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Find a basis for $ U cap W $ [duplicate]
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How to find basis of intersection of 2 vector spaces with given basis?
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Two subspaces $ W = 1+ x + x^3 , -2x+x^2+x^3 , 3+ x + x^2 + 4x ^ 3 $ and $ U = a+bx-bx^2+2ax^3 mid a,b in mathbb R $
so $ W = w_1 = (1,1,0,1), w_2 = (0,-2,1,1), w_3 = (3,1,1,4) , \ U = u_1 = (1,0,0,2), u_2 = (0,-1,1,0) $
and I need to find basis for $ U cap W $ so if $ v in U cap W $ this means $ 0 = alpha_1 w_1 + alpha_2 w_2 + alpha_3 w _3 - alpha_4 u_1 - alpha_5u_2$
so I should just rank it in matrix and so
$ beginbmatrix 1 & 0 & -1 & 0 \ 0 & 1 & -1 & 2 \ 0 & -1 & 0 & -1 \ 2 & 0 & - 1 & -1 endbmatrix Rightarrow beginbmatrix 1 & 0 & -1 & 0 \ 0 & 1 & -1 & 2 \ 0 & 0 & -1 & 1 \ 0 & 0 & 0 & 0 endbmatrix $
so it means $alpha_1 = alpha_3 = alpha_4 $ and $ alpha_2 = - alpha_3 $ so it means any $ v in U cap W $ is $ v= alpha_1u_1 -alpha_1u_2 $ or $ v = alpha_3w_3 - alpha_3 w_4 $
and how do I show by which vectors the space $ U cap W $ is spanned??
linear-algebra
edited Jul 15 at 18:55
Bernard
110k635103
asked Jul 15 at 18:38
bm1125
1216
marked as duplicate by Cameron Buie, Leucippus, Trần Thúc Minh TrÃ, Parcly Taxel, amWhy linear-algebra
Users with the  linear-algebra badge can single-handedly close linear-algebra questions as duplicates and reopen them as needed.
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Jul 16 at 11:07
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |Â
up vote
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favorite
This question already has an answer here:
How to find basis of intersection of 2 vector spaces with given basis?
2 answers
Two subspaces $ W = 1+ x + x^3 , -2x+x^2+x^3 , 3+ x + x^2 + 4x ^ 3 $ and $ U = a+bx-bx^2+2ax^3 mid a,b in mathbb R $
so $ W = w_1 = (1,1,0,1), w_2 = (0,-2,1,1), w_3 = (3,1,1,4) , \ U = u_1 = (1,0,0,2), u_2 = (0,-1,1,0) $
and I need to find basis for $ U cap W $ so if $ v in U cap W $ this means $ 0 = alpha_1 w_1 + alpha_2 w_2 + alpha_3 w _3 - alpha_4 u_1 - alpha_5u_2$
so I should just rank it in matrix and so
$ beginbmatrix 1 & 0 & -1 & 0 \ 0 & 1 & -1 & 2 \ 0 & -1 & 0 & -1 \ 2 & 0 & - 1 & -1 endbmatrix Rightarrow beginbmatrix 1 & 0 & -1 & 0 \ 0 & 1 & -1 & 2 \ 0 & 0 & -1 & 1 \ 0 & 0 & 0 & 0 endbmatrix $
so it means $alpha_1 = alpha_3 = alpha_4 $ and $ alpha_2 = - alpha_3 $ so it means any $ v in U cap W $ is $ v= alpha_1u_1 -alpha_1u_2 $ or $ v = alpha_3w_3 - alpha_3 w_4 $
and how do I show by which vectors the space $ U cap W $ is spanned??
linear-algebra
edited Jul 15 at 18:55
Bernard
110k635103
asked Jul 15 at 18:38
bm1125
1216
marked as duplicate by Cameron Buie, Leucippus, Trần Thúc Minh TrÃ, Parcly Taxel, amWhy linear-algebra
Users with the  linear-algebra badge can single-handedly close linear-algebra questions as duplicates and reopen them as needed.
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Jul 16 at 11:07
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
A vector space can have more than one basis that can be used to span the whole space...assuming your calculations are correct either is the correct answer.
