Mixing
Geometrical interpretation of systems of linear equations with infinitely many solutions
$begingroup$
The system of linear equations $AX=B$ such as
$A=pmatrix{1 ;;;1 ;;;2 \ 1 ;;;1 ;;;1 \ 2;;;2 ;;;2}, X=pmatrix{x\y\z}, B=pmatrix{3\1\2}$ has the following solutions (infinitely many):
${x=t,y=-1-t,z=2}, t in mathbb R.$
My question is about geometrical understanding/intuition.
I understand that, since the vector $pmatrix{t\-1-t\2}$ is a solution to $AX=B$, we can say that $tpmatrix{1\1\2} + (-1-t)pmatrix{1\1\2} + 2pmatrix{2\1\2}$ which can be further written as $(-1)pmatrix{1\1\2} + 2pmatrix{2\1\2}$ (since $pmatrix{-1\-1\-2} + pmatrix{4\2\4} = pmatrix{3\1\2}$). Therefore, we have a linear combination of 2 vectors of the matrix A.
Please, can you confirm or extend my reasoning:
1) This means that the vector $pmatrix{3\1\2}$ is in the span of $pmatrix{1\1\2}$ and $pmatrix{2\1\2}$.
2)The vector $pmatrix{3\1\2}$ is in the plane formed by $pmatrix{1\1\2}$ and $pmatrix{2\1\2}$; if it were not, then there would not be any solution.
3) 2 3D-vectors can only span a plane in 3D.
4) Now in terms of intersection of lines/hyperplanes, etc. I think we can say that the solution is a line (but why, really (in terms of geometrical interpretation I mean).. it is not the 'intersection of something' here?). Since the line's equation is y=mx+b (or in vector form $vec{OQ} = vec{OP}+lambda vec{d}$, with $vec{d}$ a vector in the direction of the line and $vec{OP}$ a point on this line), is it correct then to say that $vec{OP}=pmatrix{4\2\4}$ and $vec{d}=pmatrix{1\1\2}$ and $lambda=t$ ? (I sometimes have trouble "seeing" the parameters of the line or hyperplane when there are infinitely many solutions; this example is a way of clarifying that. (Indeed, if we set $lambda = -1$, then we get the vector $pmatrix{3\1\2}$ out of the vectorial line equation... but I would like the "explanation" part of how/why?)
Please note that I plotted them in Matlab as
[X,Y]=meshgrid(-10:0.1:10);
surf(X,Y,1-X-Y);
hold on;
surf(X,Y,(3/2)-X/2-Y/2);
hold on;
plot3(4+t,2+t,4+2*t)
However, the line as plotted on the figure is not there where I would have thought... maybe I made a mistake?
linear-algebra euclidean-geometry
edited Jan 28 at 16:50
J. W. Tanner
4,0171320
asked Jan 28 at 15:26
MachupicchuMachupicchu
239
$endgroup$
add a comment |
$begingroup$
The system of linear equations $AX=B$ such as
$A=pmatrix{1 ;;;1 ;;;2 \ 1 ;;;1 ;;;1 \ 2;;;2 ;;;2}, X=pmatrix{x\y\z}, B=pmatrix{3\1\2}$ has the following solutions (infinitely many):
${x=t,y=-1-t,z=2}, t in mathbb R.$
My question is about geometrical understanding/intuition.
I understand that, since the vector $pmatrix{t\-1-t\2}$ is a solution to $AX=B$, we can say that $tpmatrix{1\1\2} + (-1-t)pmatrix{1\1\2} + 2pmatrix{2\1\2}$ which can be further written as $(-1)pmatrix{1\1\2} + 2pmatrix{2\1\2}$ (since $pmatrix{-1\-1\-2} + pmatrix{4\2\4} = pmatrix{3\1\2}$). Therefore, we have a linear combination of 2 vectors of the matrix A.
Please, can you confirm or extend my reasoning:
1) This means that the vector $pmatrix{3\1\2}$ is in the span of $pmatrix{1\1\2}$ and $pmatrix{2\1\2}$.
