How can I check if one from these vector spaces equal to other?
$begingroup$
Given the following vector spaces:
$$ U = Span( (1,-1,-1) , (2,4,1) ) $$
$$ V = Span( (0,2,1) , (-1,3,2) ) $$
$$ W = Span( (2,6,2) , (1,3,1) ) $$
How can I check which from these vector spaces are equal? And how can I check if one from these vector spaces is sub-vector space of other vector space ?
linear-algebra vector-spaces
asked Jan 2 at 15:38
Software_tSoftware_t
1276
$endgroup$
add a comment |
$begingroup$
Given the following vector spaces:
$$ U = Span( (1,-1,-1) , (2,4,1) ) $$
$$ V = Span( (0,2,1) , (-1,3,2) ) $$
$$ W = Span( (2,6,2) , (1,3,1) ) $$
How can I check which from these vector spaces are equal? And how can I check if one from these vector spaces is sub-vector space of other vector space ?
linear-algebra vector-spaces
asked Jan 2 at 15:38
Software_tSoftware_t
1276
$endgroup$
$begingroup$
In this case, you know that the vectors span either a plane or a line. You can find the three equations and compare them.
$endgroup$
– Bernard Massé
Jan 2 at 16:02
add a comment |
$begingroup$
Given the following vector spaces:
$$ U = Span( (1,-1,-1) , (2,4,1) ) $$
$$ V = Span( (0,2,1) , (-1,3,2) ) $$
$$ W = Span( (2,6,2) , (1,3,1) ) $$
How can I check which from these vector spaces are equal? And how can I check if one from these vector spaces is sub-vector space of other vector space ?
linear-algebra vector-spaces
asked Jan 2 at 15:38
Software_tSoftware_t
1276
$endgroup$
Given the following vector spaces:
$$ U = Span( (1,-1,-1) , (2,4,1) ) $$
$$ V = Span( (0,2,1) , (-1,3,2) ) $$
$$ W = Span( (2,6,2) , (1,3,1) ) $$
How can I check which from these vector spaces are equal? And how can I check if one from these vector spaces is sub-vector space of other vector space ?
linear-algebra vector-spaces
linear-algebra vector-spaces
asked Jan 2 at 15:38
Software_tSoftware_t
1276
asked Jan 2 at 15:38
Software_tSoftware_t
1276
asked Jan 2 at 15:38
Software_tSoftware_t
1276
asked Jan 2 at 15:38
Software_tSoftware_t
1276
asked Jan 2 at 15:38
Software_tSoftware_t
1276
1276
$begingroup$
In this case, you know that the vectors span either a plane or a line. You can find the three equations and compare them.
$endgroup$
– Bernard Massé
Jan 2 at 16:02
add a comment |
$begingroup$
In this case, you know that the vectors span either a plane or a line. You can find the three equations and compare them.
$endgroup$
– Bernard Massé
Jan 2 at 16:02
$begingroup$
In this case, you know that the vectors span either a plane or a line. You can find the three equations and compare them.
$endgroup$
– Bernard Massé
Jan 2 at 16:02
$begingroup$
In this case, you know that the vectors span either a plane or a line. You can find the three equations and compare them.
$endgroup$
– Bernard Massé
Jan 2 at 16:02
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
To check if a vector space is a subspace of another, you must check that each vector in the first space is also contained in the second space. If you can find a spanning set for the first subspace, then it is enough to check that each of the vectors in the spanning set are in the other space, since each vector in the first space can be written as a combination of the vectors in the spanning set.
Two vector spaces are equal if they are both subsets of one another.
Lets look at a concrete example. We'll check if $Wsubset V$. This means we need to see if we can write each of $(2,6,2)$ and $(1,3,1)$ as a linear combination of vectors in $V$.
Writing $(1,3,1)$ as a linear combination of vectors in $V$ means finding scalars $a$ and $b$ so that $(1,3,1) = acdot(0,2,1) + bcdot(-1,3,2) $. This gives the system,
$$
0-b = 1, ~~ 2a+3b=3, ~~ a+2b = 1
$$
If we solve this we have $b=-1$, so plugging this into the second equation we have $2a-3=3$ so $a = 3$. Finally, the last equation gives $3-2=1$ which is true. Therefore $(1,3,1)in V$.
Similarly, $(2,6,2)in V$ since $(2,6,2) = 2cdot(1,3,1)$ so we can double the coefficients of our linear combination above.
This proves that $Wsubset V$. Now check if $Vsubset W$. If so, then $W=V$, and if not, then $Wneq V$.
answered Jan 2 at 15:49
tchtch
833310
$endgroup$
add a comment |
$begingroup$
It is clear that $dim(U)=dim(V)=2$ since the two spanning vectors for each subspace are linearly independent (one is not a multiple of the other). However $(2,6,2)=2(1,3,1)$ so $dim(W)=1$.