– Pi_die_die
Jul 15 at 18:41
What do you denote $w_4$?
– Bernard
Jul 15 at 18:56
add a comment |Â
up vote
0
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up vote
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down vote
favorite
This question already has an answer here:
How to find basis of intersection of 2 vector spaces with given basis?
2 answers
Two subspaces $ W = 1+ x + x^3 , -2x+x^2+x^3 , 3+ x + x^2 + 4x ^ 3 $ and $ U = a+bx-bx^2+2ax^3 mid a,b in mathbb R $
so $ W = w_1 = (1,1,0,1), w_2 = (0,-2,1,1), w_3 = (3,1,1,4) , \ U = u_1 = (1,0,0,2), u_2 = (0,-1,1,0) $
and I need to find basis for $ U cap W $ so if $ v in U cap W $ this means $ 0 = alpha_1 w_1 + alpha_2 w_2 + alpha_3 w _3 - alpha_4 u_1 - alpha_5u_2$
so I should just rank it in matrix and so
$ beginbmatrix 1 & 0 & -1 & 0 \ 0 & 1 & -1 & 2 \ 0 & -1 & 0 & -1 \ 2 & 0 & - 1 & -1 endbmatrix Rightarrow beginbmatrix 1 & 0 & -1 & 0 \ 0 & 1 & -1 & 2 \ 0 & 0 & -1 & 1 \ 0 & 0 & 0 & 0 endbmatrix $
so it means $alpha_1 = alpha_3 = alpha_4 $ and $ alpha_2 = - alpha_3 $ so it means any $ v in U cap W $ is $ v= alpha_1u_1 -alpha_1u_2 $ or $ v = alpha_3w_3 - alpha_3 w_4 $
and how do I show by which vectors the space $ U cap W $ is spanned??
linear-algebra
edited Jul 15 at 18:55
Bernard
110k635103
asked Jul 15 at 18:38
bm1125
1216
This question already has an answer here:
How to find basis of intersection of 2 vector spaces with given basis?
2 answers
Two subspaces $ W = 1+ x + x^3 , -2x+x^2+x^3 , 3+ x + x^2 + 4x ^ 3 $ and $ U = a+bx-bx^2+2ax^3 mid a,b in mathbb R $
so $ W = w_1 = (1,1,0,1), w_2 = (0,-2,1,1), w_3 = (3,1,1,4) , \ U = u_1 = (1,0,0,2), u_2 = (0,-1,1,0) $
and I need to find basis for $ U cap W $ so if $ v in U cap W $ this means $ 0 = alpha_1 w_1 + alpha_2 w_2 + alpha_3 w _3 - alpha_4 u_1 - alpha_5u_2$
so I should just rank it in matrix and so
$ beginbmatrix 1 & 0 & -1 & 0 \ 0 & 1 & -1 & 2 \ 0 & -1 & 0 & -1 \ 2 & 0 & - 1 & -1 endbmatrix Rightarrow beginbmatrix 1 & 0 & -1 & 0 \ 0 & 1 & -1 & 2 \ 0 & 0 & -1 & 1 \ 0 & 0 & 0 & 0 endbmatrix $
so it means $alpha_1 = alpha_3 = alpha_4 $ and $ alpha_2 = - alpha_3 $ so it means any $ v in U cap W $ is $ v= alpha_1u_1 -alpha_1u_2 $ or $ v = alpha_3w_3 - alpha_3 w_4 $
and how do I show by which vectors the space $ U cap W $ is spanned??
This question already has an answer here:
How to find basis of intersection of 2 vector spaces with given basis?
2 answers
linear-algebra
edited Jul 15 at 18:55
Bernard
110k635103
asked Jul 15 at 18:38
bm1125
1216
edited Jul 15 at 18:55
Bernard
110k635103
edited Jul 15 at 18:55
Bernard
110k635103
edited Jul 15 at 18:55
Bernard
110k635103
110k635103
asked Jul 15 at 18:38
bm1125
1216
asked Jul 15 at 18:38
bm1125
1216
asked Jul 15 at 18:38
bm1125
1216
1216
marked as duplicate by Cameron Buie, Leucippus, Trần Thúc Minh TrÃ, Parcly Taxel, amWhy linear-algebra
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Jul 16 at 11:07
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marked as duplicate by Cameron Buie, Leucippus, Trần Thúc Minh TrÃ, Parcly Taxel, amWhy linear-algebra
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Jul 16 at 11:07
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
A vector space can have more than one basis that can be used to span the whole space...assuming your calculations are correct either is the correct answer.