2)The vector $pmatrix{3\1\2}$ is in the plane formed by $pmatrix{1\1\2}$ and $pmatrix{2\1\2}$; if it were not, then there would not be any solution.
3) 2 3D-vectors can only span a plane in 3D.
4) Now in terms of intersection of lines/hyperplanes, etc. I think we can say that the solution is a line (but why, really (in terms of geometrical interpretation I mean).. it is not the 'intersection of something' here?). Since the line's equation is y=mx+b (or in vector form $vec{OQ} = vec{OP}+lambda vec{d}$, with $vec{d}$ a vector in the direction of the line and $vec{OP}$ a point on this line), is it correct then to say that $vec{OP}=pmatrix{4\2\4}$ and $vec{d}=pmatrix{1\1\2}$ and $lambda=t$ ? (I sometimes have trouble "seeing" the parameters of the line or hyperplane when there are infinitely many solutions; this example is a way of clarifying that. (Indeed, if we set $lambda = -1$, then we get the vector $pmatrix{3\1\2}$ out of the vectorial line equation... but I would like the "explanation" part of how/why?)
Please note that I plotted them in Matlab as
[X,Y]=meshgrid(-10:0.1:10);
surf(X,Y,1-X-Y);
hold on;
surf(X,Y,(3/2)-X/2-Y/2);
hold on;
plot3(4+t,2+t,4+2*t)
However, the line as plotted on the figure is not there where I would have thought... maybe I made a mistake?
linear-algebra euclidean-geometry
edited Jan 28 at 16:50
J. W. Tanner
4,0171320
asked Jan 28 at 15:26
MachupicchuMachupicchu
239
$endgroup$
add a comment |
$begingroup$
The system of linear equations $AX=B$ such as
$A=pmatrix{1 ;;;1 ;;;2 \ 1 ;;;1 ;;;1 \ 2;;;2 ;;;2}, X=pmatrix{x\y\z}, B=pmatrix{3\1\2}$ has the following solutions (infinitely many):
${x=t,y=-1-t,z=2}, t in mathbb R.$
My question is about geometrical understanding/intuition.
I understand that, since the vector $pmatrix{t\-1-t\2}$ is a solution to $AX=B$, we can say that $tpmatrix{1\1\2} + (-1-t)pmatrix{1\1\2} + 2pmatrix{2\1\2}$ which can be further written as $(-1)pmatrix{1\1\2} + 2pmatrix{2\1\2}$ (since $pmatrix{-1\-1\-2} + pmatrix{4\2\4} = pmatrix{3\1\2}$). Therefore, we have a linear combination of 2 vectors of the matrix A.
Please, can you confirm or extend my reasoning:
1) This means that the vector $pmatrix{3\1\2}$ is in the span of $pmatrix{1\1\2}$ and $pmatrix{2\1\2}$.
2)The vector $pmatrix{3\1\2}$ is in the plane formed by $pmatrix{1\1\2}$ and $pmatrix{2\1\2}$; if it were not, then there would not be any solution.
3) 2 3D-vectors can only span a plane in 3D.
4) Now in terms of intersection of lines/hyperplanes, etc. I think we can say that the solution is a line (but why, really (in terms of geometrical interpretation I mean).. it is not the 'intersection of something' here?). Since the line's equation is y=mx+b (or in vector form $vec{OQ} = vec{OP}+lambda vec{d}$, with $vec{d}$ a vector in the direction of the line and $vec{OP}$ a point on this line), is it correct then to say that $vec{OP}=pmatrix{4\2\4}$ and $vec{d}=pmatrix{1\1\2}$ and $lambda=t$ ? (I sometimes have trouble "seeing" the parameters of the line or hyperplane when there are infinitely many solutions; this example is a way of clarifying that. (Indeed, if we set $lambda = -1$, then we get the vector $pmatrix{3\1\2}$ out of the vectorial line equation... but I would like the "explanation" part of how/why?)
