Equivalent characterization of inner direct sum of subspaces

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Let $V$ be a vector space and let $A, B$ be subspaces of V.
The sum of $A, B$ is the subspace of $V$ given by
$$
A + B = a+b : ain A, bin B
$$
Moreover if $A cap B=0$ then the sum of $A,B$ is called the inner direct sum of $A,B$ and is denoted with $A oplus B$.
Let $A, B, C$ be subspaces of $V$ with $C = A + B$.
Then
$$
C = A oplus B
iff
forall c in C, exists! ain A, exists! b in B : c = a + b
$$

If we replace on the right hand side $forall cin C$ with $exists cin C$ then does the equivalence still hold?

linear-algebra vector-spaces

share|cite|improve this question

edited Jul 15 at 15:31

Jendrik Stelzner

7,55211037

asked Jul 15 at 14:37

Perpendicular

596

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your question does not seem to be clear.
  – nature1729
  Jul 15 at 14:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Take $A cap B = 0$ and $C = A$, then for any $c in C$ (in particular: $exists c in C$ s.t. ...) there are unique $a in A$ and $b in B$ s.t. $c = a + b$, namely $a = c$ and $b = 0$, but $C neq A + B$, hence $C = A oplus B$ is not true.
  – qwertz
  Jul 15 at 14:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @qwertz i restated my quastion because it was not clear enough. Thank u very much for your answer
  – Perpendicular
  Jul 15 at 15:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @nature1729 u are right. I restated my question.
  – Perpendicular
  Jul 15 at 15:14
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @AnyAD unique. So $exists !$ means there is EXACTLY ONE
  – Perpendicular
  Jul 15 at 15:38
  
  

  
  
  
  

 | 
show 1 more comment

up vote
0
down vote

favorite

1

Let $V$ be a vector space and let $A, B$ be subspaces of V.
The sum of $A, B$ is the subspace of $V$ given by
$$
A + B = a+b : ain A, bin B
$$
Moreover if $A cap B=0$ then the sum of $A,B$ is called the inner direct sum of $A,B$ and is denoted with $A oplus B$.
Let $A, B, C$ be subspaces of $V$ with $C = A + B$.
Then
$$
C = A oplus B
iff
forall c in C, exists! ain A, exists! b in B : c = a + b
$$

If we replace on the right hand side $forall cin C$ with $exists cin C$ then does the equivalence still hold?

linear-algebra vector-spaces

share|cite|improve this question

edited Jul 15 at 15:31

Jendrik Stelzner

7,55211037

asked Jul 15 at 14:37

Perpendicular

596

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your question does not seem to be clear.
  – nature1729
  Jul 15 at 14:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Take $A cap B = 0$ and $C = A$, then for any $c in C$ (in particular: $exists c in C$ s.t. ...) there are unique $a in A$ and $b in B$ s.t. $c = a + b$, namely $a = c$ and $b = 0$, but $C neq A + B$, hence $C = A oplus B$ is not true.
  – qwertz
  Jul 15 at 14:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @qwertz i restated my quastion because it was not clear enough. Thank u very much for your answer
  – Perpendicular
  Jul 15 at 15:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @nature1729 u are right. I restated my question.
  – Perpendicular
  Jul 15 at 15:14
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @AnyAD unique. So $exists !$ means there is EXACTLY ONE
  – Perpendicular
  Jul 15 at 15:38
  
  

  
  
  
  

 | 
show 1 more comment

up vote
0
down vote

favorite

1

up vote
0
down vote

favorite

1

1

Let $V$ be a vector space and let $A, B$ be subspaces of V.
The sum of $A, B$ is the subspace of $V$ given by
$$
A + B = a+b : ain A, bin B
$$
Moreover if $A cap B=0$ then the sum of $A,B$ is called the inner direct sum of $A,B$ and is denoted with $A oplus B$.
Let $A, B, C$ be subspaces of $V$ with $C = A + B$.
Then
$$
C = A oplus B
iff
forall c in C, exists! ain A, exists! b in B : c = a + b
$$

If we replace on the right hand side $forall cin C$ with $exists cin C$ then does the equivalence still hold?

linear-algebra vector-spaces

share|cite|improve this question

edited Jul 15 at 15:31

Jendrik Stelzner

7,55211037

asked Jul 15 at 14:37

Perpendicular

596

Let $V$ be a vector space and let $A, B$ be subspaces of V.
The sum of $A, B$ is the subspace of $V$ given by
$$
A + B = a+b : ain A, bin B
$$
Moreover if $A cap B=0$ then the sum of $A,B$ is called the inner direct sum of $A,B$ and is denoted with $A oplus B$.
Let $A, B, C$ be subspaces of $V$ with $C = A + B$.
Then
$$
C = A oplus B
iff
forall c in C, exists! ain A, exists! b in B : c = a + b
$$

