Proof that sum of two subspaces is another subspace
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$U_1,U_2$$⊂V$ be subspaces of V (a vector space). Define the subspace sum of $U_1,$ and $U_2$ be defined as the set:
$U_1 + U_2$ $=$ {$u_1 + u_2 : u_1 ∈ U_1, u_2 ∈ U_2$}.
Let $A$ denote the set $U_1+ U_2$
A is a subspace if a meets all the criteria of a subspace, that is, $0∈A$, it remains closed under addition, and it remains closed under multiplication.
Since $U_1$ is a subspace, by definition of subspaces it contains $au_1$ ($a∈R$), $0$ when a equals zero, and $u_1 + w_1$ (when $w_1∈U_1$).
Since $U_2$ is a subspace, by definition of subspaces it contains $au_2$ ($a∈R$), $0$ when a equals zero, and $u_2 + w_2$ (when $w_2∈U_2$).
$0u_1 + 0u_2 = 0(u_1 + u_2) = 0$; is an element of $A$
$au_1 + au_2 = a(u_1 + u_2)$; is an element of $A$
$(u_1 + u_2) + (w_1 + w_2) = (u_1 + w_1) + (u_2 + w_2)$ is an element of $A$
Q.E.D.
This is my proof, is it correct logically, symbolically etc., does it fall short of clarity, format/structure etc.?
In general what tips would you give to a young (a.k.a. not very mathematically mature) self-learner to improve their proofs. More specifically, what should I work on based on my proof.
proof-verification vector-spaces soft-question self-learning
asked Nov 23 at 14:11
oypus
518
add a comment |
up vote
1
down vote
favorite
$U_1,U_2$$⊂V$ be subspaces of V (a vector space). Define the subspace sum of $U_1,$ and $U_2$ be defined as the set:
$U_1 + U_2$ $=$ {$u_1 + u_2 : u_1 ∈ U_1, u_2 ∈ U_2$}.
Let $A$ denote the set $U_1+ U_2$
A is a subspace if a meets all the criteria of a subspace, that is, $0∈A$, it remains closed under addition, and it remains closed under multiplication.
Since $U_1$ is a subspace, by definition of subspaces it contains $au_1$ ($a∈R$), $0$ when a equals zero, and $u_1 + w_1$ (when $w_1∈U_1$).
Since $U_2$ is a subspace, by definition of subspaces it contains $au_2$ ($a∈R$), $0$ when a equals zero, and $u_2 + w_2$ (when $w_2∈U_2$).
$0u_1 + 0u_2 = 0(u_1 + u_2) = 0$; is an element of $A$
$au_1 + au_2 = a(u_1 + u_2)$; is an element of $A$
$(u_1 + u_2) + (w_1 + w_2) = (u_1 + w_1) + (u_2 + w_2)$ is an element of $A$
Q.E.D.
This is my proof, is it correct logically, symbolically etc., does it fall short of clarity, format/structure etc.?
In general what tips would you give to a young (a.k.a. not very mathematically mature) self-learner to improve their proofs. More specifically, what should I work on based on my proof.
proof-verification vector-spaces soft-question self-learning
asked Nov 23 at 14:11
oypus
518
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
$U_1,U_2$$⊂V$ be subspaces of V (a vector space). Define the subspace sum of $U_1,$ and $U_2$ be defined as the set:
$U_1 + U_2$ $=$ {$u_1 + u_2 : u_1 ∈ U_1, u_2 ∈ U_2$}.
Let $A$ denote the set $U_1+ U_2$
A is a subspace if a meets all the criteria of a subspace, that is, $0∈A$, it remains closed under addition, and it remains closed under multiplication.
Since $U_1$ is a subspace, by definition of subspaces it contains $au_1$ ($a∈R$), $0$ when a equals zero, and $u_1 + w_1$ (when $w_1∈U_1$).
Since $U_2$ is a subspace, by definition of subspaces it contains $au_2$ ($a∈R$), $0$ when a equals zero, and $u_2 + w_2$ (when $w_2∈U_2$).