Please note that I plotted them in Matlab as
[X,Y]=meshgrid(-10:0.1:10);
surf(X,Y,1-X-Y);
hold on;
surf(X,Y,(3/2)-X/2-Y/2);
hold on;
plot3(4+t,2+t,4+2*t)
However, the line as plotted on the figure is not there where I would have thought... maybe I made a mistake?
linear-algebra euclidean-geometry
edited Jan 28 at 16:50
J. W. Tanner
4,0171320
asked Jan 28 at 15:26
MachupicchuMachupicchu
239
$endgroup$
The system of linear equations $AX=B$ such as
$A=pmatrix{1 ;;;1 ;;;2 \ 1 ;;;1 ;;;1 \ 2;;;2 ;;;2}, X=pmatrix{x\y\z}, B=pmatrix{3\1\2}$ has the following solutions (infinitely many):
${x=t,y=-1-t,z=2}, t in mathbb R.$
My question is about geometrical understanding/intuition.
I understand that, since the vector $pmatrix{t\-1-t\2}$ is a solution to $AX=B$, we can say that $tpmatrix{1\1\2} + (-1-t)pmatrix{1\1\2} + 2pmatrix{2\1\2}$ which can be further written as $(-1)pmatrix{1\1\2} + 2pmatrix{2\1\2}$ (since $pmatrix{-1\-1\-2} + pmatrix{4\2\4} = pmatrix{3\1\2}$). Therefore, we have a linear combination of 2 vectors of the matrix A.
Please, can you confirm or extend my reasoning:
1) This means that the vector $pmatrix{3\1\2}$ is in the span of $pmatrix{1\1\2}$ and $pmatrix{2\1\2}$.
2)The vector $pmatrix{3\1\2}$ is in the plane formed by $pmatrix{1\1\2}$ and $pmatrix{2\1\2}$; if it were not, then there would not be any solution.
3) 2 3D-vectors can only span a plane in 3D.
4) Now in terms of intersection of lines/hyperplanes, etc. I think we can say that the solution is a line (but why, really (in terms of geometrical interpretation I mean).. it is not the 'intersection of something' here?). Since the line's equation is y=mx+b (or in vector form $vec{OQ} = vec{OP}+lambda vec{d}$, with $vec{d}$ a vector in the direction of the line and $vec{OP}$ a point on this line), is it correct then to say that $vec{OP}=pmatrix{4\2\4}$ and $vec{d}=pmatrix{1\1\2}$ and $lambda=t$ ? (I sometimes have trouble "seeing" the parameters of the line or hyperplane when there are infinitely many solutions; this example is a way of clarifying that. (Indeed, if we set $lambda = -1$, then we get the vector $pmatrix{3\1\2}$ out of the vectorial line equation... but I would like the "explanation" part of how/why?)
Please note that I plotted them in Matlab as
[X,Y]=meshgrid(-10:0.1:10);
surf(X,Y,1-X-Y);
hold on;
surf(X,Y,(3/2)-X/2-Y/2);
hold on;
plot3(4+t,2+t,4+2*t)
However, the line as plotted on the figure is not there where I would have thought... maybe I made a mistake?
linear-algebra euclidean-geometry
linear-algebra euclidean-geometry
edited Jan 28 at 16:50
J. W. Tanner
4,0171320
asked Jan 28 at 15:26
MachupicchuMachupicchu
239
edited Jan 28 at 16:50
J. W. Tanner
4,0171320
asked Jan 28 at 15:26
MachupicchuMachupicchu
239
edited Jan 28 at 16:50
J. W. Tanner
4,0171320
edited Jan 28 at 16:50
J. W. Tanner
4,0171320
edited Jan 28 at 16:50
J. W. Tanner
4,0171320
4,0171320
asked Jan 28 at 15:26
MachupicchuMachupicchu
239
asked Jan 28 at 15:26
MachupicchuMachupicchu
239
asked Jan 28 at 15:26
MachupicchuMachupicchu
239
239
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If you want to obtain the solution set as the intersection of planes, you can look at what the system $Ax=b$ means. We have,
$$
begin{cases}
1 x + 1y + 2z = 3 \
1 x + 1 y + 1z = 1 \
2 x + 2 y + 2z = 2
end{cases}
$$
These are all planes in $mathbb{R}^3$. In this case, the bottom two equation gives the same plane, which intersects the top plane as a line. This means there are infinitely many points (all on a line) which lie on all three planes.