If we replace on the right hand side $forall cin C$ with $exists cin C$ then does the equivalence still hold?

linear-algebra vector-spaces

share|cite|improve this question

edited Jul 15 at 15:31

Jendrik Stelzner

7,55211037

asked Jul 15 at 14:37

Perpendicular

596

share|cite|improve this question

share|cite|improve this question

edited Jul 15 at 15:31

Jendrik Stelzner

7,55211037

edited Jul 15 at 15:31

Jendrik Stelzner

7,55211037

edited Jul 15 at 15:31

Jendrik Stelzner

7,55211037

7,55211037

asked Jul 15 at 14:37

Perpendicular

596

asked Jul 15 at 14:37

Perpendicular

596

asked Jul 15 at 14:37

Perpendicular

596

596

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your question does not seem to be clear.
  – nature1729
  Jul 15 at 14:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Take $A cap B = 0$ and $C = A$, then for any $c in C$ (in particular: $exists c in C$ s.t. ...) there are unique $a in A$ and $b in B$ s.t. $c = a + b$, namely $a = c$ and $b = 0$, but $C neq A + B$, hence $C = A oplus B$ is not true.
  – qwertz
  Jul 15 at 14:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @qwertz i restated my quastion because it was not clear enough. Thank u very much for your answer
  – Perpendicular
  Jul 15 at 15:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @nature1729 u are right. I restated my question.
  – Perpendicular
  Jul 15 at 15:14
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @AnyAD unique. So $exists !$ means there is EXACTLY ONE
  – Perpendicular
  Jul 15 at 15:38
  
  

  
  
  
  

 | 
show 1 more comment

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your question does not seem to be clear.
  – nature1729
  Jul 15 at 14:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Take $A cap B = 0$ and $C = A$, then for any $c in C$ (in particular: $exists c in C$ s.t. ...) there are unique $a in A$ and $b in B$ s.t. $c = a + b$, namely $a = c$ and $b = 0$, but $C neq A + B$, hence $C = A oplus B$ is not true.
  – qwertz
  Jul 15 at 14:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @qwertz i restated my quastion because it was not clear enough. Thank u very much for your answer
  – Perpendicular
  Jul 15 at 15:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @nature1729 u are right. I restated my question.
  – Perpendicular
  Jul 15 at 15:14
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @AnyAD unique. So $exists !$ means there is EXACTLY ONE
  – Perpendicular
  Jul 15 at 15:38
  
  

  
  
  
  

Your question does not seem to be clear.
– nature1729
Jul 15 at 14:51

Your question does not seem to be clear.
– nature1729
Jul 15 at 14:51

Take $A cap B = 0$ and $C = A$, then for any $c in C$ (in particular: $exists c in C$ s.t. ...) there are unique $a in A$ and $b in B$ s.t. $c = a + b$, namely $a = c$ and $b = 0$, but $C neq A + B$, hence $C = A oplus B$ is not true.
– qwertz
Jul 15 at 14:54

Take $A cap B = 0$ and $C = A$, then for any $c in C$ (in particular: $exists c in C$ s.t. ...) there are unique $a in A$ and $b in B$ s.t. $c = a + b$, namely $a = c$ and $b = 0$, but $C neq A + B$, hence $C = A oplus B$ is not true.
– qwertz
Jul 15 at 14:54

@qwertz i restated my quastion because it was not clear enough. Thank u very much for your answer
– Perpendicular
Jul 15 at 15:14

@qwertz i restated my quastion because it was not clear enough. Thank u very much for your answer
– Perpendicular
Jul 15 at 15:14

@nature1729 u are right. I restated my question.
– Perpendicular
Jul 15 at 15:14

@nature1729 u are right. I restated my question.
– Perpendicular
Jul 15 at 15:14

1

1

@AnyAD unique. So $exists !$ means there is EXACTLY ONE
– Perpendicular
Jul 15 at 15:38

@AnyAD unique. So $exists !$ means there is EXACTLY ONE
– Perpendicular
Jul 15 at 15:38

 | 
show 1 more comment

1 Answer
1

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up vote
1
down vote

The equivalence still holds:
We have that
beginalign*
C = A oplus B
&iff
forall c in C, exists! ain A, exists! b in B : c = a + b
\
&,implies
exists c in C, exists! ain A, exists! b in B : c = a + b
endalign*
because we can choose $c = 0$.
To show the other implication
$$
exists c in C, exists! ain A, exists! b in B : c = a + b
implies
forall c in C, exists! ain A, exists! b in B : c = a + b
$$
let $c in C$ such that for $c = a + b$ with $a in A$, $b in B$ the summands $a, b$ are unique.