– Pi_die_die
Jul 15 at 18:41
What do you denote $w_4$?
– Bernard
Jul 15 at 18:56
add a comment |Â
A vector space can have more than one basis that can be used to span the whole space...assuming your calculations are correct either is the correct answer.
– Pi_die_die
Jul 15 at 18:41
What do you denote $w_4$?
– Bernard
Jul 15 at 18:56
A vector space can have more than one basis that can be used to span the whole space...assuming your calculations are correct either is the correct answer.
– Pi_die_die
Jul 15 at 18:41
A vector space can have more than one basis that can be used to span the whole space...assuming your calculations are correct either is the correct answer.
– Pi_die_die
Jul 15 at 18:41
What do you denote $w_4$?
– Bernard
Jul 15 at 18:56
What do you denote $w_4$?
– Bernard
Jul 15 at 18:56
add a comment |Â
2 Answers
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We have $3w_1+w_2 = w_3$ so the basis for $W$ is actually $w_1, w_2$.
On the other hand, check that $w_1, w_2, u_1$ is linearly independent while
$$u_2 = w_1 + w_2 - u_1$$
This implies that $dim (W + U) = 3$ so
$$dim (W cap U) = dim W + dim U - dim (W + U) = 2 + 2 - 3 = 1$$
Finally notice that $$W ni w_1 + w_2 = u_1 + u_2 in U$$
so $w_1 + w_2 = (1,-1,1,2)$ is a basis for $Wcap U$.
answered Jul 15 at 19:00
mechanodroid
22.3k52041
add a comment |Â
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0
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You can consider the augmented matrix with columns $w_1,w_2,w_3$ and a generic vector in $U$ to determine the compatibility conditions for this generic vector to be in $W$:
beginalign
left[!
beginarrayc<mkern-20mu
1 & 0 & 3 & a \
1 & -2& 1 & b \
0 & 1& 1 & -b\
1 & 1 & 4 & 2a
endarray!
right] &rightsquigarrow
left[!
beginarrayc<mkern-20mu
1 & 0 & 3 & a \
0 & -2 & -2 & b-a \
0 & 1 & 1 & -b\
0 & 1 & 1 & a
endarray!
right] rightsquigarrow
left[!
beginarrayc<mkern-20mu
1 & 0 & 3 & a \
0 & 1 & 1 & fraca-b2 \
0 & 1 & 1 & -b\
0 & 1 & 1 & a
endarray!
right] rightsquigarrow
left[!
beginarrayc<mkern-20mu
1 & 0 & 3 & a \
0 & 1 & 1 & fraca-b2 \
0 & 0 & 0 & -fraca+b2\
0 & 0 & 0 & fraca+b2
endarray!
right]
endalign
so the compatibility condition is $colorredb=-a$, and a vector in $Ucap W$ can be written as
$$u=a-ax+ax^2+2ax^3=a(underbrace1-x+x^2+2x^3_textbasis of :Ucap W)$$
answered Jul 15 at 19:23
Bernard
110k635103
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
We have $3w_1+w_2 = w_3$ so the basis for $W$ is actually $w_1, w_2$.
On the other hand, check that $w_1, w_2, u_1$ is linearly independent while
$$u_2 = w_1 + w_2 - u_1$$
This implies that $dim (W + U) = 3$ so
$$dim (W cap U) = dim W + dim U - dim (W + U) = 2 + 2 - 3 = 1$$
Finally notice that $$W ni w_1 + w_2 = u_1 + u_2 in U$$
so $w_1 + w_2 = (1,-1,1,2)$ is a basis for $Wcap U$.
answered Jul 15 at 19:00
mechanodroid
22.3k52041
add a comment |Â
up vote
0
down vote
We have $3w_1+w_2 = w_3$ so the basis for $W$ is actually $w_1, w_2$.
