Mixing
Expanding on the intuitive meaning of singular matrices
$begingroup$
My question is based on "What is the geometric meaning of singular matrix" posted here some years ago.
To make this a bit more intuitive I would like to add an example.
A three-dimensional force vector $F$ applied at a point $P$ with coordinates $(x, y, z)$ creates a moment $M$ at, say, point $(0, 0, 0)$. This moment $M$ is again a vector with components $M_x$, $M_y$ and $M_z$ for which, the following hold:
$$
begin{bmatrix}
0 & F_z & -F_y \
-F_z & 0 & F_x \
F_y & -F_x & 0 \
end{bmatrix}
cdot
begin{bmatrix}
x \
y \
z \
end{bmatrix}
=
begin{bmatrix}
M_x \
M_y \
M_z \
end{bmatrix}
$$
As it can be seen, the above matrix is singular. Math tells me that as a result it cannot be inverted and consequently, given the forces and moments one cannot solve for the coordinates of $P$, aka
given the freedom to apply a given force wherever, not all moments vectors can be produced.
Two questions come to mind here:
- How does one derive what (sub)part of $mathbb R^3$ this given matrix embeds to? What points are reachable?
- How about the moments that can be produced? How would one solve for those?
One can easily construct a solution-triplet by setting $P$ to $(1, 1, 1)$ from which, $M=begin{bmatrix}
F_z - F_y \
F_x - F_z \
F_y - F_x \
end{bmatrix}$
but assuming that instead of the $P$ we knew this $M$ how could $P$ be located?
linear-algebra geometry linear-transformations intuition
asked Jan 10 at 8:43
Ev. KounisEv. Kounis
1438
$endgroup$
add a comment |
$begingroup$
My question is based on "What is the geometric meaning of singular matrix" posted here some years ago.
To make this a bit more intuitive I would like to add an example.
A three-dimensional force vector $F$ applied at a point $P$ with coordinates $(x, y, z)$ creates a moment $M$ at, say, point $(0, 0, 0)$. This moment $M$ is again a vector with components $M_x$, $M_y$ and $M_z$ for which, the following hold:
$$
begin{bmatrix}
0 & F_z & -F_y \
-F_z & 0 & F_x \
F_y & -F_x & 0 \
end{bmatrix}
cdot
begin{bmatrix}
x \
y \
z \
end{bmatrix}
=
begin{bmatrix}
M_x \
M_y \
M_z \
end{bmatrix}
$$
As it can be seen, the above matrix is singular. Math tells me that as a result it cannot be inverted and consequently, given the forces and moments one cannot solve for the coordinates of $P$, aka
given the freedom to apply a given force wherever, not all moments vectors can be produced.
Two questions come to mind here:
- How does one derive what (sub)part of $mathbb R^3$ this given matrix embeds to? What points are reachable?
- How about the moments that can be produced? How would one solve for those?
One can easily construct a solution-triplet by setting $P$ to $(1, 1, 1)$ from which, $M=begin{bmatrix}
F_z - F_y \
F_x - F_z \
F_y - F_x \
end{bmatrix}$
but assuming that instead of the $P$ we knew this $M$ how could $P$ be located?
linear-algebra geometry linear-transformations intuition
asked Jan 10 at 8:43
Ev. KounisEv. Kounis
1438
$endgroup$
add a comment |
$begingroup$
My question is based on "What is the geometric meaning of singular matrix" posted here some years ago.
To make this a bit more intuitive I would like to add an example.
A three-dimensional force vector $F$ applied at a point $P$ with coordinates $(x, y, z)$ creates a moment $M$ at, say, point $(0, 0, 0)$. This moment $M$ is again a vector with components $M_x$, $M_y$ and $M_z$ for which, the following hold:
$$
begin{bmatrix}
0 & F_z & -F_y \
-F_z & 0 & F_x \
F_y & -F_x & 0 \
end{bmatrix}
cdot
begin{bmatrix}
x \
y \
z \
end{bmatrix}
=
begin{bmatrix}
M_x \
M_y \
M_z \
end{bmatrix}
$$
As it can be seen, the above matrix is singular. Math tells me that as a result it cannot be inverted and consequently, given the forces and moments one cannot solve for the coordinates of $P$, aka
given the freedom to apply a given force wherever, not all moments vectors can be produced.
Two questions come to mind here:
- How does one derive what (sub)part of $mathbb R^3$ this given matrix embeds to? What points are reachable?
- How about the moments that can be produced? How would one solve for those?
One can easily construct a solution-triplet by setting $P$ to $(1, 1, 1)$ from which, $M=begin{bmatrix}
F_z - F_y \
F_x - F_z \
F_y - F_x \
end{bmatrix}$
but assuming that instead of the $P$ we knew this $M$ how could $P$ be located?
linear-algebra geometry linear-transformations intuition
asked Jan 10 at 8:43
Ev. KounisEv. Kounis
1438
$endgroup$
My question is based on "What is the geometric meaning of singular matrix" posted here some years ago.
