disprove: $A$, $B$, $C$ are pairwise disjoint subsets of $V$; $Acup B$, $Acup C$ are bases of $V$; then...
$begingroup$
Prove or disprove:
If $A$, $B$ and $C$ are pairwise disjoint subsets of vector space $V$, such that $Acup B$ and $Acup C$ are bases of $V$, then $Span(B)=Span(C)$.
I consider an example:
I take $A={(1,0)}$, $B={(0,1)}$ and $C={(1,1)}$ as subsets of $mathbb{R}^2$.
I know that this one example is enough to disprove the statement. However, I want to know how to go about with proving/disproving this claim, if I were to do it using no specific examples.
Would proof by contradiction where, we take
$(1)$ $A$ and $B$ such that ${A,B}$ is linearly independent, and
$(2)$ $C$, every element of which can be expressed as a linear combination of $A$ and $B$ (satisfying other conditions)
be enough/correct?
linear-algebra vector-spaces
asked Dec 24 '18 at 10:17
Za IraZa Ira
161115
$endgroup$
add a comment |
$begingroup$
Prove or disprove:
If $A$, $B$ and $C$ are pairwise disjoint subsets of vector space $V$, such that $Acup B$ and $Acup C$ are bases of $V$, then $Span(B)=Span(C)$.
I consider an example:
I take $A={(1,0)}$, $B={(0,1)}$ and $C={(1,1)}$ as subsets of $mathbb{R}^2$.
I know that this one example is enough to disprove the statement. However, I want to know how to go about with proving/disproving this claim, if I were to do it using no specific examples.
Would proof by contradiction where, we take
$(1)$ $A$ and $B$ such that ${A,B}$ is linearly independent, and
$(2)$ $C$, every element of which can be expressed as a linear combination of $A$ and $B$ (satisfying other conditions)
be enough/correct?
linear-algebra vector-spaces
asked Dec 24 '18 at 10:17
Za IraZa Ira
161115
$endgroup$
add a comment |
$begingroup$
Prove or disprove:
If $A$, $B$ and $C$ are pairwise disjoint subsets of vector space $V$, such that $Acup B$ and $Acup C$ are bases of $V$, then $Span(B)=Span(C)$.
I consider an example:
I take $A={(1,0)}$, $B={(0,1)}$ and $C={(1,1)}$ as subsets of $mathbb{R}^2$.
I know that this one example is enough to disprove the statement. However, I want to know how to go about with proving/disproving this claim, if I were to do it using no specific examples.
Would proof by contradiction where, we take
$(1)$ $A$ and $B$ such that ${A,B}$ is linearly independent, and
$(2)$ $C$, every element of which can be expressed as a linear combination of $A$ and $B$ (satisfying other conditions)
be enough/correct?
linear-algebra vector-spaces
asked Dec 24 '18 at 10:17
Za IraZa Ira
161115
$endgroup$
Prove or disprove:
If $A$, $B$ and $C$ are pairwise disjoint subsets of vector space $V$, such that $Acup B$ and $Acup C$ are bases of $V$, then $Span(B)=Span(C)$.
I consider an example:
I take $A={(1,0)}$, $B={(0,1)}$ and $C={(1,1)}$ as subsets of $mathbb{R}^2$.
I know that this one example is enough to disprove the statement. However, I want to know how to go about with proving/disproving this claim, if I were to do it using no specific examples.
Would proof by contradiction where, we take
$(1)$ $A$ and $B$ such that ${A,B}$ is linearly independent, and
$(2)$ $C$, every element of which can be expressed as a linear combination of $A$ and $B$ (satisfying other conditions)
be enough/correct?
linear-algebra vector-spaces
linear-algebra vector-spaces
asked Dec 24 '18 at 10:17
Za IraZa Ira
161115
asked Dec 24 '18 at 10:17
Za IraZa Ira
161115
asked Dec 24 '18 at 10:17
Za IraZa Ira
161115
asked Dec 24 '18 at 10:17
Za IraZa Ira
161115
asked Dec 24 '18 at 10:17
Za IraZa Ira
161115
161115
add a comment |
add a comment |
1 Answer
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$begingroup$
I think that in your case you can not do a general proof of contradiction as there exist choices of $A,B,C$ and $V$ where the statement is true.
So at some point you would need to make some assumptions as for example if $B,C$ contain exactly one element that those are not multiples of each other.
In total, I can not think of a way to disprove this statement generally as it holds in some cases.
But as you mentioned, your solution is perfectly fine.
answered Dec 24 '18 at 10:27
Jonas LenzJonas Lenz
694215
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
I think that in your case you can not do a general proof of contradiction as there exist choices of $A,B,C$ and $V$ where the statement is true.
So at some point you would need to make some assumptions as for example if $B,C$ contain exactly one element that those are not multiples of each other.
In total, I can not think of a way to disprove this statement generally as it holds in some cases.
But as you mentioned, your solution is perfectly fine.
answered Dec 24 '18 at 10:27
Jonas LenzJonas Lenz
694215
$endgroup$
add a comment |
$begingroup$
I think that in your case you can not do a general proof of contradiction as there exist choices of $A,B,C$ and $V$ where the statement is true.
So at some point you would need to make some assumptions as for example if $B,C$ contain exactly one element that those are not multiples of each other.
In total, I can not think of a way to disprove this statement generally as it holds in some cases.
But as you mentioned, your solution is perfectly fine.
answered Dec 24 '18 at 10:27
Jonas LenzJonas Lenz
694215
$endgroup$
add a comment |
$begingroup$
I think that in your case you can not do a general proof of contradiction as there exist choices of $A,B,C$ and $V$ where the statement is true.
So at some point you would need to make some assumptions as for example if $B,C$ contain exactly one element that those are not multiples of each other.
In total, I can not think of a way to disprove this statement generally as it holds in some cases.
But as you mentioned, your solution is perfectly fine.
answered Dec 24 '18 at 10:27
Jonas LenzJonas Lenz
694215
$endgroup$
I think that in your case you can not do a general proof of contradiction as there exist choices of $A,B,C$ and $V$ where the statement is true.
So at some point you would need to make some assumptions as for example if $B,C$ contain exactly one element that those are not multiples of each other.
In total, I can not think of a way to disprove this statement generally as it holds in some cases.
But as you mentioned, your solution is perfectly fine.
answered Dec 24 '18 at 10:27
Jonas LenzJonas Lenz
694215
answered Dec 24 '18 at 10:27
Jonas LenzJonas Lenz
694215
answered Dec 24 '18 at 10:27
Jonas LenzJonas Lenz
694215
answered Dec 24 '18 at 10:27
Jonas LenzJonas Lenz
694215
694215
add a comment |
add a comment |
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