Linear Algebra - Intersection of Affine Spaces
$begingroup$
Let V be a vector space, $W_1, W_2$ are sub-spaces of $V$.
$v_1, v_2 in V$ and $(v_1 + W_1) cap(v_2 + W_2) neq emptyset$.
Prove that $(v_1 + W_1) cap(v_2 + W_2)$ is an affine space, i.e. there exists a sub-space $W_3$ of $V$ and $v_3 in V$ so that $(v_1 + W_1) cap(v_2 + W_2) = v_3 + W_3 $.
I have found this previous question but I couldn't figure out the next steps of proving this.
We know that $exists x in (v_1 + W_1) cap(v_2 + W_2) $.
I have no clue how to go on from here. I think I can show that since the intersection is not empty, for all $w_1 in W_1, w_2 in W_2 , w_1 = w_2$.
Would appreciate some points and guidelines about how to approach this.
linear-algebra vector-spaces
asked Dec 24 '18 at 15:37
TegernakoTegernako
968
$endgroup$
add a comment |
$begingroup$
Let V be a vector space, $W_1, W_2$ are sub-spaces of $V$.
$v_1, v_2 in V$ and $(v_1 + W_1) cap(v_2 + W_2) neq emptyset$.
Prove that $(v_1 + W_1) cap(v_2 + W_2)$ is an affine space, i.e. there exists a sub-space $W_3$ of $V$ and $v_3 in V$ so that $(v_1 + W_1) cap(v_2 + W_2) = v_3 + W_3 $.
I have found this previous question but I couldn't figure out the next steps of proving this.
We know that $exists x in (v_1 + W_1) cap(v_2 + W_2) $.
I have no clue how to go on from here. I think I can show that since the intersection is not empty, for all $w_1 in W_1, w_2 in W_2 , w_1 = w_2$.
Would appreciate some points and guidelines about how to approach this.
linear-algebra vector-spaces
asked Dec 24 '18 at 15:37
TegernakoTegernako
968
$endgroup$
1
$begingroup$
It should be more or less obvious (especially if you draw a picture) that the choice of 'offset' (i.e., choice of $v_3$) is arbitrary and can be any point in the intersection.
$endgroup$
– SvanN
Dec 24 '18 at 15:44
1
$begingroup$
If $xin v_1+W_1$, then $v_1+W_1=x+W_1$ etc.
$endgroup$
– Lord Shark the Unknown
Dec 24 '18 at 15:53
add a comment |
$begingroup$
Let V be a vector space, $W_1, W_2$ are sub-spaces of $V$.
$v_1, v_2 in V$ and $(v_1 + W_1) cap(v_2 + W_2) neq emptyset$.
Prove that $(v_1 + W_1) cap(v_2 + W_2)$ is an affine space, i.e. there exists a sub-space $W_3$ of $V$ and $v_3 in V$ so that $(v_1 + W_1) cap(v_2 + W_2) = v_3 + W_3 $.
I have found this previous question but I couldn't figure out the next steps of proving this.
We know that $exists x in (v_1 + W_1) cap(v_2 + W_2) $.
I have no clue how to go on from here. I think I can show that since the intersection is not empty, for all $w_1 in W_1, w_2 in W_2 , w_1 = w_2$.
Would appreciate some points and guidelines about how to approach this.
linear-algebra vector-spaces
asked Dec 24 '18 at 15:37
TegernakoTegernako
968
$endgroup$
Let V be a vector space, $W_1, W_2$ are sub-spaces of $V$.
$v_1, v_2 in V$ and $(v_1 + W_1) cap(v_2 + W_2) neq emptyset$.
Prove that $(v_1 + W_1) cap(v_2 + W_2)$ is an affine space, i.e. there exists a sub-space $W_3$ of $V$ and $v_3 in V$ so that $(v_1 + W_1) cap(v_2 + W_2) = v_3 + W_3 $.
I have found this previous question but I couldn't figure out the next steps of proving this.
We know that $exists x in (v_1 + W_1) cap(v_2 + W_2) $.
I have no clue how to go on from here. I think I can show that since the intersection is not empty, for all $w_1 in W_1, w_2 in W_2 , w_1 = w_2$.
