Complement of a subspace
So I have this subspace $ U = span{(1,1,1)^T,(0,1,-1)^T,(1,-1,3)^T} $. I need to find the complement (W) of U such that $ mathbb{R}^3 = U + W $ and $ U bigcap W = 0$ . I'm not too sure how I'm supposed to approach this problem. Any help would be appreciated.
vector-spaces
asked Dec 9 at 14:31
Rakoon Berry
12
add a comment |
So I have this subspace $ U = span{(1,1,1)^T,(0,1,-1)^T,(1,-1,3)^T} $. I need to find the complement (W) of U such that $ mathbb{R}^3 = U + W $ and $ U bigcap W = 0$ . I'm not too sure how I'm supposed to approach this problem. Any help would be appreciated.
vector-spaces
asked Dec 9 at 14:31
Rakoon Berry
12
add a comment |
So I have this subspace $ U = span{(1,1,1)^T,(0,1,-1)^T,(1,-1,3)^T} $. I need to find the complement (W) of U such that $ mathbb{R}^3 = U + W $ and $ U bigcap W = 0$ . I'm not too sure how I'm supposed to approach this problem. Any help would be appreciated.
vector-spaces
asked Dec 9 at 14:31
Rakoon Berry
12
So I have this subspace $ U = span{(1,1,1)^T,(0,1,-1)^T,(1,-1,3)^T} $. I need to find the complement (W) of U such that $ mathbb{R}^3 = U + W $ and $ U bigcap W = 0$ . I'm not too sure how I'm supposed to approach this problem. Any help would be appreciated.
vector-spaces
vector-spaces
asked Dec 9 at 14:31
Rakoon Berry
12
asked Dec 9 at 14:31
Rakoon Berry
12
asked Dec 9 at 14:31
Rakoon Berry
12
asked Dec 9 at 14:31
Rakoon Berry
12
asked Dec 9 at 14:31
Rakoon Berry
12
12
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
This is not an answer but it provides the steps which will lead to a solution
Step 1: Find a basis for $U$. One way to do this is to form a matrix $A$ whose rows are the vectors in the span of $U$, get $A$ to it's reduced form using Gauss elimination and then remove zero vectors.
Step 2: Let $B=(v_1,v_2)$ be the basis for $U$ (the one you obtained in step $1$, it will consist of two vectors). Find a vector in $mathbb{R}^3$ that is not in $span(B)$ (check $e_1,e_2,e_3$ one of them is not in the span). You will get a basis $B'=(v_1,v_2,e_i)$ for $mathbb{R}^3$.
Step 3: Set $W=span(e_i)$, since $e_inotin span(v_1,v_2)$ you have that $Wcap U = {0}$. Moreover since $v_1,v_2,e_i$ a basis for $mathbb{R}^3$ you have that $mathbb{R}^3 = W+U$.
answered Dec 9 at 14:38
Yanko
6,104724
Hey. I found the basis to be $ {(0,1,-1)^T,(1,-1,3)^T} $ , but when I tried to see if vectors e1, e2 and e3 are not in the span. I found that all of them are not in it. Is my basis wrong?
– Rakoon Berry
Dec 9 at 15:39
@RakoonBerry It's ok you only need that one of them is not in the span. Your basis is not wrong.
– Yanko
Dec 9 at 17:03
Sorry for the dumb question, but how is that possible? What I understood is that we are trying to find a vector span that is not in U, so that when we add them together, we get the full three-dimensional plane. Going by the steps you mentioned, the basis would be (v1,v2,e1,e2,e3), which wouldn't make sense because v1 and v2 are in $ span(e1,e2,e3) $. Shouldn't the basis be the smallest number of vectors that construct a given vector space?
– Rakoon Berry
Dec 9 at 17:22
So I just thought about this a bit more. My understanding is that we only need one of the unit vectors to be not in the span, because then we can construct the other two from the new span B'. Please correct me if I'm wrong
– Rakoon Berry
Dec 9 at 18:06
About your previous comment: That actually make sense, consider this example: Take $V=mathbb{R}^2$, then clearly $e_1,e_2$ a basis for $V$. However there exists a subspace $W$ of $V$, say $W=span((1,1))$ (the line $y=x$). This is indeed a subspace of $V$ but it doesn't include $e_1$ nor it includes $e_2$.
