[ begin{matrix}
0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 \
0 & 0 & 0 & 0 & 1 & 1 & 1 & 1\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
end {matrix} right]
$$
To the following:
$$
left[ begin{matrix}
0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 \
0 & 0 & 0 & 0 & 1 & 1 & 1 & 1\
0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 \
end {matrix} right]
$$
I understand that the 8x8 transformation matrix is not unique, I'm ok with any one that works. Essentially I want to raise every other point (if every vector represents an element in $mathbb{R}^3$) to unity in the third dimension.
If the matrices were square I would solve $AX=B$ by $X=A^{-1}B$ but neither A or B are square. Any quick ways to do this?
matrices linear-transformations matrix-equations
asked Dec 28 '18 at 23:45
user1357015user1357015
22718
$endgroup$
add a comment |
$begingroup$
I want to convert the following matrix:
$$
left[ begin{matrix}
0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 \
0 & 0 & 0 & 0 & 1 & 1 & 1 & 1\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
end {matrix} right]
$$
To the following:
$$
left[ begin{matrix}
0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 \
0 & 0 & 0 & 0 & 1 & 1 & 1 & 1\
0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 \
end {matrix} right]
$$
I understand that the 8x8 transformation matrix is not unique, I'm ok with any one that works. Essentially I want to raise every other point (if every vector represents an element in $mathbb{R}^3$) to unity in the third dimension.
If the matrices were square I would solve $AX=B$ by $X=A^{-1}B$ but neither A or B are square. Any quick ways to do this?
matrices linear-transformations matrix-equations
asked Dec 28 '18 at 23:45
user1357015user1357015
22718
$endgroup$
add a comment |
$begingroup$
I want to convert the following matrix:
$$
left[ begin{matrix}
0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 \
0 & 0 & 0 & 0 & 1 & 1 & 1 & 1\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
end {matrix} right]
$$
To the following:
$$
left[ begin{matrix}
0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 \
0 & 0 & 0 & 0 & 1 & 1 & 1 & 1\
0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 \
end {matrix} right]
$$
I understand that the 8x8 transformation matrix is not unique, I'm ok with any one that works. Essentially I want to raise every other point (if every vector represents an element in $mathbb{R}^3$) to unity in the third dimension.
If the matrices were square I would solve $AX=B$ by $X=A^{-1}B$ but neither A or B are square. Any quick ways to do this?
matrices linear-transformations matrix-equations
asked Dec 28 '18 at 23:45
user1357015user1357015
22718
$endgroup$
I want to convert the following matrix:
$$
left[ begin{matrix}
0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 \
0 & 0 & 0 & 0 & 1 & 1 & 1 & 1\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
end {matrix} right]
$$
To the following:
$$
left[ begin{matrix}
0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 \
0 & 0 & 0 & 0 & 1 & 1 & 1 & 1\
0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 \
end {matrix} right]
$$
I understand that the 8x8 transformation matrix is not unique, I'm ok with any one that works. Essentially I want to raise every other point (if every vector represents an element in $mathbb{R}^3$) to unity in the third dimension.
If the matrices were square I would solve $AX=B$ by $X=A^{-1}B$ but neither A or B are square. Any quick ways to do this?
matrices linear-transformations matrix-equations
matrices linear-transformations matrix-equations
asked Dec 28 '18 at 23:45
user1357015user1357015
22718
asked Dec 28 '18 at 23:45
user1357015user1357015
22718
asked Dec 28 '18 at 23:45
user1357015user1357015
22718
asked Dec 28 '18 at 23:45
user1357015user1357015
22718
asked Dec 28 '18 at 23:45
user1357015user1357015
22718
22718
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In the following, I'm going to denote the column $i$ of matrix $M$ as $M_i$. I don't think this is standard notation, but it will make things simpler for this answer.
So, another way to think of $AX$ is by the equation $(AX)_i=AX_i$, which basically says that column $i$ of $AX$ is a linear combination of the columns of $A$, where the coefficients are determined by what column $i$ of $X$ is. Now, this means that each column of $AX$ is a linear combination of the columns of $A$, so each column of $AX$ needs to be in the column space of $A$.
However, if we look at $B$, we can clearly see the second column, $(0,0,1)^t$, has a non-zero third coordinate, while in $A$, all of the columns has a zero third coordinate. Therefore, $(0,0,1)^t$ is not in the column space of $A$, so there's no way $(0,0,1)^t$ could be a column of $AX$. Therefore, $AXneq B$ for any $X$.
