Prove sum of subspaces is the smallest containing subspace
$begingroup$
I understand that it takes three steps:
$U_1+...+U_m$ is a vector space- $U_1,...,U_m subset U_1+...+U_m$
- $U_1+...+U_m subset U_1,...,U_m$
I know how to prove point 1. And point 2 is OK since the sum must contain all its elements.
The problem is point 3. How to see that $U_1+...+U_m subset U_1,...,U_m$
linear-algebra
edited Dec 27 '18 at 23:33
Rafael Holanda
2,721722
asked Dec 27 '18 at 22:24
JOHN JOHN
4589
$endgroup$
add a comment |
$begingroup$
I understand that it takes three steps:
$U_1+...+U_m$ is a vector space- $U_1,...,U_m subset U_1+...+U_m$
- $U_1+...+U_m subset U_1,...,U_m$
I know how to prove point 1. And point 2 is OK since the sum must contain all its elements.
The problem is point 3. How to see that $U_1+...+U_m subset U_1,...,U_m$
linear-algebra
edited Dec 27 '18 at 23:33
Rafael Holanda
2,721722
asked Dec 27 '18 at 22:24
JOHN JOHN
4589
$endgroup$
1
$begingroup$
What is $V$ in this stuff?
$endgroup$
– Bernard
Dec 27 '18 at 22:27
$begingroup$
@Bernard is it better now?
$endgroup$
– JOHN
Dec 27 '18 at 22:34
add a comment |
$begingroup$
I understand that it takes three steps:
$U_1+...+U_m$ is a vector space- $U_1,...,U_m subset U_1+...+U_m$
- $U_1+...+U_m subset U_1,...,U_m$
I know how to prove point 1. And point 2 is OK since the sum must contain all its elements.
The problem is point 3. How to see that $U_1+...+U_m subset U_1,...,U_m$
linear-algebra
edited Dec 27 '18 at 23:33
Rafael Holanda
2,721722
asked Dec 27 '18 at 22:24
JOHN JOHN
4589
$endgroup$
I understand that it takes three steps:
$U_1+...+U_m$ is a vector space- $U_1,...,U_m subset U_1+...+U_m$
- $U_1+...+U_m subset U_1,...,U_m$
I know how to prove point 1. And point 2 is OK since the sum must contain all its elements.
The problem is point 3. How to see that $U_1+...+U_m subset U_1,...,U_m$
linear-algebra
linear-algebra
edited Dec 27 '18 at 23:33
Rafael Holanda
2,721722
asked Dec 27 '18 at 22:24
JOHN JOHN
4589
edited Dec 27 '18 at 23:33
Rafael Holanda
2,721722
asked Dec 27 '18 at 22:24
JOHN JOHN
4589
edited Dec 27 '18 at 23:33
Rafael Holanda
2,721722
edited Dec 27 '18 at 23:33
Rafael Holanda
2,721722
edited Dec 27 '18 at 23:33
Rafael Holanda
2,721722
2,721722
asked Dec 27 '18 at 22:24
JOHN JOHN
4589
asked Dec 27 '18 at 22:24
JOHN JOHN
4589
asked Dec 27 '18 at 22:24
JOHN JOHN
4589
4589
1
$begingroup$
What is $V$ in this stuff?
$endgroup$
– Bernard
Dec 27 '18 at 22:27
$begingroup$
@Bernard is it better now?
$endgroup$
– JOHN
Dec 27 '18 at 22:34
add a comment |
1
$begingroup$
What is $V$ in this stuff?
$endgroup$
– Bernard
Dec 27 '18 at 22:27
$begingroup$
@Bernard is it better now?
$endgroup$
– JOHN
Dec 27 '18 at 22:34
1
1
$begingroup$
What is $V$ in this stuff?
$endgroup$
– Bernard
Dec 27 '18 at 22:27
$begingroup$
What is $V$ in this stuff?
