Question with regards to logic of proof: The 2019 Stack Overflow Developer Survey Results Are InLinear combination of natural numbers with positive coefficientsLinear Algebra- independence and dependenceProof with Positive Symmetric MatricesLinear dependence of following polynomialsA question about linear combinationProof Using Gronwall's InequalityLet $a,b in mathbbZ$ and suppose that $a | b$ and $b |a$. Then $a=b$ or $a=-b$How exactly does elimination discover linearly dependent rows of $A$ (in $AX=b$)?Finding values of constants when solving linearly dependent equationhelp with a vector proof
Do these rules for Critical Successes and Critical Failures seem fair?
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Question with regards to logic of proof:
The 2019 Stack Overflow Developer Survey Results Are InLinear combination of natural numbers with positive coefficientsLinear Algebra- independence and dependenceProof with Positive Symmetric MatricesLinear dependence of following polynomialsA question about linear combinationProof Using Gronwall's InequalityLet $a,b in mathbbZ$ and suppose that $a | b$ and $b |a$. Then $a=b$ or $a=-b$How exactly does elimination discover linearly dependent rows of $A$ (in $AX=b$)?Finding values of constants when solving linearly dependent equationhelp with a vector proof
$begingroup$
Proof: Let $V$ be a vector space over $mathbbR$. Let $a,b in V$ and $a,b neq 0$ . If $a, b$ are linearly dependent, then there exists w $in mathbbR$ such that $a=wb.$
I know that if $a,b$ are linearly dependent then the equation $$K_1a+K_2b=0$$means that $K_1=K_2=0 $ is not the only solution, so $K_1$ or $K_2$ is non-zero, and if I set $$w = -K_2/K_1$$ (WLOG assuming $K_1$ is non-zero) I have showed the existence of the scalar. But does the following show the uniqueness of the scalar up to coeffiecients?
$$K_1a+K_2b=0implies K_1a=-K_2b, text and so a= frac-K_2K_1b$$
Edit:
Thank you guys for your replies. I was thinking about this in terms of logic. To show uniqueness one can show through contradiction or to show the existence of an element that satisfies the condition and then proving that for any other element x that satisfies the condition for x implies x is the existing element. What I have shown does not show uniqueness of the solutions, because $K_1=K_2=0$ is already a solution. What I have shown is that if I strictly work under the assumption that $K_1 neq 0$ then the solution is unique.
linear-algebra proof-verification logic proof-writing
asked Mar 30 at 17:44
topologicalmagiciantopologicalmagician
1249
$endgroup$
add a comment |
$begingroup$
Proof: Let $V$ be a vector space over $mathbbR$. Let $a,b in V$ and $a,b neq 0$ . If $a, b$ are linearly dependent, then there exists w $in mathbbR$ such that $a=wb.$
I know that if $a,b$ are linearly dependent then the equation $$K_1a+K_2b=0$$means that $K_1=K_2=0 $ is not the only solution, so $K_1$ or $K_2$ is non-zero, and if I set $$w = -K_2/K_1$$ (WLOG assuming $K_1$ is non-zero) I have showed the existence of the scalar. But does the following show the uniqueness of the scalar up to coeffiecients?
$$K_1a+K_2b=0implies K_1a=-K_2b, text and so a= frac-K_2K_1b$$
Edit:
Thank you guys for your replies. I was thinking about this in terms of logic. To show uniqueness one can show through contradiction or to show the existence of an element that satisfies the condition and then proving that for any other element x that satisfies the condition for x implies x is the existing element. What I have shown does not show uniqueness of the solutions, because $K_1=K_2=0$ is already a solution. What I have shown is that if I strictly work under the assumption that $K_1 neq 0$ then the solution is unique.
linear-algebra proof-verification logic proof-writing
asked Mar 30 at 17:44
topologicalmagiciantopologicalmagician
1249
$endgroup$
$begingroup$
Are you sure you have the hypothesis right? If $a neq 0$ and $b = 0$, then you can't find the $w$ you're looking for.
$endgroup$
– Michael Biro
Mar 30 at 17:47
$begingroup$
@MichaelBiro fixed it
$endgroup$
– topologicalmagician
Mar 30 at 17:48
add a comment |
$begingroup$
Proof: Let $V$ be a vector space over $mathbbR$. Let $a,b in V$ and $a,b neq 0$ . If $a, b$ are linearly dependent, then there exists w $in mathbbR$ such that $a=wb.$
I know that if $a,b$ are linearly dependent then the equation $$K_1a+K_2b=0$$means that $K_1=K_2=0 $ is not the only solution, so $K_1$ or $K_2$ is non-zero, and if I set $$w = -K_2/K_1$$ (WLOG assuming $K_1$ is non-zero) I have showed the existence of the scalar. But does the following show the uniqueness of the scalar up to coeffiecients?
$$K_1a+K_2b=0implies K_1a=-K_2b, text and so a= frac-K_2K_1b$$
Edit:
Thank you guys for your replies. I was thinking about this in terms of logic. To show uniqueness one can show through contradiction or to show the existence of an element that satisfies the condition and then proving that for any other element x that satisfies the condition for x implies x is the existing element. What I have shown does not show uniqueness of the solutions, because $K_1=K_2=0$ is already a solution. What I have shown is that if I strictly work under the assumption that $K_1 neq 0$ then the solution is unique.
linear-algebra proof-verification logic proof-writing
asked Mar 30 at 17:44
topologicalmagiciantopologicalmagician
1249
$endgroup$
Proof: Let $V$ be a vector space over $mathbbR$. Let $a,b in V$ and $a,b neq 0$ . If $a, b$ are linearly dependent, then there exists w $in mathbbR$ such that $a=wb.$
I know that if $a,b$ are linearly dependent then the equation $$K_1a+K_2b=0$$means that $K_1=K_2=0 $ is not the only solution, so $K_1$ or $K_2$ is non-zero, and if I set $$w = -K_2/K_1$$ (WLOG assuming $K_1$ is non-zero) I have showed the existence of the scalar. But does the following show the uniqueness of the scalar up to coeffiecients?