Let $x in A cap B$.
It then follows from $x in A$ that $a + x in A$ and it follows from $x in B$ that $b - x in B$.
We have that
$$
c
= a + b
= (a + x) + (b - x)
$$
so it follows from the uniqueness of $a, b$ that $a+x = a$ and $b-x = b$.
Both equations show that $x = 0$.
This shows that $A cap B = 0$ and together with $C = A+B$ that $C = A oplus B$.

For subspaces $A, B, C subseteq V$ with $C = A + B$ one can show more generally that the following conditions are all equivalent:

• $C = A oplus B$, i.e. $A cap B = 0$


• $forall c in C: exists! a in A exists! b in B: c = a + b$


• $forall c in C: exists! a in A exists b in B: c = a + b$


• $forall c in C: exists a in A exists! b in B: c = a + b$


• $exists! a in A exists! b in B: 0 = a + b$


• $exists! a in A exists b in B: 0 = a + b$


• $exists a in A exists! b in B: 0 = a + b$

share|cite|improve this answer

answered Jul 15 at 15:44

Jendrik Stelzner

7,55211037

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are excellent
  – Perpendicular
  Jul 15 at 15:47
  

  
  
  
  

add a comment | 

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1 Answer
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active

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1 Answer
1

active

oldest

votes

active

oldest

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active

oldest

votes

up vote
1
down vote

The equivalence still holds:
We have that
beginalign*
C = A oplus B
&iff
forall c in C, exists! ain A, exists! b in B : c = a + b
\
&,implies
exists c in C, exists! ain A, exists! b in B : c = a + b
endalign*
because we can choose $c = 0$.
To show the other implication
$$
exists c in C, exists! ain A, exists! b in B : c = a + b
implies
forall c in C, exists! ain A, exists! b in B : c = a + b
$$
let $c in C$ such that for $c = a + b$ with $a in A$, $b in B$ the summands $a, b$ are unique.

Let $x in A cap B$.
It then follows from $x in A$ that $a + x in A$ and it follows from $x in B$ that $b - x in B$.
We have that
$$
c
= a + b
= (a + x) + (b - x)
$$
so it follows from the uniqueness of $a, b$ that $a+x = a$ and $b-x = b$.
Both equations show that $x = 0$.
This shows that $A cap B = 0$ and together with $C = A+B$ that $C = A oplus B$.

For subspaces $A, B, C subseteq V$ with $C = A + B$ one can show more generally that the following conditions are all equivalent:

• $C = A oplus B$, i.e. $A cap B = 0$


• $forall c in C: exists! a in A exists! b in B: c = a + b$


• $forall c in C: exists! a in A exists b in B: c = a + b$


• $forall c in C: exists a in A exists! b in B: c = a + b$


• $exists! a in A exists! b in B: 0 = a + b$


• $exists! a in A exists b in B: 0 = a + b$


• $exists a in A exists! b in B: 0 = a + b$

share|cite|improve this answer

answered Jul 15 at 15:44

Jendrik Stelzner

7,55211037

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are excellent
  – Perpendicular
  Jul 15 at 15:47
  

  
  
  
  

add a comment | 

up vote
1
down vote

The equivalence still holds:
We have that
beginalign*
C = A oplus B
&iff
forall c in C, exists! ain A, exists! b in B : c = a + b
\
&,implies
exists c in C, exists! ain A, exists! b in B : c = a + b
endalign*
because we can choose $c = 0$.
To show the other implication
$$
exists c in C, exists! ain A, exists! b in B : c = a + b
implies
forall c in C, exists! ain A, exists! b in B : c = a + b
$$
let $c in C$ such that for $c = a + b$ with $a in A$, $b in B$ the summands $a, b$ are unique.

Let $x in A cap B$.
It then follows from $x in A$ that $a + x in A$ and it follows from $x in B$ that $b - x in B$.
We have that
$$
c
= a + b
= (a + x) + (b - x)
$$
so it follows from the uniqueness of $a, b$ that $a+x = a$ and $b-x = b$.
Both equations show that $x = 0$.
This shows that $A cap B = 0$ and together with $C = A+B$ that $C = A oplus B$.