– Yanko
Dec 10 at 13:58
add a comment |
There are lots of ways to do this. One way is the following:
First you should find a basis for $U$ so you know what dimension it is.
After that you can start building a basis for $W$. One way to do this is to take a vector $(a,b,c)$ and set its dot product with each basis element of $U$ to 0.
This should lead you to a linearly independent vector - note these equations are underdetermined so be sure to choose a value for the remaining variables.
Keep adding vectors orthogonal to the previous ones until $W$ is of the correct dimension. Realistically you only need one or two.
answered Dec 9 at 14:44
Ben
2,488616
add a comment |
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2 Answers
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2 Answers
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This is not an answer but it provides the steps which will lead to a solution
Step 1: Find a basis for $U$. One way to do this is to form a matrix $A$ whose rows are the vectors in the span of $U$, get $A$ to it's reduced form using Gauss elimination and then remove zero vectors.
Step 2: Let $B=(v_1,v_2)$ be the basis for $U$ (the one you obtained in step $1$, it will consist of two vectors). Find a vector in $mathbb{R}^3$ that is not in $span(B)$ (check $e_1,e_2,e_3$ one of them is not in the span). You will get a basis $B'=(v_1,v_2,e_i)$ for $mathbb{R}^3$.
Step 3: Set $W=span(e_i)$, since $e_inotin span(v_1,v_2)$ you have that $Wcap U = {0}$. Moreover since $v_1,v_2,e_i$ a basis for $mathbb{R}^3$ you have that $mathbb{R}^3 = W+U$.
answered Dec 9 at 14:38
Yanko
6,104724
Hey. I found the basis to be $ {(0,1,-1)^T,(1,-1,3)^T} $ , but when I tried to see if vectors e1, e2 and e3 are not in the span. I found that all of them are not in it. Is my basis wrong?
– Rakoon Berry
Dec 9 at 15:39
@RakoonBerry It's ok you only need that one of them is not in the span. Your basis is not wrong.
– Yanko
Dec 9 at 17:03
Sorry for the dumb question, but how is that possible? What I understood is that we are trying to find a vector span that is not in U, so that when we add them together, we get the full three-dimensional plane. Going by the steps you mentioned, the basis would be (v1,v2,e1,e2,e3), which wouldn't make sense because v1 and v2 are in $ span(e1,e2,e3) $. Shouldn't the basis be the smallest number of vectors that construct a given vector space?
– Rakoon Berry
Dec 9 at 17:22
So I just thought about this a bit more. My understanding is that we only need one of the unit vectors to be not in the span, because then we can construct the other two from the new span B'. Please correct me if I'm wrong
– Rakoon Berry
Dec 9 at 18:06
About your previous comment: That actually make sense, consider this example: Take $V=mathbb{R}^2$, then clearly $e_1,e_2$ a basis for $V$. However there exists a subspace $W$ of $V$, say $W=span((1,1))$ (the line $y=x$). This is indeed a subspace of $V$ but it doesn't include $e_1$ nor it includes $e_2$.
– Yanko
Dec 10 at 13:58
add a comment |
This is not an answer but it provides the steps which will lead to a solution
Step 1: Find a basis for $U$. One way to do this is to form a matrix $A$ whose rows are the vectors in the span of $U$, get $A$ to it's reduced form using Gauss elimination and then remove zero vectors.
Step 2: Let $B=(v_1,v_2)$ be the basis for $U$ (the one you obtained in step $1$, it will consist of two vectors). Find a vector in $mathbb{R}^3$ that is not in $span(B)$ (check $e_1,e_2,e_3$ one of them is not in the span). You will get a basis $B'=(v_1,v_2,e_i)$ for $mathbb{R}^3$.
