Mixing
Get from a transformation matrix to the resultant span of the solution set
$begingroup$
Guess that's a very basic question, but anyway:
I have the following transformation matrix:
$$begin{bmatrix}1 & 0 & 0 & -tfrac{1}{3}\ 0 & -6 & -3 & -1end{bmatrix}$$
And I know that the span of the solution set is $Sp{(2, -1, 0, 6), (0, -1, 2, 0)}$. But how do I get it?
I always thought that solving such matrix requires using parameters, so the solution would always be a general one (i.e., parameterized). But I guess that's not the case.
(If there's some missing part here, please let me know)
linear-algebra matrices linear-transformations
asked Jan 9 at 1:45
HeyJudeHeyJude
1687
$endgroup$
|
show 4 more comments
$begingroup$
Guess that's a very basic question, but anyway:
I have the following transformation matrix:
$$begin{bmatrix}1 & 0 & 0 & -tfrac{1}{3}\ 0 & -6 & -3 & -1end{bmatrix}$$
And I know that the span of the solution set is $Sp{(2, -1, 0, 6), (0, -1, 2, 0)}$. But how do I get it?
I always thought that solving such matrix requires using parameters, so the solution would always be a general one (i.e., parameterized). But I guess that's not the case.
(If there's some missing part here, please let me know)
linear-algebra matrices linear-transformations
asked Jan 9 at 1:45
HeyJudeHeyJude
1687
$endgroup$
$begingroup$
Can you give a definition of a span of a matrix? I've not heard this term used before. I know what the span of a set of vectors is. And I know what the image (or range) of a matrix is. But I'm not sure what the span of a matrix is.
$endgroup$
– NicNic8
Jan 9 at 1:51
$begingroup$
Why, of course not; I'll edit
$endgroup$
– HeyJude
Jan 9 at 1:52
$begingroup$
@NicNic8, I think it's better now
$endgroup$
– HeyJude
Jan 9 at 1:58
$begingroup$
What is a solution set of a matrix? I think part of your issue may be that you're not using correct terminology. If we have an equation Ax=b, then we might ask "What are the values of x that satisfy this equation?" Another way to ask this is "What are the solutions to this equation?" But I've never heard of a solution set for a matrix. The set that you're calling the solution set looks to be the Null space of the matrix. Is that what you're going after? That's the special set of solutions to Ax=0, where A is the matrix of interest.
$endgroup$
– NicNic8
Jan 9 at 2:00
1
$begingroup$
The span of a pair of vectors is the set of all of their linear combinations, in this case $lambda(2,-1,0,6)+mu(0,-1,2,0)$, et voilà, there’s your parameterization. The two ways of describing the set are equivalent.
$endgroup$
– amd
Jan 9 at 2:05
|
show 4 more comments
$begingroup$
Guess that's a very basic question, but anyway:
I have the following transformation matrix:
$$begin{bmatrix}1 & 0 & 0 & -tfrac{1}{3}\ 0 & -6 & -3 & -1end{bmatrix}$$
And I know that the span of the solution set is $Sp{(2, -1, 0, 6), (0, -1, 2, 0)}$. But how do I get it?
I always thought that solving such matrix requires using parameters, so the solution would always be a general one (i.e., parameterized). But I guess that's not the case.
(If there's some missing part here, please let me know)
linear-algebra matrices linear-transformations
asked Jan 9 at 1:45
HeyJudeHeyJude
1687
$endgroup$
Guess that's a very basic question, but anyway:
I have the following transformation matrix:
$$begin{bmatrix}1 & 0 & 0 & -tfrac{1}{3}\ 0 & -6 & -3 & -1end{bmatrix}$$
And I know that the span of the solution set is $Sp{(2, -1, 0, 6), (0, -1, 2, 0)}$. But how do I get it?
I always thought that solving such matrix requires using parameters, so the solution would always be a general one (i.e., parameterized). But I guess that's not the case.