Would appreciate some points and guidelines about how to approach this.
linear-algebra vector-spaces
linear-algebra vector-spaces
asked Dec 24 '18 at 15:37
TegernakoTegernako
968
asked Dec 24 '18 at 15:37
TegernakoTegernako
968
asked Dec 24 '18 at 15:37
TegernakoTegernako
968
asked Dec 24 '18 at 15:37
TegernakoTegernako
968
asked Dec 24 '18 at 15:37
TegernakoTegernako
968
968
1
$begingroup$
It should be more or less obvious (especially if you draw a picture) that the choice of 'offset' (i.e., choice of $v_3$) is arbitrary and can be any point in the intersection.
$endgroup$
– SvanN
Dec 24 '18 at 15:44
1
$begingroup$
If $xin v_1+W_1$, then $v_1+W_1=x+W_1$ etc.
$endgroup$
– Lord Shark the Unknown
Dec 24 '18 at 15:53
add a comment |
1
$begingroup$
It should be more or less obvious (especially if you draw a picture) that the choice of 'offset' (i.e., choice of $v_3$) is arbitrary and can be any point in the intersection.
$endgroup$
– SvanN
Dec 24 '18 at 15:44
1
$begingroup$
If $xin v_1+W_1$, then $v_1+W_1=x+W_1$ etc.
$endgroup$
– Lord Shark the Unknown
Dec 24 '18 at 15:53
1
1
$begingroup$
It should be more or less obvious (especially if you draw a picture) that the choice of 'offset' (i.e., choice of $v_3$) is arbitrary and can be any point in the intersection.
$endgroup$
– SvanN
Dec 24 '18 at 15:44
$begingroup$
It should be more or less obvious (especially if you draw a picture) that the choice of 'offset' (i.e., choice of $v_3$) is arbitrary and can be any point in the intersection.
$endgroup$
– SvanN
Dec 24 '18 at 15:44
1
1
$begingroup$
If $xin v_1+W_1$, then $v_1+W_1=x+W_1$ etc.
$endgroup$
– Lord Shark the Unknown
Dec 24 '18 at 15:53
$begingroup$
If $xin v_1+W_1$, then $v_1+W_1=x+W_1$ etc.
$endgroup$
– Lord Shark the Unknown
Dec 24 '18 at 15:53
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The idea is to prove that $x+W_1 = v_1+W_1$ and $x+W_2=v_2+W_2$. And then it follows that
$(x+W_1)cap (x+W_2)=x+(W_1cap W_2)$.
answered Dec 24 '18 at 15:53
MirceaMircea
1736
$endgroup$
add a comment |
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$begingroup$
The idea is to prove that $x+W_1 = v_1+W_1$ and $x+W_2=v_2+W_2$. And then it follows that
$(x+W_1)cap (x+W_2)=x+(W_1cap W_2)$.
answered Dec 24 '18 at 15:53
MirceaMircea
1736
$endgroup$
add a comment |
$begingroup$
The idea is to prove that $x+W_1 = v_1+W_1$ and $x+W_2=v_2+W_2$. And then it follows that
$(x+W_1)cap (x+W_2)=x+(W_1cap W_2)$.
answered Dec 24 '18 at 15:53
MirceaMircea
1736
$endgroup$
add a comment |
$begingroup$
The idea is to prove that $x+W_1 = v_1+W_1$ and $x+W_2=v_2+W_2$. And then it follows that
$(x+W_1)cap (x+W_2)=x+(W_1cap W_2)$.
answered Dec 24 '18 at 15:53
MirceaMircea
1736
$endgroup$
The idea is to prove that $x+W_1 = v_1+W_1$ and $x+W_2=v_2+W_2$. And then it follows that
$(x+W_1)cap (x+W_2)=x+(W_1cap W_2)$.
answered Dec 24 '18 at 15:53
MirceaMircea
1736
answered Dec 24 '18 at 15:53
MirceaMircea
1736
answered Dec 24 '18 at 15:53
MirceaMircea
1736
answered Dec 24 '18 at 15:53
MirceaMircea
1736
1736
add a comment |
add a comment |
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$begingroup$
It should be more or less obvious (especially if you draw a picture) that the choice of 'offset' (i.e., choice of $v_3$) is arbitrary and can be any point in the intersection.
$endgroup$
– SvanN
Dec 24 '18 at 15:44
1
$begingroup$
If $xin v_1+W_1$, then $v_1+W_1=x+W_1$ etc.
$endgroup$
– Lord Shark the Unknown
Dec 24 '18 at 15:53