$0u_1 + 0u_2 = 0(u_1 + u_2) = 0$; is an element of $A$
$au_1 + au_2 = a(u_1 + u_2)$; is an element of $A$
$(u_1 + u_2) + (w_1 + w_2) = (u_1 + w_1) + (u_2 + w_2)$ is an element of $A$
Q.E.D.
This is my proof, is it correct logically, symbolically etc., does it fall short of clarity, format/structure etc.?
In general what tips would you give to a young (a.k.a. not very mathematically mature) self-learner to improve their proofs. More specifically, what should I work on based on my proof.
proof-verification vector-spaces soft-question self-learning
asked Nov 23 at 14:11
oypus
518
$U_1,U_2$$⊂V$ be subspaces of V (a vector space). Define the subspace sum of $U_1,$ and $U_2$ be defined as the set:
$U_1 + U_2$ $=$ {$u_1 + u_2 : u_1 ∈ U_1, u_2 ∈ U_2$}.
Let $A$ denote the set $U_1+ U_2$
A is a subspace if a meets all the criteria of a subspace, that is, $0∈A$, it remains closed under addition, and it remains closed under multiplication.
Since $U_1$ is a subspace, by definition of subspaces it contains $au_1$ ($a∈R$), $0$ when a equals zero, and $u_1 + w_1$ (when $w_1∈U_1$).
Since $U_2$ is a subspace, by definition of subspaces it contains $au_2$ ($a∈R$), $0$ when a equals zero, and $u_2 + w_2$ (when $w_2∈U_2$).
$0u_1 + 0u_2 = 0(u_1 + u_2) = 0$; is an element of $A$
$au_1 + au_2 = a(u_1 + u_2)$; is an element of $A$
$(u_1 + u_2) + (w_1 + w_2) = (u_1 + w_1) + (u_2 + w_2)$ is an element of $A$
Q.E.D.
This is my proof, is it correct logically, symbolically etc., does it fall short of clarity, format/structure etc.?
In general what tips would you give to a young (a.k.a. not very mathematically mature) self-learner to improve their proofs. More specifically, what should I work on based on my proof.
proof-verification vector-spaces soft-question self-learning
proof-verification vector-spaces soft-question self-learning
asked Nov 23 at 14:11
oypus
518
asked Nov 23 at 14:11
oypus
518
asked Nov 23 at 14:11
oypus
518
asked Nov 23 at 14:11
oypus
518
asked Nov 23 at 14:11
oypus
518
518
add a comment |
add a comment |
1 Answer
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up vote
0
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Yes your proof is fine as a minor issue I would prefer to present the second and third properties in that way
$u_1 + u_2in U_1 + U_2 implies a(u_1 + u_2)=au_1 + au_2$ with $au_1in U_1 + U_2$ and $au_2in U_2 $
and
$(u_1 + u_2) + (w_1 + w_2)in U_1 + U_2 implies (u_1 + u_2) + (w_1 + w_2)=(u_1+w_1)+(u_2+w_2)$ and $(u_1+w_1)in U_1$, $(u_2+w_2)in U_2$
answered Nov 23 at 14:18
gimusi
93k94495
So would you classify this as a structure or clarity error?
– oypus
Nov 23 at 14:25
@oypus It is more an issue of clarity fro the presentation. We need to prove that $U_1 + U_2$ is a subspace then I would suggest to start form that set and prove that the elements satisfy the requested properties.
– gimusi
Nov 23 at 14:27
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Yes your proof is fine as a minor issue I would prefer to present the second and third properties in that way
$u_1 + u_2in U_1 + U_2 implies a(u_1 + u_2)=au_1 + au_2$ with $au_1in U_1 + U_2$ and $au_2in U_2 $
and
$(u_1 + u_2) + (w_1 + w_2)in U_1 + U_2 implies (u_1 + u_2) + (w_1 + w_2)=(u_1+w_1)+(u_2+w_2)$ and $(u_1+w_1)in U_1$, $(u_2+w_2)in U_2$
answered Nov 23 at 14:18
gimusi
93k94495
So would you classify this as a structure or clarity error?