Step 3: Set $W=span(e_i)$, since $e_inotin span(v_1,v_2)$ you have that $Wcap U = {0}$. Moreover since $v_1,v_2,e_i$ a basis for $mathbb{R}^3$ you have that $mathbb{R}^3 = W+U$.
answered Dec 9 at 14:38
Yanko
6,104724
Hey. I found the basis to be $ {(0,1,-1)^T,(1,-1,3)^T} $ , but when I tried to see if vectors e1, e2 and e3 are not in the span. I found that all of them are not in it. Is my basis wrong?
– Rakoon Berry
Dec 9 at 15:39
@RakoonBerry It's ok you only need that one of them is not in the span. Your basis is not wrong.
– Yanko
Dec 9 at 17:03
Sorry for the dumb question, but how is that possible? What I understood is that we are trying to find a vector span that is not in U, so that when we add them together, we get the full three-dimensional plane. Going by the steps you mentioned, the basis would be (v1,v2,e1,e2,e3), which wouldn't make sense because v1 and v2 are in $ span(e1,e2,e3) $. Shouldn't the basis be the smallest number of vectors that construct a given vector space?
– Rakoon Berry
Dec 9 at 17:22
So I just thought about this a bit more. My understanding is that we only need one of the unit vectors to be not in the span, because then we can construct the other two from the new span B'. Please correct me if I'm wrong
– Rakoon Berry
Dec 9 at 18:06
About your previous comment: That actually make sense, consider this example: Take $V=mathbb{R}^2$, then clearly $e_1,e_2$ a basis for $V$. However there exists a subspace $W$ of $V$, say $W=span((1,1))$ (the line $y=x$). This is indeed a subspace of $V$ but it doesn't include $e_1$ nor it includes $e_2$.
– Yanko
Dec 10 at 13:58
add a comment |
This is not an answer but it provides the steps which will lead to a solution
Step 1: Find a basis for $U$. One way to do this is to form a matrix $A$ whose rows are the vectors in the span of $U$, get $A$ to it's reduced form using Gauss elimination and then remove zero vectors.
Step 2: Let $B=(v_1,v_2)$ be the basis for $U$ (the one you obtained in step $1$, it will consist of two vectors). Find a vector in $mathbb{R}^3$ that is not in $span(B)$ (check $e_1,e_2,e_3$ one of them is not in the span). You will get a basis $B'=(v_1,v_2,e_i)$ for $mathbb{R}^3$.
Step 3: Set $W=span(e_i)$, since $e_inotin span(v_1,v_2)$ you have that $Wcap U = {0}$. Moreover since $v_1,v_2,e_i$ a basis for $mathbb{R}^3$ you have that $mathbb{R}^3 = W+U$.
answered Dec 9 at 14:38
Yanko
6,104724
This is not an answer but it provides the steps which will lead to a solution
Step 1: Find a basis for $U$. One way to do this is to form a matrix $A$ whose rows are the vectors in the span of $U$, get $A$ to it's reduced form using Gauss elimination and then remove zero vectors.
Step 2: Let $B=(v_1,v_2)$ be the basis for $U$ (the one you obtained in step $1$, it will consist of two vectors). Find a vector in $mathbb{R}^3$ that is not in $span(B)$ (check $e_1,e_2,e_3$ one of them is not in the span). You will get a basis $B'=(v_1,v_2,e_i)$ for $mathbb{R}^3$.
Step 3: Set $W=span(e_i)$, since $e_inotin span(v_1,v_2)$ you have that $Wcap U = {0}$. Moreover since $v_1,v_2,e_i$ a basis for $mathbb{R}^3$ you have that $mathbb{R}^3 = W+U$.
answered Dec 9 at 14:38
Yanko
6,104724
answered Dec 9 at 14:38
Yanko
6,104724
answered Dec 9 at 14:38
Yanko
6,104724
answered Dec 9 at 14:38
Yanko
6,104724
6,104724
Hey. I found the basis to be $ {(0,1,-1)^T,(1,-1,3)^T} $ , but when I tried to see if vectors e1, e2 and e3 are not in the span. I found that all of them are not in it. Is my basis wrong?