On the other hand, check that $w_1, w_2, u_1$ is linearly independent while
$$u_2 = w_1 + w_2 - u_1$$
This implies that $dim (W + U) = 3$ so
$$dim (W cap U) = dim W + dim U - dim (W + U) = 2 + 2 - 3 = 1$$
Finally notice that $$W ni w_1 + w_2 = u_1 + u_2 in U$$
so $w_1 + w_2 = (1,-1,1,2)$ is a basis for $Wcap U$.
answered Jul 15 at 19:00
mechanodroid
22.3k52041
add a comment |Â
up vote
0
down vote
up vote
0
down vote
We have $3w_1+w_2 = w_3$ so the basis for $W$ is actually $w_1, w_2$.
On the other hand, check that $w_1, w_2, u_1$ is linearly independent while
$$u_2 = w_1 + w_2 - u_1$$
This implies that $dim (W + U) = 3$ so
$$dim (W cap U) = dim W + dim U - dim (W + U) = 2 + 2 - 3 = 1$$
Finally notice that $$W ni w_1 + w_2 = u_1 + u_2 in U$$
so $w_1 + w_2 = (1,-1,1,2)$ is a basis for $Wcap U$.
answered Jul 15 at 19:00
mechanodroid
22.3k52041
We have $3w_1+w_2 = w_3$ so the basis for $W$ is actually $w_1, w_2$.
On the other hand, check that $w_1, w_2, u_1$ is linearly independent while
$$u_2 = w_1 + w_2 - u_1$$
This implies that $dim (W + U) = 3$ so
$$dim (W cap U) = dim W + dim U - dim (W + U) = 2 + 2 - 3 = 1$$
Finally notice that $$W ni w_1 + w_2 = u_1 + u_2 in U$$
so $w_1 + w_2 = (1,-1,1,2)$ is a basis for $Wcap U$.
answered Jul 15 at 19:00
mechanodroid
22.3k52041
answered Jul 15 at 19:00
mechanodroid
22.3k52041
answered Jul 15 at 19:00
mechanodroid
22.3k52041
answered Jul 15 at 19:00
mechanodroid
22.3k52041
22.3k52041
add a comment |Â
add a comment |Â
up vote
0
down vote
You can consider the augmented matrix with columns $w_1,w_2,w_3$ and a generic vector in $U$ to determine the compatibility conditions for this generic vector to be in $W$:
beginalign
left[!
beginarrayc<mkern-20mu
1 & 0 & 3 & a \
1 & -2& 1 & b \
0 & 1& 1 & -b\
1 & 1 & 4 & 2a
endarray!
right] &rightsquigarrow
left[!
beginarrayc<mkern-20mu
1 & 0 & 3 & a \
0 & -2 & -2 & b-a \
0 & 1 & 1 & -b\
0 & 1 & 1 & a
endarray!
right] rightsquigarrow
left[!
beginarrayc<mkern-20mu
1 & 0 & 3 & a \
0 & 1 & 1 & fraca-b2 \
0 & 1 & 1 & -b\
0 & 1 & 1 & a
endarray!
right] rightsquigarrow
left[!
beginarrayc<mkern-20mu
1 & 0 & 3 & a \
0 & 1 & 1 & fraca-b2 \
0 & 0 & 0 & -fraca+b2\
0 & 0 & 0 & fraca+b2
endarray!
right]
endalign
so the compatibility condition is $colorredb=-a$, and a vector in $Ucap W$ can be written as
$$u=a-ax+ax^2+2ax^3=a(underbrace1-x+x^2+2x^3_textbasis of :Ucap W)$$
answered Jul 15 at 19:23
Bernard
110k635103
add a comment |Â
up vote
0
down vote
You can consider the augmented matrix with columns $w_1,w_2,w_3$ and a generic vector in $U$ to determine the compatibility conditions for this generic vector to be in $W$:
beginalign
left[!
beginarrayc<mkern-20mu
1 & 0 & 3 & a \
1 & -2& 1 & b \
0 & 1& 1 & -b\
1 & 1 & 4 & 2a
endarray!
right] &rightsquigarrow
left[!
beginarrayc<mkern-20mu
1 & 0 & 3 & a \
0 & -2 & -2 & b-a \
0 & 1 & 1 & -b\
0 & 1 & 1 & a
endarray!