For subspaces $A, B, C subseteq V$ with $C = A + B$ one can show more generally that the following conditions are all equivalent:

• $C = A oplus B$, i.e. $A cap B = 0$


• $forall c in C: exists! a in A exists! b in B: c = a + b$


• $forall c in C: exists! a in A exists b in B: c = a + b$


• $forall c in C: exists a in A exists! b in B: c = a + b$


• $exists! a in A exists! b in B: 0 = a + b$


• $exists! a in A exists b in B: 0 = a + b$


• $exists a in A exists! b in B: 0 = a + b$

share|cite|improve this answer

answered Jul 15 at 15:44

Jendrik Stelzner

7,55211037

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are excellent
  – Perpendicular
  Jul 15 at 15:47
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

The equivalence still holds:
We have that
beginalign*
C = A oplus B
&iff
forall c in C, exists! ain A, exists! b in B : c = a + b
\
&,implies
exists c in C, exists! ain A, exists! b in B : c = a + b
endalign*
because we can choose $c = 0$.
To show the other implication
$$
exists c in C, exists! ain A, exists! b in B : c = a + b
implies
forall c in C, exists! ain A, exists! b in B : c = a + b
$$
let $c in C$ such that for $c = a + b$ with $a in A$, $b in B$ the summands $a, b$ are unique.

Let $x in A cap B$.
It then follows from $x in A$ that $a + x in A$ and it follows from $x in B$ that $b - x in B$.
We have that
$$
c
= a + b
= (a + x) + (b - x)
$$
so it follows from the uniqueness of $a, b$ that $a+x = a$ and $b-x = b$.
Both equations show that $x = 0$.
This shows that $A cap B = 0$ and together with $C = A+B$ that $C = A oplus B$.

For subspaces $A, B, C subseteq V$ with $C = A + B$ one can show more generally that the following conditions are all equivalent:

• $C = A oplus B$, i.e. $A cap B = 0$


• $forall c in C: exists! a in A exists! b in B: c = a + b$


• $forall c in C: exists! a in A exists b in B: c = a + b$


• $forall c in C: exists a in A exists! b in B: c = a + b$


• $exists! a in A exists! b in B: 0 = a + b$


• $exists! a in A exists b in B: 0 = a + b$


• $exists a in A exists! b in B: 0 = a + b$

share|cite|improve this answer

answered Jul 15 at 15:44

Jendrik Stelzner

7,55211037

The equivalence still holds:
We have that
beginalign*
C = A oplus B
&iff
forall c in C, exists! ain A, exists! b in B : c = a + b
\
&,implies
exists c in C, exists! ain A, exists! b in B : c = a + b
endalign*
because we can choose $c = 0$.
To show the other implication
$$
exists c in C, exists! ain A, exists! b in B : c = a + b
implies
forall c in C, exists! ain A, exists! b in B : c = a + b
$$
let $c in C$ such that for $c = a + b$ with $a in A$, $b in B$ the summands $a, b$ are unique.

Let $x in A cap B$.
It then follows from $x in A$ that $a + x in A$ and it follows from $x in B$ that $b - x in B$.
We have that
$$
c
= a + b
= (a + x) + (b - x)
$$
so it follows from the uniqueness of $a, b$ that $a+x = a$ and $b-x = b$.
Both equations show that $x = 0$.
This shows that $A cap B = 0$ and together with $C = A+B$ that $C = A oplus B$.

For subspaces $A, B, C subseteq V$ with $C = A + B$ one can show more generally that the following conditions are all equivalent:

• $C = A oplus B$, i.e. $A cap B = 0$


• $forall c in C: exists! a in A exists! b in B: c = a + b$


• $forall c in C: exists! a in A exists b in B: c = a + b$


• $forall c in C: exists a in A exists! b in B: c = a + b$


• $exists! a in A exists! b in B: 0 = a + b$


• $exists! a in A exists b in B: 0 = a + b$


• $exists a in A exists! b in B: 0 = a + b$

share|cite|improve this answer

answered Jul 15 at 15:44

Jendrik Stelzner

7,55211037

share|cite|improve this answer

share|cite|improve this answer

answered Jul 15 at 15:44

Jendrik Stelzner

7,55211037

answered Jul 15 at 15:44

Jendrik Stelzner

7,55211037

answered Jul 15 at 15:44

Jendrik Stelzner

7,55211037

7,55211037

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are excellent
  – Perpendicular
  Jul 15 at 15:47
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are excellent
  – Perpendicular
  Jul 15 at 15:47
  

  
  
  
  

You are excellent
– Perpendicular
Jul 15 at 15:47

You are excellent
– Perpendicular
Jul 15 at 15:47

add a comment | 

 

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