(If there's some missing part here, please let me know)
linear-algebra matrices linear-transformations
linear-algebra matrices linear-transformations
asked Jan 9 at 1:45
HeyJudeHeyJude
1687
asked Jan 9 at 1:45
HeyJudeHeyJude
1687
edited Jan 9 at 1:58
HeyJude
asked Jan 9 at 1:45
HeyJudeHeyJude
1687
asked Jan 9 at 1:45
HeyJudeHeyJude
1687
asked Jan 9 at 1:45
HeyJudeHeyJude
1687
1687
$begingroup$
Can you give a definition of a span of a matrix? I've not heard this term used before. I know what the span of a set of vectors is. And I know what the image (or range) of a matrix is. But I'm not sure what the span of a matrix is.
$endgroup$
– NicNic8
Jan 9 at 1:51
$begingroup$
Why, of course not; I'll edit
$endgroup$
– HeyJude
Jan 9 at 1:52
$begingroup$
@NicNic8, I think it's better now
$endgroup$
– HeyJude
Jan 9 at 1:58
$begingroup$
What is a solution set of a matrix? I think part of your issue may be that you're not using correct terminology. If we have an equation Ax=b, then we might ask "What are the values of x that satisfy this equation?" Another way to ask this is "What are the solutions to this equation?" But I've never heard of a solution set for a matrix. The set that you're calling the solution set looks to be the Null space of the matrix. Is that what you're going after? That's the special set of solutions to Ax=0, where A is the matrix of interest.
$endgroup$
– NicNic8
Jan 9 at 2:00
1
$begingroup$
The span of a pair of vectors is the set of all of their linear combinations, in this case $lambda(2,-1,0,6)+mu(0,-1,2,0)$, et voilà, there’s your parameterization. The two ways of describing the set are equivalent.
$endgroup$
– amd
Jan 9 at 2:05
|
show 4 more comments
$begingroup$
Can you give a definition of a span of a matrix? I've not heard this term used before. I know what the span of a set of vectors is. And I know what the image (or range) of a matrix is. But I'm not sure what the span of a matrix is.
$endgroup$
– NicNic8
Jan 9 at 1:51
$begingroup$
Why, of course not; I'll edit
$endgroup$
– HeyJude
Jan 9 at 1:52
$begingroup$
@NicNic8, I think it's better now
$endgroup$
– HeyJude
Jan 9 at 1:58
$begingroup$
What is a solution set of a matrix? I think part of your issue may be that you're not using correct terminology. If we have an equation Ax=b, then we might ask "What are the values of x that satisfy this equation?" Another way to ask this is "What are the solutions to this equation?" But I've never heard of a solution set for a matrix. The set that you're calling the solution set looks to be the Null space of the matrix. Is that what you're going after? That's the special set of solutions to Ax=0, where A is the matrix of interest.
$endgroup$
– NicNic8
Jan 9 at 2:00
1
$begingroup$
The span of a pair of vectors is the set of all of their linear combinations, in this case $lambda(2,-1,0,6)+mu(0,-1,2,0)$, et voilà, there’s your parameterization. The two ways of describing the set are equivalent.
$endgroup$
– amd
Jan 9 at 2:05
$begingroup$
Can you give a definition of a span of a matrix? I've not heard this term used before. I know what the span of a set of vectors is. And I know what the image (or range) of a matrix is. But I'm not sure what the span of a matrix is.
$endgroup$
– NicNic8
Jan 9 at 1:51
$begingroup$
Can you give a definition of a span of a matrix? I've not heard this term used before. I know what the span of a set of vectors is. And I know what the image (or range) of a matrix is. But I'm not sure what the span of a matrix is.