Another way to think of the solution set as a line is using the solution you found. Write $u = (0,-1,2)^T$ and $v=(1,-1,0)^T$. Then,
$$
begin{bmatrix}
t \ -1 - t \ 2
end{bmatrix}
=
begin{bmatrix}
0 \ -1 \ 2
end{bmatrix}
+ t
begin{bmatrix}
1 \ -1 \ 0
end{bmatrix}
= u + t v
$$
Note that no matter what other representation we use for the line, the direction part, $v=(1,-1,0)^T$ cannot change.
What is special about this vector? It turns out that $Av=0$ and $Au = b$. Then, no matter what we pick as $t$,
$$
A(u+tv) = Au + tAv = b + tcdot 0 = b
$$
So we see that once we have found a point which gives the solution (in this case $u$), that adding anything in the null space of $A$ to this point will still give a solution. In the case of this matrix, the null space is one dimensional, so the solution set is a line.
If the null space were two dimensional, then it would be a plane. That means once we find a solution (if a solution exists) we could add anything in the null space to it and still have a solution. In this case, the solution set would be a plane.
answered Jan 28 at 16:03
tchtch
833310
$endgroup$
$begingroup$
Thanks for the nice answer, but the part i don't understand is exactly how you get to $begin{bmatrix} 0 \ -1 \ 2 end{bmatrix} + t begin{bmatrix} 1 \ -1 \ 0 end{bmatrix} = u + t v$ ... I do understand that you 'decompose' the solution vector $pmatrix{t\-1-t\ 2} = pmatrix{0\-1\2} + pmatrix{t\-t\ 0} = pmatrix{0 \-1\2} +tpmatrix{1 \ -1 \ 0}$ OK, but how do you know to do it like that ?(what is the logic behind that)
$endgroup$
– Machupicchu
Jan 28 at 16:11
$begingroup$
The equation for a line can always be written as $P + t D$ where $P$ is some offset, and $D$ is a direction. The direction part is the part which scales with $t$, so I pulled out each of the pieces with a $t$ in it. I made an edit to the original post to try to clarify this.
$endgroup$
– tch
Jan 28 at 16:14
$begingroup$
thanks, nice explanations, one more thing... Was i mixing something with the linear combination of the vectors of A's columnspace? It seems to me that these are 2 different approaches: either you think of it as the solution vector which is a linear combination of the columns of A OR as the intersection of planes ... but it's like 2 different ways of seeing it?
$endgroup$
– Machupicchu
Jan 28 at 16:23
$begingroup$
Also thanks for the explanation about the nullspace. It that a general rule or specific to this case? If general, could you develop the intuition behind,?
$endgroup$
– Machupicchu
Jan 28 at 16:59
$begingroup$
Yeah, its two different ways of looking at it. The part about the null space is true, because by the linearity of $A$ the part from the null space will always go to zero.
$endgroup$
– tch
Jan 28 at 18:28
add a comment |
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1 Answer
1
active
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1 Answer
1
active
oldest
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oldest
votes
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oldest
votes
$begingroup$
If you want to obtain the solution set as the intersection of planes, you can look at what the system $Ax=b$ means. We have,
$$
begin{cases}
1 x + 1y + 2z = 3 \
1 x + 1 y + 1z = 1 \
2 x + 2 y + 2z = 2
end{cases}
$$
These are all planes in $mathbb{R}^3$. In this case, the bottom two equation gives the same plane, which intersects the top plane as a line. This means there are infinitely many points (all on a line) which lie on all three planes.
Another way to think of the solution set as a line is using the solution you found. Write $u = (0,-1,2)^T$ and $v=(1,-1,0)^T$. Then,
$$
begin{bmatrix}
t \ -1 - t \ 2
end{bmatrix}
=
begin{bmatrix}
0 \ -1 \ 2
end{bmatrix}
+ t
begin{bmatrix}
1 \ -1 \ 0
end{bmatrix}
= u + t v
$$
Note that no matter what other representation we use for the line, the direction part, $v=(1,-1,0)^T$ cannot change.
