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How to show that the set $W=(x,y,z):x+y=0$ is a subspace of the vector space $V_3(mathbbR)$? [closed]
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Let alpha=(x,y,z) and beta=(u,v,w) be any two elements of W.
Then x,y,z,u,v,w are the elements of R and are such that
($x+u=0).....(1)
($y+v=0)......(2)
($z+w=0).....(3)
$If a,b be any two elements of R,we have
a $alpha+ b beta=a(x,y,z)+b(u,v,w)
=(ax+bu),(ay+bv),(az+bw)
Now,. ax+bu+ay+bv+az+bw=a(x+y+z)+b(u+v+w)??
This is the way I did this question,where I went wrong?
linear-algebra
asked Jul 17 at 6:24
Anurag Malik
12
closed as off-topic by Alex Francisco, Leucippus, Shailesh, Delta-u, max_zorn Jul 17 at 7:36
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, Leucippus, Shailesh, Delta-u, max_zorn
 |Â
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up vote
-1
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Let alpha=(x,y,z) and beta=(u,v,w) be any two elements of W.
Then x,y,z,u,v,w are the elements of R and are such that
($x+u=0).....(1)
($y+v=0)......(2)
($z+w=0).....(3)
$If a,b be any two elements of R,we have
a $alpha+ b beta=a(x,y,z)+b(u,v,w)
=(ax+bu),(ay+bv),(az+bw)
Now,. ax+bu+ay+bv+az+bw=a(x+y+z)+b(u+v+w)??
This is the way I did this question,where I went wrong?
linear-algebra
asked Jul 17 at 6:24
Anurag Malik
12
closed as off-topic by Alex Francisco, Leucippus, Shailesh, Delta-u, max_zorn Jul 17 at 7:36
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, Leucippus, Shailesh, Delta-u, max_zorn
A subset of a vector space needs to fulfill three simple rules in order to be a subspace. Do you know what those rules are?
– Arthur
Jul 17 at 6:30
1
Zero vector should be in the set W.
– Anurag Malik
Jul 17 at 6:31
Good, that's rule 1. Now, is the zero vector in the set $W$?
– Arthur
Jul 17 at 6:33
Scalar multiplication and vector addition.
– Anurag Malik
Jul 17 at 6:33
That's rule 2 and 3. So, 2) if you add two vectors from $W$, is the result still always in $W$, no matter which two vectors you added? And 3) if you scale a vector from $W$, is the result in $W$ no matter which vector you started with and which scalar you used? Those are the questions you need to answer. And we would really appreciate it if you edited your question post to show what you have tried for each rule.
– Arthur
Jul 17 at 6:35
 |Â
show 1 more comment
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Let alpha=(x,y,z) and beta=(u,v,w) be any two elements of W.
Then x,y,z,u,v,w are the elements of R and are such that
($x+u=0).....(1)
($y+v=0)......(2)
($z+w=0).....(3)
$If a,b be any two elements of R,we have
a $alpha+ b beta=a(x,y,z)+b(u,v,w)
=(ax+bu),(ay+bv),(az+bw)
Now,. ax+bu+ay+bv+az+bw=a(x+y+z)+b(u+v+w)??
This is the way I did this question,where I went wrong?
linear-algebra
asked Jul 17 at 6:24
Anurag Malik
12
Let alpha=(x,y,z) and beta=(u,v,w) be any two elements of W.
Then x,y,z,u,v,w are the elements of R and are such that
($x+u=0).....(1)
($y+v=0)......(2)
($z+w=0).....(3)
$If a,b be any two elements of R,we have
a $alpha+ b beta=a(x,y,z)+b(u,v,w)
=(ax+bu),(ay+bv),(az+bw)
Now,. ax+bu+ay+bv+az+bw=a(x+y+z)+b(u+v+w)??