$endgroup$
– NicNic8
Jan 9 at 1:51
$begingroup$
Why, of course not; I'll edit
$endgroup$
– HeyJude
Jan 9 at 1:52
$begingroup$
Why, of course not; I'll edit
$endgroup$
– HeyJude
Jan 9 at 1:52
$begingroup$
@NicNic8, I think it's better now
$endgroup$
– HeyJude
Jan 9 at 1:58
$begingroup$
@NicNic8, I think it's better now
$endgroup$
– HeyJude
Jan 9 at 1:58
$begingroup$
What is a solution set of a matrix? I think part of your issue may be that you're not using correct terminology. If we have an equation Ax=b, then we might ask "What are the values of x that satisfy this equation?" Another way to ask this is "What are the solutions to this equation?" But I've never heard of a solution set for a matrix. The set that you're calling the solution set looks to be the Null space of the matrix. Is that what you're going after? That's the special set of solutions to Ax=0, where A is the matrix of interest.
$endgroup$
– NicNic8
Jan 9 at 2:00
$begingroup$
What is a solution set of a matrix? I think part of your issue may be that you're not using correct terminology. If we have an equation Ax=b, then we might ask "What are the values of x that satisfy this equation?" Another way to ask this is "What are the solutions to this equation?" But I've never heard of a solution set for a matrix. The set that you're calling the solution set looks to be the Null space of the matrix. Is that what you're going after? That's the special set of solutions to Ax=0, where A is the matrix of interest.
$endgroup$
– NicNic8
Jan 9 at 2:00
1
1
$begingroup$
The span of a pair of vectors is the set of all of their linear combinations, in this case $lambda(2,-1,0,6)+mu(0,-1,2,0)$, et voilà, there’s your parameterization. The two ways of describing the set are equivalent.
$endgroup$
– amd
Jan 9 at 2:05
$begingroup$
The span of a pair of vectors is the set of all of their linear combinations, in this case $lambda(2,-1,0,6)+mu(0,-1,2,0)$, et voilà, there’s your parameterization. The two ways of describing the set are equivalent.
$endgroup$
– amd
Jan 9 at 2:05
|
show 4 more comments
1 Answer
1
active
oldest
votes
$begingroup$
To find the null space of a matrix, one wants to solve the equation $Ax=0$, where $A$ is the matrix of interest. Let's do this for your case.
We want to find all the $x=(x_1,x_2,x_3,x_4)$ such that
$$underbrace{begin{bmatrix}1 & 0 & 0 & -tfrac{1}{3}\ 0 & -6 & -3 & -1end{bmatrix}}_{A} begin{bmatrix} x_1 \ x_2 \ x_3 \ x_4 end{bmatrix} = begin{bmatrix}0\ 0end{bmatrix}.$$
Therefore we know
begin{aligned}
x_1 - (1/3) x_4 &= 0 \
-6 x_2 - 3 x_3 - x_4 &= 0
end{aligned}
We have two equations, but we have four variables. Therefore, if the equations are not redundant, we'll end up with a two dimensional vector space as the null set. Let's find this vector space.
Let $x_4=s$ Then we know that $x_1=(1/3)s$. Let $x_3=t$ Then we know that
$$-6x_2 -3t -s=0.$$
Equivalently, this shows us that $x_2 = t/2 - s/6$. This shows us that any vector in the null space of A is of the form
$$begin{bmatrix}
1/3 s \ t/2 - s/6 \ t \ s
end{bmatrix} =
sbegin{bmatrix}1/3 \ -1/6 \ 0 \ 1end{bmatrix} + tbegin{bmatrix}0 \ 1/2 \ 1 \ 0end{bmatrix},$$
where $s$ and $t$ are real numbers. Another way to state that is
$$text{null}(A) = text{span}left(begin{bmatrix}1/3 \ -1/6 \ 0 \ 1end{bmatrix},begin{bmatrix}0\ 1/2 \ 1 \ 0end{bmatrix}right)= text{span}left(begin{bmatrix}2 \ -1 \ 0 \ 6end{bmatrix},
begin{bmatrix}0\ 1 \ 2 \ 0end{bmatrix}right).$$
answered Jan 9 at 2:23
NicNic8NicNic8
4,33531023
$endgroup$
1
$begingroup$
Thank you very much, learned some new things from this discussion.