What is special about this vector? It turns out that $Av=0$ and $Au = b$. Then, no matter what we pick as $t$,
$$
A(u+tv) = Au + tAv = b + tcdot 0 = b
$$
So we see that once we have found a point which gives the solution (in this case $u$), that adding anything in the null space of $A$ to this point will still give a solution. In the case of this matrix, the null space is one dimensional, so the solution set is a line.
If the null space were two dimensional, then it would be a plane. That means once we find a solution (if a solution exists) we could add anything in the null space to it and still have a solution. In this case, the solution set would be a plane.
answered Jan 28 at 16:03
tchtch
833310
$endgroup$
$begingroup$
Thanks for the nice answer, but the part i don't understand is exactly how you get to $begin{bmatrix} 0 \ -1 \ 2 end{bmatrix} + t begin{bmatrix} 1 \ -1 \ 0 end{bmatrix} = u + t v$ ... I do understand that you 'decompose' the solution vector $pmatrix{t\-1-t\ 2} = pmatrix{0\-1\2} + pmatrix{t\-t\ 0} = pmatrix{0 \-1\2} +tpmatrix{1 \ -1 \ 0}$ OK, but how do you know to do it like that ?(what is the logic behind that)
$endgroup$
– Machupicchu
Jan 28 at 16:11
$begingroup$
The equation for a line can always be written as $P + t D$ where $P$ is some offset, and $D$ is a direction. The direction part is the part which scales with $t$, so I pulled out each of the pieces with a $t$ in it. I made an edit to the original post to try to clarify this.
$endgroup$
– tch
Jan 28 at 16:14
$begingroup$
thanks, nice explanations, one more thing... Was i mixing something with the linear combination of the vectors of A's columnspace? It seems to me that these are 2 different approaches: either you think of it as the solution vector which is a linear combination of the columns of A OR as the intersection of planes ... but it's like 2 different ways of seeing it?
$endgroup$
– Machupicchu
Jan 28 at 16:23
$begingroup$
Also thanks for the explanation about the nullspace. It that a general rule or specific to this case? If general, could you develop the intuition behind,?
$endgroup$
– Machupicchu
Jan 28 at 16:59
$begingroup$
Yeah, its two different ways of looking at it. The part about the null space is true, because by the linearity of $A$ the part from the null space will always go to zero.
$endgroup$
– tch
Jan 28 at 18:28
add a comment |
$begingroup$
If you want to obtain the solution set as the intersection of planes, you can look at what the system $Ax=b$ means. We have,
$$
begin{cases}
1 x + 1y + 2z = 3 \
1 x + 1 y + 1z = 1 \
2 x + 2 y + 2z = 2
end{cases}
$$
These are all planes in $mathbb{R}^3$. In this case, the bottom two equation gives the same plane, which intersects the top plane as a line. This means there are infinitely many points (all on a line) which lie on all three planes.
Another way to think of the solution set as a line is using the solution you found. Write $u = (0,-1,2)^T$ and $v=(1,-1,0)^T$. Then,
$$
begin{bmatrix}
t \ -1 - t \ 2
end{bmatrix}
=
begin{bmatrix}
0 \ -1 \ 2
end{bmatrix}
+ t
begin{bmatrix}
1 \ -1 \ 0
end{bmatrix}
= u + t v
$$
Note that no matter what other representation we use for the line, the direction part, $v=(1,-1,0)^T$ cannot change.
What is special about this vector? It turns out that $Av=0$ and $Au = b$. Then, no matter what we pick as $t$,
$$
A(u+tv) = Au + tAv = b + tcdot 0 = b
$$
So we see that once we have found a point which gives the solution (in this case $u$), that adding anything in the null space of $A$ to this point will still give a solution. In the case of this matrix, the null space is one dimensional, so the solution set is a line.