This is the way I did this question,where I went wrong?
linear-algebra
asked Jul 17 at 6:24
Anurag Malik
12
edited Jul 19 at 5:34
asked Jul 17 at 6:24
Anurag Malik
12
asked Jul 17 at 6:24
Anurag Malik
12
asked Jul 17 at 6:24
Anurag Malik
12
12
closed as off-topic by Alex Francisco, Leucippus, Shailesh, Delta-u, max_zorn Jul 17 at 7:36
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, Leucippus, Shailesh, Delta-u, max_zorn
closed as off-topic by Alex Francisco, Leucippus, Shailesh, Delta-u, max_zorn Jul 17 at 7:36
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, Leucippus, Shailesh, Delta-u, max_zorn
A subset of a vector space needs to fulfill three simple rules in order to be a subspace. Do you know what those rules are?
– Arthur
Jul 17 at 6:30
1
Zero vector should be in the set W.
– Anurag Malik
Jul 17 at 6:31
Good, that's rule 1. Now, is the zero vector in the set $W$?
– Arthur
Jul 17 at 6:33
Scalar multiplication and vector addition.
– Anurag Malik
Jul 17 at 6:33
That's rule 2 and 3. So, 2) if you add two vectors from $W$, is the result still always in $W$, no matter which two vectors you added? And 3) if you scale a vector from $W$, is the result in $W$ no matter which vector you started with and which scalar you used? Those are the questions you need to answer. And we would really appreciate it if you edited your question post to show what you have tried for each rule.
– Arthur
Jul 17 at 6:35
 |Â
show 1 more comment
A subset of a vector space needs to fulfill three simple rules in order to be a subspace. Do you know what those rules are?
– Arthur
Jul 17 at 6:30
1
Zero vector should be in the set W.
– Anurag Malik
Jul 17 at 6:31
Good, that's rule 1. Now, is the zero vector in the set $W$?
– Arthur
Jul 17 at 6:33
Scalar multiplication and vector addition.
– Anurag Malik
Jul 17 at 6:33
That's rule 2 and 3. So, 2) if you add two vectors from $W$, is the result still always in $W$, no matter which two vectors you added? And 3) if you scale a vector from $W$, is the result in $W$ no matter which vector you started with and which scalar you used? Those are the questions you need to answer. And we would really appreciate it if you edited your question post to show what you have tried for each rule.
– Arthur
Jul 17 at 6:35
A subset of a vector space needs to fulfill three simple rules in order to be a subspace. Do you know what those rules are?
– Arthur
Jul 17 at 6:30
A subset of a vector space needs to fulfill three simple rules in order to be a subspace. Do you know what those rules are?
– Arthur
Jul 17 at 6:30
1
1
Zero vector should be in the set W.
– Anurag Malik
Jul 17 at 6:31
Zero vector should be in the set W.
– Anurag Malik
Jul 17 at 6:31
Good, that's rule 1. Now, is the zero vector in the set $W$?
– Arthur
Jul 17 at 6:33
Good, that's rule 1. Now, is the zero vector in the set $W$?
– Arthur
Jul 17 at 6:33
Scalar multiplication and vector addition.
– Anurag Malik
Jul 17 at 6:33
Scalar multiplication and vector addition.
– Anurag Malik
Jul 17 at 6:33
That's rule 2 and 3. So, 2) if you add two vectors from $W$, is the result still always in $W$, no matter which two vectors you added? And 3) if you scale a vector from $W$, is the result in $W$ no matter which vector you started with and which scalar you used? Those are the questions you need to answer. And we would really appreciate it if you edited your question post to show what you have tried for each rule.
– Arthur
Jul 17 at 6:35
That's rule 2 and 3. So, 2) if you add two vectors from $W$, is the result still always in $W$, no matter which two vectors you added? And 3) if you scale a vector from $W$, is the result in $W$ no matter which vector you started with and which scalar you used? Those are the questions you need to answer. And we would really appreciate it if you edited your question post to show what you have tried for each rule.