$endgroup$
– HeyJude
Jan 9 at 2:31
$begingroup$
@HeyJude My pleasure!
$endgroup$
– NicNic8
Jan 9 at 2:33
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
To find the null space of a matrix, one wants to solve the equation $Ax=0$, where $A$ is the matrix of interest. Let's do this for your case.
We want to find all the $x=(x_1,x_2,x_3,x_4)$ such that
$$underbrace{begin{bmatrix}1 & 0 & 0 & -tfrac{1}{3}\ 0 & -6 & -3 & -1end{bmatrix}}_{A} begin{bmatrix} x_1 \ x_2 \ x_3 \ x_4 end{bmatrix} = begin{bmatrix}0\ 0end{bmatrix}.$$
Therefore we know
begin{aligned}
x_1 - (1/3) x_4 &= 0 \
-6 x_2 - 3 x_3 - x_4 &= 0
end{aligned}
We have two equations, but we have four variables. Therefore, if the equations are not redundant, we'll end up with a two dimensional vector space as the null set. Let's find this vector space.
Let $x_4=s$ Then we know that $x_1=(1/3)s$. Let $x_3=t$ Then we know that
$$-6x_2 -3t -s=0.$$
Equivalently, this shows us that $x_2 = t/2 - s/6$. This shows us that any vector in the null space of A is of the form
$$begin{bmatrix}
1/3 s \ t/2 - s/6 \ t \ s
end{bmatrix} =
sbegin{bmatrix}1/3 \ -1/6 \ 0 \ 1end{bmatrix} + tbegin{bmatrix}0 \ 1/2 \ 1 \ 0end{bmatrix},$$
where $s$ and $t$ are real numbers. Another way to state that is
$$text{null}(A) = text{span}left(begin{bmatrix}1/3 \ -1/6 \ 0 \ 1end{bmatrix},begin{bmatrix}0\ 1/2 \ 1 \ 0end{bmatrix}right)= text{span}left(begin{bmatrix}2 \ -1 \ 0 \ 6end{bmatrix},
begin{bmatrix}0\ 1 \ 2 \ 0end{bmatrix}right).$$
answered Jan 9 at 2:23
NicNic8NicNic8
4,33531023
$endgroup$
1
$begingroup$
Thank you very much, learned some new things from this discussion.
$endgroup$
– HeyJude
Jan 9 at 2:31
$begingroup$
@HeyJude My pleasure!
$endgroup$
– NicNic8
Jan 9 at 2:33
add a comment |
$begingroup$
To find the null space of a matrix, one wants to solve the equation $Ax=0$, where $A$ is the matrix of interest. Let's do this for your case.
We want to find all the $x=(x_1,x_2,x_3,x_4)$ such that
$$underbrace{begin{bmatrix}1 & 0 & 0 & -tfrac{1}{3}\ 0 & -6 & -3 & -1end{bmatrix}}_{A} begin{bmatrix} x_1 \ x_2 \ x_3 \ x_4 end{bmatrix} = begin{bmatrix}0\ 0end{bmatrix}.$$
Therefore we know
begin{aligned}
x_1 - (1/3) x_4 &= 0 \
-6 x_2 - 3 x_3 - x_4 &= 0
end{aligned}
We have two equations, but we have four variables. Therefore, if the equations are not redundant, we'll end up with a two dimensional vector space as the null set. Let's find this vector space.