– oypus
Nov 23 at 14:25
@oypus It is more an issue of clarity fro the presentation. We need to prove that $U_1 + U_2$ is a subspace then I would suggest to start form that set and prove that the elements satisfy the requested properties.
– gimusi
Nov 23 at 14:27
add a comment |
up vote
0
down vote
accepted
Yes your proof is fine as a minor issue I would prefer to present the second and third properties in that way
$u_1 + u_2in U_1 + U_2 implies a(u_1 + u_2)=au_1 + au_2$ with $au_1in U_1 + U_2$ and $au_2in U_2 $
and
$(u_1 + u_2) + (w_1 + w_2)in U_1 + U_2 implies (u_1 + u_2) + (w_1 + w_2)=(u_1+w_1)+(u_2+w_2)$ and $(u_1+w_1)in U_1$, $(u_2+w_2)in U_2$
answered Nov 23 at 14:18
gimusi
93k94495
So would you classify this as a structure or clarity error?
– oypus
Nov 23 at 14:25
@oypus It is more an issue of clarity fro the presentation. We need to prove that $U_1 + U_2$ is a subspace then I would suggest to start form that set and prove that the elements satisfy the requested properties.
– gimusi
Nov 23 at 14:27
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Yes your proof is fine as a minor issue I would prefer to present the second and third properties in that way
$u_1 + u_2in U_1 + U_2 implies a(u_1 + u_2)=au_1 + au_2$ with $au_1in U_1 + U_2$ and $au_2in U_2 $
and
$(u_1 + u_2) + (w_1 + w_2)in U_1 + U_2 implies (u_1 + u_2) + (w_1 + w_2)=(u_1+w_1)+(u_2+w_2)$ and $(u_1+w_1)in U_1$, $(u_2+w_2)in U_2$
answered Nov 23 at 14:18
gimusi
93k94495
Yes your proof is fine as a minor issue I would prefer to present the second and third properties in that way
$u_1 + u_2in U_1 + U_2 implies a(u_1 + u_2)=au_1 + au_2$ with $au_1in U_1 + U_2$ and $au_2in U_2 $
and
$(u_1 + u_2) + (w_1 + w_2)in U_1 + U_2 implies (u_1 + u_2) + (w_1 + w_2)=(u_1+w_1)+(u_2+w_2)$ and $(u_1+w_1)in U_1$, $(u_2+w_2)in U_2$
answered Nov 23 at 14:18
gimusi
93k94495
edited Nov 23 at 14:29
answered Nov 23 at 14:18
gimusi
93k94495
answered Nov 23 at 14:18
gimusi
93k94495
answered Nov 23 at 14:18
gimusi
93k94495
93k94495
So would you classify this as a structure or clarity error?
– oypus
Nov 23 at 14:25
@oypus It is more an issue of clarity fro the presentation. We need to prove that $U_1 + U_2$ is a subspace then I would suggest to start form that set and prove that the elements satisfy the requested properties.
– gimusi
Nov 23 at 14:27
add a comment |
So would you classify this as a structure or clarity error?
– oypus
Nov 23 at 14:25
@oypus It is more an issue of clarity fro the presentation. We need to prove that $U_1 + U_2$ is a subspace then I would suggest to start form that set and prove that the elements satisfy the requested properties.
– gimusi
Nov 23 at 14:27
So would you classify this as a structure or clarity error?
– oypus
Nov 23 at 14:25
So would you classify this as a structure or clarity error?
– oypus
Nov 23 at 14:25
@oypus It is more an issue of clarity fro the presentation. We need to prove that $U_1 + U_2$ is a subspace then I would suggest to start form that set and prove that the elements satisfy the requested properties.
– gimusi
Nov 23 at 14:27
@oypus It is more an issue of clarity fro the presentation. We need to prove that $U_1 + U_2$ is a subspace then I would suggest to start form that set and prove that the elements satisfy the requested properties.
– gimusi
Nov 23 at 14:27
add a comment |
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