right] rightsquigarrow
left[!
beginarrayc<mkern-20mu
1 & 0 & 3 & a \
0 & 1 & 1 & fraca-b2 \
0 & 1 & 1 & -b\
0 & 1 & 1 & a
endarray!
right] rightsquigarrow
left[!
beginarrayc<mkern-20mu
1 & 0 & 3 & a \
0 & 1 & 1 & fraca-b2 \
0 & 0 & 0 & -fraca+b2\
0 & 0 & 0 & fraca+b2
endarray!
right]
endalign
so the compatibility condition is $colorredb=-a$, and a vector in $Ucap W$ can be written as
$$u=a-ax+ax^2+2ax^3=a(underbrace1-x+x^2+2x^3_textbasis of :Ucap W)$$
answered Jul 15 at 19:23
Bernard
110k635103
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You can consider the augmented matrix with columns $w_1,w_2,w_3$ and a generic vector in $U$ to determine the compatibility conditions for this generic vector to be in $W$:
beginalign
left[!
beginarrayc<mkern-20mu
1 & 0 & 3 & a \
1 & -2& 1 & b \
0 & 1& 1 & -b\
1 & 1 & 4 & 2a
endarray!
right] &rightsquigarrow
left[!
beginarrayc<mkern-20mu
1 & 0 & 3 & a \
0 & -2 & -2 & b-a \
0 & 1 & 1 & -b\
0 & 1 & 1 & a
endarray!
right] rightsquigarrow
left[!
beginarrayc<mkern-20mu
1 & 0 & 3 & a \
0 & 1 & 1 & fraca-b2 \
0 & 1 & 1 & -b\
0 & 1 & 1 & a
endarray!
right] rightsquigarrow
left[!
beginarrayc<mkern-20mu
1 & 0 & 3 & a \
0 & 1 & 1 & fraca-b2 \
0 & 0 & 0 & -fraca+b2\
0 & 0 & 0 & fraca+b2
endarray!
right]
endalign
so the compatibility condition is $colorredb=-a$, and a vector in $Ucap W$ can be written as
$$u=a-ax+ax^2+2ax^3=a(underbrace1-x+x^2+2x^3_textbasis of :Ucap W)$$
answered Jul 15 at 19:23
Bernard
110k635103
You can consider the augmented matrix with columns $w_1,w_2,w_3$ and a generic vector in $U$ to determine the compatibility conditions for this generic vector to be in $W$:
beginalign
left[!
beginarrayc<mkern-20mu
1 & 0 & 3 & a \
1 & -2& 1 & b \
0 & 1& 1 & -b\
1 & 1 & 4 & 2a
endarray!
right] &rightsquigarrow
left[!
beginarrayc<mkern-20mu
1 & 0 & 3 & a \
0 & -2 & -2 & b-a \
0 & 1 & 1 & -b\
0 & 1 & 1 & a
endarray!
right] rightsquigarrow
left[!
beginarrayc<mkern-20mu
1 & 0 & 3 & a \
0 & 1 & 1 & fraca-b2 \
0 & 1 & 1 & -b\
0 & 1 & 1 & a
endarray!
right] rightsquigarrow
left[!
beginarrayc<mkern-20mu
1 & 0 & 3 & a \
0 & 1 & 1 & fraca-b2 \
0 & 0 & 0 & -fraca+b2\
0 & 0 & 0 & fraca+b2
endarray!
right]
endalign
so the compatibility condition is $colorredb=-a$, and a vector in $Ucap W$ can be written as
$$u=a-ax+ax^2+2ax^3=a(underbrace1-x+x^2+2x^3_textbasis of :Ucap W)$$
answered Jul 15 at 19:23
Bernard
110k635103
answered Jul 15 at 19:23
Bernard
110k635103
answered Jul 15 at 19:23
Bernard
110k635103
answered Jul 15 at 19:23
Bernard
110k635103
110k635103
add a comment |Â
add a comment |Â
A vector space can have more than one basis that can be used to span the whole space...assuming your calculations are correct either is the correct answer.
– Pi_die_die
Jul 15 at 18:41
What do you denote $w_4$?
– Bernard
Jul 15 at 18:56