Let $x_4=s$ Then we know that $x_1=(1/3)s$. Let $x_3=t$ Then we know that
$$-6x_2 -3t -s=0.$$
Equivalently, this shows us that $x_2 = t/2 - s/6$. This shows us that any vector in the null space of A is of the form
$$begin{bmatrix}
1/3 s \ t/2 - s/6 \ t \ s
end{bmatrix} =
sbegin{bmatrix}1/3 \ -1/6 \ 0 \ 1end{bmatrix} + tbegin{bmatrix}0 \ 1/2 \ 1 \ 0end{bmatrix},$$
where $s$ and $t$ are real numbers. Another way to state that is
$$text{null}(A) = text{span}left(begin{bmatrix}1/3 \ -1/6 \ 0 \ 1end{bmatrix},begin{bmatrix}0\ 1/2 \ 1 \ 0end{bmatrix}right)= text{span}left(begin{bmatrix}2 \ -1 \ 0 \ 6end{bmatrix},
begin{bmatrix}0\ 1 \ 2 \ 0end{bmatrix}right).$$
answered Jan 9 at 2:23
NicNic8NicNic8
4,33531023
$endgroup$
1
$begingroup$
Thank you very much, learned some new things from this discussion.
$endgroup$
– HeyJude
Jan 9 at 2:31
$begingroup$
@HeyJude My pleasure!
$endgroup$
– NicNic8
Jan 9 at 2:33
add a comment |
$begingroup$
To find the null space of a matrix, one wants to solve the equation $Ax=0$, where $A$ is the matrix of interest. Let's do this for your case.
We want to find all the $x=(x_1,x_2,x_3,x_4)$ such that
$$underbrace{begin{bmatrix}1 & 0 & 0 & -tfrac{1}{3}\ 0 & -6 & -3 & -1end{bmatrix}}_{A} begin{bmatrix} x_1 \ x_2 \ x_3 \ x_4 end{bmatrix} = begin{bmatrix}0\ 0end{bmatrix}.$$
Therefore we know
begin{aligned}
x_1 - (1/3) x_4 &= 0 \
-6 x_2 - 3 x_3 - x_4 &= 0
end{aligned}
We have two equations, but we have four variables. Therefore, if the equations are not redundant, we'll end up with a two dimensional vector space as the null set. Let's find this vector space.
Let $x_4=s$ Then we know that $x_1=(1/3)s$. Let $x_3=t$ Then we know that
$$-6x_2 -3t -s=0.$$
Equivalently, this shows us that $x_2 = t/2 - s/6$. This shows us that any vector in the null space of A is of the form
$$begin{bmatrix}
1/3 s \ t/2 - s/6 \ t \ s
end{bmatrix} =
sbegin{bmatrix}1/3 \ -1/6 \ 0 \ 1end{bmatrix} + tbegin{bmatrix}0 \ 1/2 \ 1 \ 0end{bmatrix},$$
where $s$ and $t$ are real numbers. Another way to state that is
$$text{null}(A) = text{span}left(begin{bmatrix}1/3 \ -1/6 \ 0 \ 1end{bmatrix},begin{bmatrix}0\ 1/2 \ 1 \ 0end{bmatrix}right)= text{span}left(begin{bmatrix}2 \ -1 \ 0 \ 6end{bmatrix},
begin{bmatrix}0\ 1 \ 2 \ 0end{bmatrix}right).$$
answered Jan 9 at 2:23
NicNic8NicNic8
4,33531023
$endgroup$
To find the null space of a matrix, one wants to solve the equation $Ax=0$, where $A$ is the matrix of interest. Let's do this for your case.
We want to find all the $x=(x_1,x_2,x_3,x_4)$ such that
$$underbrace{begin{bmatrix}1 & 0 & 0 & -tfrac{1}{3}\ 0 & -6 & -3 & -1end{bmatrix}}_{A} begin{bmatrix} x_1 \ x_2 \ x_3 \ x_4 end{bmatrix} = begin{bmatrix}0\ 0end{bmatrix}.$$
Therefore we know
begin{aligned}
x_1 - (1/3) x_4 &= 0 \
-6 x_2 - 3 x_3 - x_4 &= 0
end{aligned}
We have two equations, but we have four variables. Therefore, if the equations are not redundant, we'll end up with a two dimensional vector space as the null set. Let's find this vector space.