Since equal vector spaces must have the same dimension we can see that $U ne W$ and $V ne W$. The only remaining question is whether $U=V$. You can proceed as in tch's answer above. Alternatively, as we are working in $mathbb{R}^3$ you can find a vector that is perpendicular to the plane $U$ using the cross-product, and then determine whether the same vector is perpendicular to the plane $V$ - if it is then $U=V$.
answered Jan 2 at 17:06
gandalf61gandalf61
9,184825
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
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active
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$begingroup$
To check if a vector space is a subspace of another, you must check that each vector in the first space is also contained in the second space. If you can find a spanning set for the first subspace, then it is enough to check that each of the vectors in the spanning set are in the other space, since each vector in the first space can be written as a combination of the vectors in the spanning set.
Two vector spaces are equal if they are both subsets of one another.
Lets look at a concrete example. We'll check if $Wsubset V$. This means we need to see if we can write each of $(2,6,2)$ and $(1,3,1)$ as a linear combination of vectors in $V$.
Writing $(1,3,1)$ as a linear combination of vectors in $V$ means finding scalars $a$ and $b$ so that $(1,3,1) = acdot(0,2,1) + bcdot(-1,3,2) $. This gives the system,
$$
0-b = 1, ~~ 2a+3b=3, ~~ a+2b = 1
$$
If we solve this we have $b=-1$, so plugging this into the second equation we have $2a-3=3$ so $a = 3$. Finally, the last equation gives $3-2=1$ which is true. Therefore $(1,3,1)in V$.
Similarly, $(2,6,2)in V$ since $(2,6,2) = 2cdot(1,3,1)$ so we can double the coefficients of our linear combination above.
This proves that $Wsubset V$. Now check if $Vsubset W$. If so, then $W=V$, and if not, then $Wneq V$.
answered Jan 2 at 15:49
tchtch
833310
$endgroup$
add a comment |
$begingroup$
To check if a vector space is a subspace of another, you must check that each vector in the first space is also contained in the second space. If you can find a spanning set for the first subspace, then it is enough to check that each of the vectors in the spanning set are in the other space, since each vector in the first space can be written as a combination of the vectors in the spanning set.
Two vector spaces are equal if they are both subsets of one another.
Lets look at a concrete example. We'll check if $Wsubset V$. This means we need to see if we can write each of $(2,6,2)$ and $(1,3,1)$ as a linear combination of vectors in $V$.
Writing $(1,3,1)$ as a linear combination of vectors in $V$ means finding scalars $a$ and $b$ so that $(1,3,1) = acdot(0,2,1) + bcdot(-1,3,2) $. This gives the system,
$$
0-b = 1, ~~ 2a+3b=3, ~~ a+2b = 1
$$
If we solve this we have $b=-1$, so plugging this into the second equation we have $2a-3=3$ so $a = 3$. Finally, the last equation gives $3-2=1$ which is true. Therefore $(1,3,1)in V$.
Similarly, $(2,6,2)in V$ since $(2,6,2) = 2cdot(1,3,1)$ so we can double the coefficients of our linear combination above.
This proves that $Wsubset V$. Now check if $Vsubset W$. If so, then $W=V$, and if not, then $Wneq V$.
answered Jan 2 at 15:49
tchtch
833310
$endgroup$
add a comment |
$begingroup$
To check if a vector space is a subspace of another, you must check that each vector in the first space is also contained in the second space. If you can find a spanning set for the first subspace, then it is enough to check that each of the vectors in the spanning set are in the other space, since each vector in the first space can be written as a combination of the vectors in the spanning set.
Two vector spaces are equal if they are both subsets of one another.
Lets look at a concrete example. We'll check if $Wsubset V$. This means we need to see if we can write each of $(2,6,2)$ and $(1,3,1)$ as a linear combination of vectors in $V$.
Writing $(1,3,1)$ as a linear combination of vectors in $V$ means finding scalars $a$ and $b$ so that $(1,3,1) = acdot(0,2,1) + bcdot(-1,3,2) $. This gives the system,
$$
0-b = 1, ~~ 2a+3b=3, ~~ a+2b = 1
$$
If we solve this we have $b=-1$, so plugging this into the second equation we have $2a-3=3$ so $a = 3$. Finally, the last equation gives $3-2=1$ which is true. Therefore $(1,3,1)in V$.
Similarly, $(2,6,2)in V$ since $(2,6,2) = 2cdot(1,3,1)$ so we can double the coefficients of our linear combination above.
This proves that $Wsubset V$. Now check if $Vsubset W$. If so, then $W=V$, and if not, then $Wneq V$.
answered Jan 2 at 15:49
tchtch
833310
$endgroup$
To check if a vector space is a subspace of another, you must check that each vector in the first space is also contained in the second space. If you can find a spanning set for the first subspace, then it is enough to check that each of the vectors in the spanning set are in the other space, since each vector in the first space can be written as a combination of the vectors in the spanning set.