$endgroup$
– Bernard
Dec 27 '18 at 22:27
$begingroup$
@Bernard is it better now?
$endgroup$
– JOHN
Dec 27 '18 at 22:34
$begingroup$
@Bernard is it better now?
$endgroup$
– JOHN
Dec 27 '18 at 22:34
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You can't prove point $3$: it is false in general (except if all $U_i$s are contained in one of them).
Actually what remains to prove is that if a subspace $V$ contains $U_1,dots,U_m$, it contains their sum
$$U_1+dots+U_m=bigl{u_1+dots+u_mmid forall i=1,dots, m,;u_iin U_ibigr}.$$
This is clear, since if each $u_iin U_i$, it also belongs to $V$, which is a subspace, so their sum belongs to $V$. This proves that
$$U_1+dots+U_msubset V.$$
answered Dec 27 '18 at 22:48
BernardBernard
124k742117
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add a comment |
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$begingroup$
You can't prove point $3$: it is false in general (except if all $U_i$s are contained in one of them).
Actually what remains to prove is that if a subspace $V$ contains $U_1,dots,U_m$, it contains their sum
$$U_1+dots+U_m=bigl{u_1+dots+u_mmid forall i=1,dots, m,;u_iin U_ibigr}.$$
This is clear, since if each $u_iin U_i$, it also belongs to $V$, which is a subspace, so their sum belongs to $V$. This proves that
$$U_1+dots+U_msubset V.$$
answered Dec 27 '18 at 22:48
BernardBernard
124k742117
$endgroup$
add a comment |
$begingroup$
You can't prove point $3$: it is false in general (except if all $U_i$s are contained in one of them).
Actually what remains to prove is that if a subspace $V$ contains $U_1,dots,U_m$, it contains their sum
$$U_1+dots+U_m=bigl{u_1+dots+u_mmid forall i=1,dots, m,;u_iin U_ibigr}.$$
This is clear, since if each $u_iin U_i$, it also belongs to $V$, which is a subspace, so their sum belongs to $V$. This proves that
$$U_1+dots+U_msubset V.$$
answered Dec 27 '18 at 22:48
BernardBernard
124k742117
$endgroup$
add a comment |
$begingroup$
You can't prove point $3$: it is false in general (except if all $U_i$s are contained in one of them).
Actually what remains to prove is that if a subspace $V$ contains $U_1,dots,U_m$, it contains their sum
$$U_1+dots+U_m=bigl{u_1+dots+u_mmid forall i=1,dots, m,;u_iin U_ibigr}.$$
This is clear, since if each $u_iin U_i$, it also belongs to $V$, which is a subspace, so their sum belongs to $V$. This proves that
$$U_1+dots+U_msubset V.$$
answered Dec 27 '18 at 22:48
BernardBernard
124k742117
$endgroup$
You can't prove point $3$: it is false in general (except if all $U_i$s are contained in one of them).
Actually what remains to prove is that if a subspace $V$ contains $U_1,dots,U_m$, it contains their sum
$$U_1+dots+U_m=bigl{u_1+dots+u_mmid forall i=1,dots, m,;u_iin U_ibigr}.$$
This is clear, since if each $u_iin U_i$, it also belongs to $V$, which is a subspace, so their sum belongs to $V$. This proves that
$$U_1+dots+U_msubset V.$$
answered Dec 27 '18 at 22:48
BernardBernard
124k742117
answered Dec 27 '18 at 22:48
BernardBernard
124k742117
answered Dec 27 '18 at 22:48
BernardBernard
124k742117
answered Dec 27 '18 at 22:48
BernardBernard
124k742117
124k742117
add a comment |
add a comment |
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1
$begingroup$
What is $V$ in this stuff?
$endgroup$
– Bernard
Dec 27 '18 at 22:27
$begingroup$
@Bernard is it better now?
$endgroup$
– JOHN
Dec 27 '18 at 22:34