Let $x_4=s$ Then we know that $x_1=(1/3)s$. Let $x_3=t$ Then we know that
$$-6x_2 -3t -s=0.$$
Equivalently, this shows us that $x_2 = t/2 - s/6$. This shows us that any vector in the null space of A is of the form
$$begin{bmatrix}
1/3 s \ t/2 - s/6 \ t \ s
end{bmatrix} =
sbegin{bmatrix}1/3 \ -1/6 \ 0 \ 1end{bmatrix} + tbegin{bmatrix}0 \ 1/2 \ 1 \ 0end{bmatrix},$$
where $s$ and $t$ are real numbers. Another way to state that is
$$text{null}(A) = text{span}left(begin{bmatrix}1/3 \ -1/6 \ 0 \ 1end{bmatrix},begin{bmatrix}0\ 1/2 \ 1 \ 0end{bmatrix}right)= text{span}left(begin{bmatrix}2 \ -1 \ 0 \ 6end{bmatrix},
begin{bmatrix}0\ 1 \ 2 \ 0end{bmatrix}right).$$
answered Jan 9 at 2:23
NicNic8NicNic8
4,33531023
edited Jan 9 at 2:30
answered Jan 9 at 2:23
NicNic8NicNic8
4,33531023
answered Jan 9 at 2:23
NicNic8NicNic8
4,33531023
answered Jan 9 at 2:23
NicNic8NicNic8
4,33531023
4,33531023
1
$begingroup$
Thank you very much, learned some new things from this discussion.
$endgroup$
– HeyJude
Jan 9 at 2:31
$begingroup$
@HeyJude My pleasure!
$endgroup$
– NicNic8
Jan 9 at 2:33
add a comment |
1
$begingroup$
Thank you very much, learned some new things from this discussion.
$endgroup$
– HeyJude
Jan 9 at 2:31
$begingroup$
@HeyJude My pleasure!
$endgroup$
– NicNic8
Jan 9 at 2:33
1
1
$begingroup$
Thank you very much, learned some new things from this discussion.
$endgroup$
– HeyJude
Jan 9 at 2:31
$begingroup$
Thank you very much, learned some new things from this discussion.
$endgroup$
– HeyJude
Jan 9 at 2:31
$begingroup$
@HeyJude My pleasure!
$endgroup$
– NicNic8
Jan 9 at 2:33
$begingroup$
@HeyJude My pleasure!
$endgroup$
– NicNic8
Jan 9 at 2:33
add a comment |
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$begingroup$
Can you give a definition of a span of a matrix? I've not heard this term used before. I know what the span of a set of vectors is. And I know what the image (or range) of a matrix is. But I'm not sure what the span of a matrix is.
$endgroup$
– NicNic8
Jan 9 at 1:51
$begingroup$
Why, of course not; I'll edit
$endgroup$
– HeyJude
Jan 9 at 1:52
$begingroup$
@NicNic8, I think it's better now
$endgroup$
– HeyJude
Jan 9 at 1:58
$begingroup$
What is a solution set of a matrix? I think part of your issue may be that you're not using correct terminology. If we have an equation Ax=b, then we might ask "What are the values of x that satisfy this equation?" Another way to ask this is "What are the solutions to this equation?" But I've never heard of a solution set for a matrix. The set that you're calling the solution set looks to be the Null space of the matrix. Is that what you're going after? That's the special set of solutions to Ax=0, where A is the matrix of interest.
$endgroup$
– NicNic8
Jan 9 at 2:00
1
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The span of a pair of vectors is the set of all of their linear combinations, in this case $lambda(2,-1,0,6)+mu(0,-1,2,0)$, et voilà, there’s your parameterization. The two ways of describing the set are equivalent.
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– amd
Jan 9 at 2:05