Two vector spaces are equal if they are both subsets of one another.
Lets look at a concrete example. We'll check if $Wsubset V$. This means we need to see if we can write each of $(2,6,2)$ and $(1,3,1)$ as a linear combination of vectors in $V$.
Writing $(1,3,1)$ as a linear combination of vectors in $V$ means finding scalars $a$ and $b$ so that $(1,3,1) = acdot(0,2,1) + bcdot(-1,3,2) $. This gives the system,
$$
0-b = 1, ~~ 2a+3b=3, ~~ a+2b = 1
$$
If we solve this we have $b=-1$, so plugging this into the second equation we have $2a-3=3$ so $a = 3$. Finally, the last equation gives $3-2=1$ which is true. Therefore $(1,3,1)in V$.
Similarly, $(2,6,2)in V$ since $(2,6,2) = 2cdot(1,3,1)$ so we can double the coefficients of our linear combination above.
This proves that $Wsubset V$. Now check if $Vsubset W$. If so, then $W=V$, and if not, then $Wneq V$.
answered Jan 2 at 15:49
tchtch
833310
edited Jan 2 at 15:55
answered Jan 2 at 15:49
tchtch
833310
answered Jan 2 at 15:49
tchtch
833310
answered Jan 2 at 15:49
tchtch
833310
833310
add a comment |
add a comment |
$begingroup$
It is clear that $dim(U)=dim(V)=2$ since the two spanning vectors for each subspace are linearly independent (one is not a multiple of the other). However $(2,6,2)=2(1,3,1)$ so $dim(W)=1$.
Since equal vector spaces must have the same dimension we can see that $U ne W$ and $V ne W$. The only remaining question is whether $U=V$. You can proceed as in tch's answer above. Alternatively, as we are working in $mathbb{R}^3$ you can find a vector that is perpendicular to the plane $U$ using the cross-product, and then determine whether the same vector is perpendicular to the plane $V$ - if it is then $U=V$.
answered Jan 2 at 17:06
gandalf61gandalf61
9,184825
$endgroup$
add a comment |
$begingroup$
It is clear that $dim(U)=dim(V)=2$ since the two spanning vectors for each subspace are linearly independent (one is not a multiple of the other). However $(2,6,2)=2(1,3,1)$ so $dim(W)=1$.
Since equal vector spaces must have the same dimension we can see that $U ne W$ and $V ne W$. The only remaining question is whether $U=V$. You can proceed as in tch's answer above. Alternatively, as we are working in $mathbb{R}^3$ you can find a vector that is perpendicular to the plane $U$ using the cross-product, and then determine whether the same vector is perpendicular to the plane $V$ - if it is then $U=V$.
answered Jan 2 at 17:06
gandalf61gandalf61
9,184825
$endgroup$
add a comment |
$begingroup$
It is clear that $dim(U)=dim(V)=2$ since the two spanning vectors for each subspace are linearly independent (one is not a multiple of the other). However $(2,6,2)=2(1,3,1)$ so $dim(W)=1$.
Since equal vector spaces must have the same dimension we can see that $U ne W$ and $V ne W$. The only remaining question is whether $U=V$. You can proceed as in tch's answer above. Alternatively, as we are working in $mathbb{R}^3$ you can find a vector that is perpendicular to the plane $U$ using the cross-product, and then determine whether the same vector is perpendicular to the plane $V$ - if it is then $U=V$.
answered Jan 2 at 17:06
gandalf61gandalf61
9,184825
$endgroup$
It is clear that $dim(U)=dim(V)=2$ since the two spanning vectors for each subspace are linearly independent (one is not a multiple of the other). However $(2,6,2)=2(1,3,1)$ so $dim(W)=1$.
Since equal vector spaces must have the same dimension we can see that $U ne W$ and $V ne W$. The only remaining question is whether $U=V$. You can proceed as in tch's answer above. Alternatively, as we are working in $mathbb{R}^3$ you can find a vector that is perpendicular to the plane $U$ using the cross-product, and then determine whether the same vector is perpendicular to the plane $V$ - if it is then $U=V$.
answered Jan 2 at 17:06
gandalf61gandalf61
9,184825
answered Jan 2 at 17:06
gandalf61gandalf61
9,184825
answered Jan 2 at 17:06
gandalf61gandalf61
9,184825
answered Jan 2 at 17:06
gandalf61gandalf61
9,184825
9,184825
add a comment |
add a comment |
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$begingroup$
In this case, you know that the vectors span either a plane or a line. You can find the three equations and compare them.
$endgroup$
– Bernard Massé
Jan 2 at 16:02