– Arthur
Jul 17 at 6:35
 |Â
show 1 more comment
1 Answer
1
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oldest
votes
up vote
0
down vote
accepted
You have to show: if $(x,y,z), (u,v,w) in W$ and $ alpha in mathbb R$ then
$(x+u,y+v,z+w) in W$ and $(alpha x,alpha y,alpha z) in W$.
answered Jul 17 at 6:39
Fred
37.6k1237
You're forgetting $0in W$. Which assuming your properties is really equivalent to $Wneqvarnothing$, so it's quite trivial. Nevertheless it must be pointed out of a proof is to be complete.
– Arthur
Jul 17 at 6:41
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
You have to show: if $(x,y,z), (u,v,w) in W$ and $ alpha in mathbb R$ then
$(x+u,y+v,z+w) in W$ and $(alpha x,alpha y,alpha z) in W$.
answered Jul 17 at 6:39
Fred
37.6k1237
You're forgetting $0in W$. Which assuming your properties is really equivalent to $Wneqvarnothing$, so it's quite trivial. Nevertheless it must be pointed out of a proof is to be complete.
– Arthur
Jul 17 at 6:41
add a comment |Â
up vote
0
down vote
accepted
You have to show: if $(x,y,z), (u,v,w) in W$ and $ alpha in mathbb R$ then
$(x+u,y+v,z+w) in W$ and $(alpha x,alpha y,alpha z) in W$.
answered Jul 17 at 6:39
Fred
37.6k1237
You're forgetting $0in W$. Which assuming your properties is really equivalent to $Wneqvarnothing$, so it's quite trivial. Nevertheless it must be pointed out of a proof is to be complete.
– Arthur
Jul 17 at 6:41
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
You have to show: if $(x,y,z), (u,v,w) in W$ and $ alpha in mathbb R$ then
$(x+u,y+v,z+w) in W$ and $(alpha x,alpha y,alpha z) in W$.
answered Jul 17 at 6:39
Fred
37.6k1237
You have to show: if $(x,y,z), (u,v,w) in W$ and $ alpha in mathbb R$ then
$(x+u,y+v,z+w) in W$ and $(alpha x,alpha y,alpha z) in W$.
answered Jul 17 at 6:39
Fred
37.6k1237
answered Jul 17 at 6:39
Fred
37.6k1237
answered Jul 17 at 6:39
Fred
37.6k1237
answered Jul 17 at 6:39
Fred
37.6k1237
37.6k1237
You're forgetting $0in W$. Which assuming your properties is really equivalent to $Wneqvarnothing$, so it's quite trivial. Nevertheless it must be pointed out of a proof is to be complete.
– Arthur
Jul 17 at 6:41
add a comment |Â
You're forgetting $0in W$. Which assuming your properties is really equivalent to $Wneqvarnothing$, so it's quite trivial. Nevertheless it must be pointed out of a proof is to be complete.
– Arthur
Jul 17 at 6:41
You're forgetting $0in W$. Which assuming your properties is really equivalent to $Wneqvarnothing$, so it's quite trivial. Nevertheless it must be pointed out of a proof is to be complete.
– Arthur
Jul 17 at 6:41
You're forgetting $0in W$. Which assuming your properties is really equivalent to $Wneqvarnothing$, so it's quite trivial. Nevertheless it must be pointed out of a proof is to be complete.
– Arthur
Jul 17 at 6:41
add a comment |Â
A subset of a vector space needs to fulfill three simple rules in order to be a subspace. Do you know what those rules are?
– Arthur
Jul 17 at 6:30
1
Zero vector should be in the set W.
– Anurag Malik
Jul 17 at 6:31
Good, that's rule 1. Now, is the zero vector in the set $W$?
– Arthur
Jul 17 at 6:33
Scalar multiplication and vector addition.
– Anurag Malik
Jul 17 at 6:33
That's rule 2 and 3. So, 2) if you add two vectors from $W$, is the result still always in $W$, no matter which two vectors you added? And 3) if you scale a vector from $W$, is the result in $W$ no matter which vector you started with and which scalar you used? Those are the questions you need to answer. And we would really appreciate it if you edited your question post to show what you have tried for each rule.
– Arthur
Jul 17 at 6:35