If the null space were two dimensional, then it would be a plane. That means once we find a solution (if a solution exists) we could add anything in the null space to it and still have a solution. In this case, the solution set would be a plane.
answered Jan 28 at 16:03
tchtch
833310
$endgroup$
$begingroup$
Thanks for the nice answer, but the part i don't understand is exactly how you get to $begin{bmatrix} 0 \ -1 \ 2 end{bmatrix} + t begin{bmatrix} 1 \ -1 \ 0 end{bmatrix} = u + t v$ ... I do understand that you 'decompose' the solution vector $pmatrix{t\-1-t\ 2} = pmatrix{0\-1\2} + pmatrix{t\-t\ 0} = pmatrix{0 \-1\2} +tpmatrix{1 \ -1 \ 0}$ OK, but how do you know to do it like that ?(what is the logic behind that)
$endgroup$
– Machupicchu
Jan 28 at 16:11
$begingroup$
The equation for a line can always be written as $P + t D$ where $P$ is some offset, and $D$ is a direction. The direction part is the part which scales with $t$, so I pulled out each of the pieces with a $t$ in it. I made an edit to the original post to try to clarify this.
$endgroup$
– tch
Jan 28 at 16:14
$begingroup$
thanks, nice explanations, one more thing... Was i mixing something with the linear combination of the vectors of A's columnspace? It seems to me that these are 2 different approaches: either you think of it as the solution vector which is a linear combination of the columns of A OR as the intersection of planes ... but it's like 2 different ways of seeing it?
$endgroup$
– Machupicchu
Jan 28 at 16:23
$begingroup$
Also thanks for the explanation about the nullspace. It that a general rule or specific to this case? If general, could you develop the intuition behind,?
$endgroup$
– Machupicchu
Jan 28 at 16:59
$begingroup$
Yeah, its two different ways of looking at it. The part about the null space is true, because by the linearity of $A$ the part from the null space will always go to zero.
$endgroup$
– tch
Jan 28 at 18:28
add a comment |
$begingroup$
If you want to obtain the solution set as the intersection of planes, you can look at what the system $Ax=b$ means. We have,
$$
begin{cases}
1 x + 1y + 2z = 3 \
1 x + 1 y + 1z = 1 \
2 x + 2 y + 2z = 2
end{cases}
$$
These are all planes in $mathbb{R}^3$. In this case, the bottom two equation gives the same plane, which intersects the top plane as a line. This means there are infinitely many points (all on a line) which lie on all three planes.
Another way to think of the solution set as a line is using the solution you found. Write $u = (0,-1,2)^T$ and $v=(1,-1,0)^T$. Then,
$$
begin{bmatrix}
t \ -1 - t \ 2
end{bmatrix}
=
begin{bmatrix}
0 \ -1 \ 2
end{bmatrix}
+ t
begin{bmatrix}
1 \ -1 \ 0
end{bmatrix}
= u + t v
$$
Note that no matter what other representation we use for the line, the direction part, $v=(1,-1,0)^T$ cannot change.
What is special about this vector? It turns out that $Av=0$ and $Au = b$. Then, no matter what we pick as $t$,
$$
A(u+tv) = Au + tAv = b + tcdot 0 = b
$$
So we see that once we have found a point which gives the solution (in this case $u$), that adding anything in the null space of $A$ to this point will still give a solution. In the case of this matrix, the null space is one dimensional, so the solution set is a line.
If the null space were two dimensional, then it would be a plane. That means once we find a solution (if a solution exists) we could add anything in the null space to it and still have a solution. In this case, the solution set would be a plane.
answered Jan 28 at 16:03
tchtch
833310
$endgroup$
If you want to obtain the solution set as the intersection of planes, you can look at what the system $Ax=b$ means. We have,
$$
begin{cases}
1 x + 1y + 2z = 3 \
1 x + 1 y + 1z = 1 \
2 x + 2 y + 2z = 2
end{cases}
$$
These are all planes in $mathbb{R}^3$. In this case, the bottom two equation gives the same plane, which intersects the top plane as a line. This means there are infinitely many points (all on a line) which lie on all three planes.