To make this a bit more intuitive I would like to add an example.
A three-dimensional force vector $F$ applied at a point $P$ with coordinates $(x, y, z)$ creates a moment $M$ at, say, point $(0, 0, 0)$. This moment $M$ is again a vector with components $M_x$, $M_y$ and $M_z$ for which, the following hold:
$$
begin{bmatrix}
0 & F_z & -F_y \
-F_z & 0 & F_x \
F_y & -F_x & 0 \
end{bmatrix}
cdot
begin{bmatrix}
x \
y \
z \
end{bmatrix}
=
begin{bmatrix}
M_x \
M_y \
M_z \
end{bmatrix}
$$
As it can be seen, the above matrix is singular. Math tells me that as a result it cannot be inverted and consequently, given the forces and moments one cannot solve for the coordinates of $P$, aka
given the freedom to apply a given force wherever, not all moments vectors can be produced.
Two questions come to mind here:
- How does one derive what (sub)part of $mathbb R^3$ this given matrix embeds to? What points are reachable?
- How about the moments that can be produced? How would one solve for those?
One can easily construct a solution-triplet by setting $P$ to $(1, 1, 1)$ from which, $M=begin{bmatrix}
F_z - F_y \
F_x - F_z \
F_y - F_x \
end{bmatrix}$
but assuming that instead of the $P$ we knew this $M$ how could $P$ be located?
linear-algebra geometry linear-transformations intuition
linear-algebra geometry linear-transformations intuition
asked Jan 10 at 8:43
Ev. KounisEv. Kounis
1438
asked Jan 10 at 8:43
Ev. KounisEv. Kounis
1438
edited Jan 10 at 8:53
Ev. Kounis
asked Jan 10 at 8:43
Ev. KounisEv. Kounis
1438
asked Jan 10 at 8:43
Ev. KounisEv. Kounis
1438
asked Jan 10 at 8:43
Ev. KounisEv. Kounis
1438
1438
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The moment at $O = (0, 0, 0)$ is $vec{M} = vec{P} times vec{F}$. Geometrically, we can think of the moment as a vector that is perpendicular to both the force $vec{F}$ and the vector $vec{P}$ (right-hand rule), and length equal to the area of the parallelogram with sides $OF$ and $OP$. Thus, if the force is fixed, but you can vary $P$, you can achieve any point on the plane through 0, perpendicular to $vec{F}$.
On the other hand, infinitely many points $P$ can give the same moment. We know $P$ must be on the plane through 0, perpendicular to $vec{M}$. Further, we want the parallelogram with sides $OP$ and $OF$ to have area equal to $||vec{M}||$. So $P$ must be on a line parallel to $vec{F}$, at a distance $frac{||vec{M}||}{||vec{F}||}$ from $O$. There are two of them (on the appropriate plane), and which one comes from the direction of the moment.
answered Jan 10 at 10:03
Todor MarkovTodor Markov
2,176411
$endgroup$
$begingroup$
Did you mean to write $vec{M} = vec{P} times vec{F}$?
$endgroup$
– Ev. Kounis
Jan 14 at 12:20
$begingroup$
@Ev.Kounis Yes, indeed
$endgroup$
– Todor Markov
Jan 14 at 12:36
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
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oldest
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active
oldest
votes
$begingroup$
The moment at $O = (0, 0, 0)$ is $vec{M} = vec{P} times vec{F}$. Geometrically, we can think of the moment as a vector that is perpendicular to both the force $vec{F}$ and the vector $vec{P}$ (right-hand rule), and length equal to the area of the parallelogram with sides $OF$ and $OP$. Thus, if the force is fixed, but you can vary $P$, you can achieve any point on the plane through 0, perpendicular to $vec{F}$.
On the other hand, infinitely many points $P$ can give the same moment. We know $P$ must be on the plane through 0, perpendicular to $vec{M}$. Further, we want the parallelogram with sides $OP$ and $OF$ to have area equal to $||vec{M}||$. So $P$ must be on a line parallel to $vec{F}$, at a distance $frac{||vec{M}||}{||vec{F}||}$ from $O$. There are two of them (on the appropriate plane), and which one comes from the direction of the moment.
answered Jan 10 at 10:03
Todor MarkovTodor Markov
2,176411
$endgroup$
$begingroup$
Did you mean to write $vec{M} = vec{P} times vec{F}$?