– Rakoon Berry
Dec 9 at 15:39
@RakoonBerry It's ok you only need that one of them is not in the span. Your basis is not wrong.
– Yanko
Dec 9 at 17:03
Sorry for the dumb question, but how is that possible? What I understood is that we are trying to find a vector span that is not in U, so that when we add them together, we get the full three-dimensional plane. Going by the steps you mentioned, the basis would be (v1,v2,e1,e2,e3), which wouldn't make sense because v1 and v2 are in $ span(e1,e2,e3) $. Shouldn't the basis be the smallest number of vectors that construct a given vector space?
– Rakoon Berry
Dec 9 at 17:22
So I just thought about this a bit more. My understanding is that we only need one of the unit vectors to be not in the span, because then we can construct the other two from the new span B'. Please correct me if I'm wrong
– Rakoon Berry
Dec 9 at 18:06
About your previous comment: That actually make sense, consider this example: Take $V=mathbb{R}^2$, then clearly $e_1,e_2$ a basis for $V$. However there exists a subspace $W$ of $V$, say $W=span((1,1))$ (the line $y=x$). This is indeed a subspace of $V$ but it doesn't include $e_1$ nor it includes $e_2$.
– Yanko
Dec 10 at 13:58
add a comment |
Hey. I found the basis to be $ {(0,1,-1)^T,(1,-1,3)^T} $ , but when I tried to see if vectors e1, e2 and e3 are not in the span. I found that all of them are not in it. Is my basis wrong?
– Rakoon Berry
Dec 9 at 15:39
@RakoonBerry It's ok you only need that one of them is not in the span. Your basis is not wrong.
– Yanko
Dec 9 at 17:03
Sorry for the dumb question, but how is that possible? What I understood is that we are trying to find a vector span that is not in U, so that when we add them together, we get the full three-dimensional plane. Going by the steps you mentioned, the basis would be (v1,v2,e1,e2,e3), which wouldn't make sense because v1 and v2 are in $ span(e1,e2,e3) $. Shouldn't the basis be the smallest number of vectors that construct a given vector space?
– Rakoon Berry
Dec 9 at 17:22
So I just thought about this a bit more. My understanding is that we only need one of the unit vectors to be not in the span, because then we can construct the other two from the new span B'. Please correct me if I'm wrong
– Rakoon Berry
Dec 9 at 18:06
About your previous comment: That actually make sense, consider this example: Take $V=mathbb{R}^2$, then clearly $e_1,e_2$ a basis for $V$. However there exists a subspace $W$ of $V$, say $W=span((1,1))$ (the line $y=x$). This is indeed a subspace of $V$ but it doesn't include $e_1$ nor it includes $e_2$.
– Yanko
Dec 10 at 13:58
Hey. I found the basis to be $ {(0,1,-1)^T,(1,-1,3)^T} $ , but when I tried to see if vectors e1, e2 and e3 are not in the span. I found that all of them are not in it. Is my basis wrong?
– Rakoon Berry
Dec 9 at 15:39
Hey. I found the basis to be $ {(0,1,-1)^T,(1,-1,3)^T} $ , but when I tried to see if vectors e1, e2 and e3 are not in the span. I found that all of them are not in it. Is my basis wrong?
– Rakoon Berry
Dec 9 at 15:39
@RakoonBerry It's ok you only need that one of them is not in the span. Your basis is not wrong.
– Yanko
Dec 9 at 17:03
@RakoonBerry It's ok you only need that one of them is not in the span. Your basis is not wrong.
– Yanko
Dec 9 at 17:03
Sorry for the dumb question, but how is that possible? What I understood is that we are trying to find a vector span that is not in U, so that when we add them together, we get the full three-dimensional plane. Going by the steps you mentioned, the basis would be (v1,v2,e1,e2,e3), which wouldn't make sense because v1 and v2 are in $ span(e1,e2,e3) $. Shouldn't the basis be the smallest number of vectors that construct a given vector space?