Another way to think of the solution set as a line is using the solution you found. Write $u = (0,-1,2)^T$ and $v=(1,-1,0)^T$. Then,
$$
begin{bmatrix}
t \ -1 - t \ 2
end{bmatrix}
=
begin{bmatrix}
0 \ -1 \ 2
end{bmatrix}
+ t
begin{bmatrix}
1 \ -1 \ 0
end{bmatrix}
= u + t v
$$
Note that no matter what other representation we use for the line, the direction part, $v=(1,-1,0)^T$ cannot change.
What is special about this vector? It turns out that $Av=0$ and $Au = b$. Then, no matter what we pick as $t$,
$$
A(u+tv) = Au + tAv = b + tcdot 0 = b
$$
So we see that once we have found a point which gives the solution (in this case $u$), that adding anything in the null space of $A$ to this point will still give a solution. In the case of this matrix, the null space is one dimensional, so the solution set is a line.
If the null space were two dimensional, then it would be a plane. That means once we find a solution (if a solution exists) we could add anything in the null space to it and still have a solution. In this case, the solution set would be a plane.
answered Jan 28 at 16:03
tchtch
833310
edited Jan 28 at 16:11
answered Jan 28 at 16:03
tchtch
833310
answered Jan 28 at 16:03
tchtch
833310
answered Jan 28 at 16:03
tchtch
833310
833310
$begingroup$
Thanks for the nice answer, but the part i don't understand is exactly how you get to $begin{bmatrix} 0 \ -1 \ 2 end{bmatrix} + t begin{bmatrix} 1 \ -1 \ 0 end{bmatrix} = u + t v$ ... I do understand that you 'decompose' the solution vector $pmatrix{t\-1-t\ 2} = pmatrix{0\-1\2} + pmatrix{t\-t\ 0} = pmatrix{0 \-1\2} +tpmatrix{1 \ -1 \ 0}$ OK, but how do you know to do it like that ?(what is the logic behind that)
$endgroup$
– Machupicchu
Jan 28 at 16:11
$begingroup$
The equation for a line can always be written as $P + t D$ where $P$ is some offset, and $D$ is a direction. The direction part is the part which scales with $t$, so I pulled out each of the pieces with a $t$ in it. I made an edit to the original post to try to clarify this.
$endgroup$
– tch
Jan 28 at 16:14
$begingroup$
thanks, nice explanations, one more thing... Was i mixing something with the linear combination of the vectors of A's columnspace? It seems to me that these are 2 different approaches: either you think of it as the solution vector which is a linear combination of the columns of A OR as the intersection of planes ... but it's like 2 different ways of seeing it?
$endgroup$
– Machupicchu
Jan 28 at 16:23
$begingroup$
Also thanks for the explanation about the nullspace. It that a general rule or specific to this case? If general, could you develop the intuition behind,?
$endgroup$
– Machupicchu
Jan 28 at 16:59
$begingroup$
Yeah, its two different ways of looking at it. The part about the null space is true, because by the linearity of $A$ the part from the null space will always go to zero.
$endgroup$
– tch
Jan 28 at 18:28
add a comment |
$begingroup$
Thanks for the nice answer, but the part i don't understand is exactly how you get to $begin{bmatrix} 0 \ -1 \ 2 end{bmatrix} + t begin{bmatrix} 1 \ -1 \ 0 end{bmatrix} = u + t v$ ... I do understand that you 'decompose' the solution vector $pmatrix{t\-1-t\ 2} = pmatrix{0\-1\2} + pmatrix{t\-t\ 0} = pmatrix{0 \-1\2} +tpmatrix{1 \ -1 \ 0}$ OK, but how do you know to do it like that ?(what is the logic behind that)
$endgroup$
– Machupicchu
Jan 28 at 16:11
$begingroup$
The equation for a line can always be written as $P + t D$ where $P$ is some offset, and $D$ is a direction. The direction part is the part which scales with $t$, so I pulled out each of the pieces with a $t$ in it. I made an edit to the original post to try to clarify this.
$endgroup$
– tch
Jan 28 at 16:14
$begingroup$
thanks, nice explanations, one more thing... Was i mixing something with the linear combination of the vectors of A's columnspace? It seems to me that these are 2 different approaches: either you think of it as the solution vector which is a linear combination of the columns of A OR as the intersection of planes ... but it's like 2 different ways of seeing it?