$endgroup$
– Ev. Kounis
Jan 14 at 12:20
$begingroup$
@Ev.Kounis Yes, indeed
$endgroup$
– Todor Markov
Jan 14 at 12:36
add a comment |
$begingroup$
The moment at $O = (0, 0, 0)$ is $vec{M} = vec{P} times vec{F}$. Geometrically, we can think of the moment as a vector that is perpendicular to both the force $vec{F}$ and the vector $vec{P}$ (right-hand rule), and length equal to the area of the parallelogram with sides $OF$ and $OP$. Thus, if the force is fixed, but you can vary $P$, you can achieve any point on the plane through 0, perpendicular to $vec{F}$.
On the other hand, infinitely many points $P$ can give the same moment. We know $P$ must be on the plane through 0, perpendicular to $vec{M}$. Further, we want the parallelogram with sides $OP$ and $OF$ to have area equal to $||vec{M}||$. So $P$ must be on a line parallel to $vec{F}$, at a distance $frac{||vec{M}||}{||vec{F}||}$ from $O$. There are two of them (on the appropriate plane), and which one comes from the direction of the moment.
answered Jan 10 at 10:03
Todor MarkovTodor Markov
2,176411
$endgroup$
$begingroup$
Did you mean to write $vec{M} = vec{P} times vec{F}$?
$endgroup$
– Ev. Kounis
Jan 14 at 12:20
$begingroup$
@Ev.Kounis Yes, indeed
$endgroup$
– Todor Markov
Jan 14 at 12:36
add a comment |
$begingroup$
The moment at $O = (0, 0, 0)$ is $vec{M} = vec{P} times vec{F}$. Geometrically, we can think of the moment as a vector that is perpendicular to both the force $vec{F}$ and the vector $vec{P}$ (right-hand rule), and length equal to the area of the parallelogram with sides $OF$ and $OP$. Thus, if the force is fixed, but you can vary $P$, you can achieve any point on the plane through 0, perpendicular to $vec{F}$.
On the other hand, infinitely many points $P$ can give the same moment. We know $P$ must be on the plane through 0, perpendicular to $vec{M}$. Further, we want the parallelogram with sides $OP$ and $OF$ to have area equal to $||vec{M}||$. So $P$ must be on a line parallel to $vec{F}$, at a distance $frac{||vec{M}||}{||vec{F}||}$ from $O$. There are two of them (on the appropriate plane), and which one comes from the direction of the moment.
answered Jan 10 at 10:03
Todor MarkovTodor Markov
2,176411
$endgroup$
The moment at $O = (0, 0, 0)$ is $vec{M} = vec{P} times vec{F}$. Geometrically, we can think of the moment as a vector that is perpendicular to both the force $vec{F}$ and the vector $vec{P}$ (right-hand rule), and length equal to the area of the parallelogram with sides $OF$ and $OP$. Thus, if the force is fixed, but you can vary $P$, you can achieve any point on the plane through 0, perpendicular to $vec{F}$.
On the other hand, infinitely many points $P$ can give the same moment. We know $P$ must be on the plane through 0, perpendicular to $vec{M}$. Further, we want the parallelogram with sides $OP$ and $OF$ to have area equal to $||vec{M}||$. So $P$ must be on a line parallel to $vec{F}$, at a distance $frac{||vec{M}||}{||vec{F}||}$ from $O$. There are two of them (on the appropriate plane), and which one comes from the direction of the moment.
answered Jan 10 at 10:03
Todor MarkovTodor Markov
2,176411
edited Jan 14 at 12:36
answered Jan 10 at 10:03
Todor MarkovTodor Markov
2,176411
answered Jan 10 at 10:03
Todor MarkovTodor Markov
2,176411
answered Jan 10 at 10:03
Todor MarkovTodor Markov
2,176411
2,176411
$begingroup$
Did you mean to write $vec{M} = vec{P} times vec{F}$?
$endgroup$
– Ev. Kounis
Jan 14 at 12:20
$begingroup$
@Ev.Kounis Yes, indeed
$endgroup$
– Todor Markov
Jan 14 at 12:36
add a comment |
$begingroup$
Did you mean to write $vec{M} = vec{P} times vec{F}$?
$endgroup$
– Ev. Kounis
Jan 14 at 12:20
$begingroup$
@Ev.Kounis Yes, indeed
$endgroup$
– Todor Markov
Jan 14 at 12:36
$begingroup$
Did you mean to write $vec{M} = vec{P} times vec{F}$?
$endgroup$
– Ev. Kounis
Jan 14 at 12:20
$begingroup$
Did you mean to write $vec{M} = vec{P} times vec{F}$?
$endgroup$
– Ev. Kounis
Jan 14 at 12:20
$begingroup$
@Ev.Kounis Yes, indeed
$endgroup$
– Todor Markov
Jan 14 at 12:36
$begingroup$
@Ev.Kounis Yes, indeed
$endgroup$
– Todor Markov
Jan 14 at 12:36
add a comment |
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