– Rakoon Berry
Dec 9 at 17:22
Sorry for the dumb question, but how is that possible? What I understood is that we are trying to find a vector span that is not in U, so that when we add them together, we get the full three-dimensional plane. Going by the steps you mentioned, the basis would be (v1,v2,e1,e2,e3), which wouldn't make sense because v1 and v2 are in $ span(e1,e2,e3) $. Shouldn't the basis be the smallest number of vectors that construct a given vector space?
– Rakoon Berry
Dec 9 at 17:22
So I just thought about this a bit more. My understanding is that we only need one of the unit vectors to be not in the span, because then we can construct the other two from the new span B'. Please correct me if I'm wrong
– Rakoon Berry
Dec 9 at 18:06
So I just thought about this a bit more. My understanding is that we only need one of the unit vectors to be not in the span, because then we can construct the other two from the new span B'. Please correct me if I'm wrong
– Rakoon Berry
Dec 9 at 18:06
About your previous comment: That actually make sense, consider this example: Take $V=mathbb{R}^2$, then clearly $e_1,e_2$ a basis for $V$. However there exists a subspace $W$ of $V$, say $W=span((1,1))$ (the line $y=x$). This is indeed a subspace of $V$ but it doesn't include $e_1$ nor it includes $e_2$.
– Yanko
Dec 10 at 13:58
About your previous comment: That actually make sense, consider this example: Take $V=mathbb{R}^2$, then clearly $e_1,e_2$ a basis for $V$. However there exists a subspace $W$ of $V$, say $W=span((1,1))$ (the line $y=x$). This is indeed a subspace of $V$ but it doesn't include $e_1$ nor it includes $e_2$.
– Yanko
Dec 10 at 13:58
add a comment |
There are lots of ways to do this. One way is the following:
First you should find a basis for $U$ so you know what dimension it is.
After that you can start building a basis for $W$. One way to do this is to take a vector $(a,b,c)$ and set its dot product with each basis element of $U$ to 0.
This should lead you to a linearly independent vector - note these equations are underdetermined so be sure to choose a value for the remaining variables.
Keep adding vectors orthogonal to the previous ones until $W$ is of the correct dimension. Realistically you only need one or two.
answered Dec 9 at 14:44
Ben
2,488616
add a comment |
There are lots of ways to do this. One way is the following:
First you should find a basis for $U$ so you know what dimension it is.
After that you can start building a basis for $W$. One way to do this is to take a vector $(a,b,c)$ and set its dot product with each basis element of $U$ to 0.
This should lead you to a linearly independent vector - note these equations are underdetermined so be sure to choose a value for the remaining variables.
Keep adding vectors orthogonal to the previous ones until $W$ is of the correct dimension. Realistically you only need one or two.
answered Dec 9 at 14:44
Ben
2,488616
add a comment |
There are lots of ways to do this. One way is the following:
First you should find a basis for $U$ so you know what dimension it is.
After that you can start building a basis for $W$. One way to do this is to take a vector $(a,b,c)$ and set its dot product with each basis element of $U$ to 0.
This should lead you to a linearly independent vector - note these equations are underdetermined so be sure to choose a value for the remaining variables.
Keep adding vectors orthogonal to the previous ones until $W$ is of the correct dimension. Realistically you only need one or two.
answered Dec 9 at 14:44
Ben
2,488616
There are lots of ways to do this. One way is the following:
First you should find a basis for $U$ so you know what dimension it is.
After that you can start building a basis for $W$. One way to do this is to take a vector $(a,b,c)$ and set its dot product with each basis element of $U$ to 0.
This should lead you to a linearly independent vector - note these equations are underdetermined so be sure to choose a value for the remaining variables.
Keep adding vectors orthogonal to the previous ones until $W$ is of the correct dimension. Realistically you only need one or two.
answered Dec 9 at 14:44
Ben
2,488616
answered Dec 9 at 14:44
Ben
2,488616
answered Dec 9 at 14:44
Ben
2,488616
answered Dec 9 at 14:44
Ben
2,488616
2,488616
add a comment |
add a comment |
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