$endgroup$
– Machupicchu
Jan 28 at 16:23
$begingroup$
Also thanks for the explanation about the nullspace. It that a general rule or specific to this case? If general, could you develop the intuition behind,?
$endgroup$
– Machupicchu
Jan 28 at 16:59
$begingroup$
Yeah, its two different ways of looking at it. The part about the null space is true, because by the linearity of $A$ the part from the null space will always go to zero.
$endgroup$
– tch
Jan 28 at 18:28
$begingroup$
Thanks for the nice answer, but the part i don't understand is exactly how you get to $begin{bmatrix} 0 \ -1 \ 2 end{bmatrix} + t begin{bmatrix} 1 \ -1 \ 0 end{bmatrix} = u + t v$ ... I do understand that you 'decompose' the solution vector $pmatrix{t\-1-t\ 2} = pmatrix{0\-1\2} + pmatrix{t\-t\ 0} = pmatrix{0 \-1\2} +tpmatrix{1 \ -1 \ 0}$ OK, but how do you know to do it like that ?(what is the logic behind that)
$endgroup$
– Machupicchu
Jan 28 at 16:11
$begingroup$
Thanks for the nice answer, but the part i don't understand is exactly how you get to $begin{bmatrix} 0 \ -1 \ 2 end{bmatrix} + t begin{bmatrix} 1 \ -1 \ 0 end{bmatrix} = u + t v$ ... I do understand that you 'decompose' the solution vector $pmatrix{t\-1-t\ 2} = pmatrix{0\-1\2} + pmatrix{t\-t\ 0} = pmatrix{0 \-1\2} +tpmatrix{1 \ -1 \ 0}$ OK, but how do you know to do it like that ?(what is the logic behind that)
$endgroup$
– Machupicchu
Jan 28 at 16:11
$begingroup$
The equation for a line can always be written as $P + t D$ where $P$ is some offset, and $D$ is a direction. The direction part is the part which scales with $t$, so I pulled out each of the pieces with a $t$ in it. I made an edit to the original post to try to clarify this.
$endgroup$
– tch
Jan 28 at 16:14
$begingroup$
The equation for a line can always be written as $P + t D$ where $P$ is some offset, and $D$ is a direction. The direction part is the part which scales with $t$, so I pulled out each of the pieces with a $t$ in it. I made an edit to the original post to try to clarify this.
$endgroup$
– tch
Jan 28 at 16:14
$begingroup$
thanks, nice explanations, one more thing... Was i mixing something with the linear combination of the vectors of A's columnspace? It seems to me that these are 2 different approaches: either you think of it as the solution vector which is a linear combination of the columns of A OR as the intersection of planes ... but it's like 2 different ways of seeing it?
$endgroup$
– Machupicchu
Jan 28 at 16:23
$begingroup$
thanks, nice explanations, one more thing... Was i mixing something with the linear combination of the vectors of A's columnspace? It seems to me that these are 2 different approaches: either you think of it as the solution vector which is a linear combination of the columns of A OR as the intersection of planes ... but it's like 2 different ways of seeing it?
$endgroup$
– Machupicchu
Jan 28 at 16:23
$begingroup$
Also thanks for the explanation about the nullspace. It that a general rule or specific to this case? If general, could you develop the intuition behind,?
$endgroup$
– Machupicchu
Jan 28 at 16:59
$begingroup$
Also thanks for the explanation about the nullspace. It that a general rule or specific to this case? If general, could you develop the intuition behind,?
$endgroup$
– Machupicchu
Jan 28 at 16:59
$begingroup$
Yeah, its two different ways of looking at it. The part about the null space is true, because by the linearity of $A$ the part from the null space will always go to zero.
$endgroup$
– tch
Jan 28 at 18:28
$begingroup$
Yeah, its two different ways of looking at it. The part about the null space is true, because by the linearity of $A$ the part from the null space will always go to zero.
$endgroup$
– tch
Jan 28 at 18:28
add a comment |
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