In short, there is no way to solve the equation $AX=B$.
answered Dec 28 '18 at 23:58
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In the following, I'm going to denote the column $i$ of matrix $M$ as $M_i$. I don't think this is standard notation, but it will make things simpler for this answer.
So, another way to think of $AX$ is by the equation $(AX)_i=AX_i$, which basically says that column $i$ of $AX$ is a linear combination of the columns of $A$, where the coefficients are determined by what column $i$ of $X$ is. Now, this means that each column of $AX$ is a linear combination of the columns of $A$, so each column of $AX$ needs to be in the column space of $A$.
However, if we look at $B$, we can clearly see the second column, $(0,0,1)^t$, has a non-zero third coordinate, while in $A$, all of the columns has a zero third coordinate. Therefore, $(0,0,1)^t$ is not in the column space of $A$, so there's no way $(0,0,1)^t$ could be a column of $AX$. Therefore, $AXneq B$ for any $X$.
In short, there is no way to solve the equation $AX=B$.
answered Dec 28 '18 at 23:58
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
add a comment |
$begingroup$
In the following, I'm going to denote the column $i$ of matrix $M$ as $M_i$. I don't think this is standard notation, but it will make things simpler for this answer.
So, another way to think of $AX$ is by the equation $(AX)_i=AX_i$, which basically says that column $i$ of $AX$ is a linear combination of the columns of $A$, where the coefficients are determined by what column $i$ of $X$ is. Now, this means that each column of $AX$ is a linear combination of the columns of $A$, so each column of $AX$ needs to be in the column space of $A$.
However, if we look at $B$, we can clearly see the second column, $(0,0,1)^t$, has a non-zero third coordinate, while in $A$, all of the columns has a zero third coordinate. Therefore, $(0,0,1)^t$ is not in the column space of $A$, so there's no way $(0,0,1)^t$ could be a column of $AX$. Therefore, $AXneq B$ for any $X$.
In short, there is no way to solve the equation $AX=B$.
answered Dec 28 '18 at 23:58
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
add a comment |
$begingroup$
In the following, I'm going to denote the column $i$ of matrix $M$ as $M_i$. I don't think this is standard notation, but it will make things simpler for this answer.
So, another way to think of $AX$ is by the equation $(AX)_i=AX_i$, which basically says that column $i$ of $AX$ is a linear combination of the columns of $A$, where the coefficients are determined by what column $i$ of $X$ is. Now, this means that each column of $AX$ is a linear combination of the columns of $A$, so each column of $AX$ needs to be in the column space of $A$.
However, if we look at $B$, we can clearly see the second column, $(0,0,1)^t$, has a non-zero third coordinate, while in $A$, all of the columns has a zero third coordinate. Therefore, $(0,0,1)^t$ is not in the column space of $A$, so there's no way $(0,0,1)^t$ could be a column of $AX$. Therefore, $AXneq B$ for any $X$.
In short, there is no way to solve the equation $AX=B$.
answered Dec 28 '18 at 23:58
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
In the following, I'm going to denote the column $i$ of matrix $M$ as $M_i$. I don't think this is standard notation, but it will make things simpler for this answer.
So, another way to think of $AX$ is by the equation $(AX)_i=AX_i$, which basically says that column $i$ of $AX$ is a linear combination of the columns of $A$, where the coefficients are determined by what column $i$ of $X$ is. Now, this means that each column of $AX$ is a linear combination of the columns of $A$, so each column of $AX$ needs to be in the column space of $A$.
However, if we look at $B$, we can clearly see the second column, $(0,0,1)^t$, has a non-zero third coordinate, while in $A$, all of the columns has a zero third coordinate. Therefore, $(0,0,1)^t$ is not in the column space of $A$, so there's no way $(0,0,1)^t$ could be a column of $AX$. Therefore, $AXneq B$ for any $X$.
In short, there is no way to solve the equation $AX=B$.
answered Dec 28 '18 at 23:58
Noble MushtakNoble Mushtak
15.3k1835
answered Dec 28 '18 at 23:58
Noble MushtakNoble Mushtak
15.3k1835
answered Dec 28 '18 at 23:58
Noble MushtakNoble Mushtak
15.3k1835
answered Dec 28 '18 at 23:58
Noble MushtakNoble Mushtak
15.3k1835
15.3k1835
add a comment |
add a comment |
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