$$K_1a+K_2b=0implies K_1a=-K_2b, text and so a= frac-K_2K_1b$$
Edit:
Thank you guys for your replies. I was thinking about this in terms of logic. To show uniqueness one can show through contradiction or to show the existence of an element that satisfies the condition and then proving that for any other element x that satisfies the condition for x implies x is the existing element. What I have shown does not show uniqueness of the solutions, because $K_1=K_2=0$ is already a solution. What I have shown is that if I strictly work under the assumption that $K_1 neq 0$ then the solution is unique.
linear-algebra proof-verification logic proof-writing
linear-algebra proof-verification logic proof-writing
asked Mar 30 at 17:44
topologicalmagiciantopologicalmagician
1249
asked Mar 30 at 17:44
topologicalmagiciantopologicalmagician
1249
edited Mar 30 at 18:11
topologicalmagician
asked Mar 30 at 17:44
topologicalmagiciantopologicalmagician
1249
asked Mar 30 at 17:44
topologicalmagiciantopologicalmagician
1249
asked Mar 30 at 17:44
topologicalmagiciantopologicalmagician
1249
1249
$begingroup$
Are you sure you have the hypothesis right? If $a neq 0$ and $b = 0$, then you can't find the $w$ you're looking for.
$endgroup$
– Michael Biro
Mar 30 at 17:47
$begingroup$
@MichaelBiro fixed it
$endgroup$
– topologicalmagician
Mar 30 at 17:48
add a comment |
$begingroup$
Are you sure you have the hypothesis right? If $a neq 0$ and $b = 0$, then you can't find the $w$ you're looking for.
$endgroup$
– Michael Biro
Mar 30 at 17:47
$begingroup$
@MichaelBiro fixed it
$endgroup$
– topologicalmagician
Mar 30 at 17:48
$begingroup$
Are you sure you have the hypothesis right? If $a neq 0$ and $b = 0$, then you can't find the $w$ you're looking for.
$endgroup$
– Michael Biro
Mar 30 at 17:47
$begingroup$
Are you sure you have the hypothesis right? If $a neq 0$ and $b = 0$, then you can't find the $w$ you're looking for.
$endgroup$
– Michael Biro
Mar 30 at 17:47
$begingroup$
@MichaelBiro fixed it
$endgroup$
– topologicalmagician
Mar 30 at 17:48
$begingroup$
@MichaelBiro fixed it
$endgroup$
– topologicalmagician
Mar 30 at 17:48
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Yes, for two non-zero vectors, being dependent means that each is a multiple of the other, and uniqueness of multiplication means that there is only one multiple that works. You do want to be a little careful in not assuming that your $K_1 neq 0$ in the operation $-fracK_2K_1$ (but you can easily verify that separately).
In other words, if $a = w_1 b$ and $a = w_2b$ for $w_1 neq w_2$ then
$$b = fracw_1 - w_2w_1 - w_2 b = fracw_1b - w_2bw_1-w_2 = fraca - aw_1 - w_2 = 0$$
answered Mar 30 at 17:57
Michael BiroMichael Biro
11.7k21831
$endgroup$
$begingroup$
so I have shown that the solution is unique if i'm strictly working under the assumption that $K_1$ is non-zero, right?
$endgroup$
– topologicalmagician
Mar 30 at 18:17
1
$begingroup$
Yes, $K_1 neq 0$ is enough since $a, b neq 0$.
$endgroup$
– Michael Biro
Mar 30 at 18:22
add a comment |
$begingroup$
No, because the other solution is always $$K_1 = K_2 = 0.$$
(Yes, there must be at least one other, which you selected.)
By the way, at the end you expressed unambiguosly the vector $a$, NOT the pair of scalars $K_1, K_2.$
answered Mar 30 at 18:00
MarianDMarianD
2,2611618
$endgroup$
$begingroup$
But have I shown it is unique if i'm strictly working under the assumption that $K_1 neq 0$?
$endgroup$
– topologicalmagician
Mar 30 at 18:12
$begingroup$
Unique must be the pair of coefficients $K_1, K_2$ to vectors $a, b$ be linearly independent. As they are NOT independent, you had a chance to choose a nonzero pair of coefficients (from many such pairs - coefficients $c.K_1, c.K_2$ for $c ne 0$ are examples of other such pairs).
$endgroup$
– MarianD
Mar 30 at 21:20
add a comment |
$begingroup$
It looks alright except that you may be dividing by zero.
Note that if $K_1$ is zero then $K_2b = 0$ too, which implies $K_2 = 0$, in contradiction. From here your proof is fine.
answered Mar 30 at 17:57
MariahMariah
2,1471718
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes, for two non-zero vectors, being dependent means that each is a multiple of the other, and uniqueness of multiplication means that there is only one multiple that works. You do want to be a little careful in not assuming that your $K_1 neq 0$ in the operation $-fracK_2K_1$ (but you can easily verify that separately).
In other words, if $a = w_1 b$ and $a = w_2b$ for $w_1 neq w_2$ then
$$b = fracw_1 - w_2w_1 - w_2 b = fracw_1b - w_2bw_1-w_2 = fraca - aw_1 - w_2 = 0$$
answered Mar 30 at 17:57
Michael BiroMichael Biro
11.7k21831
$endgroup$
$begingroup$
so I have shown that the solution is unique if i'm strictly working under the assumption that $K_1$ is non-zero, right?
$endgroup$
– topologicalmagician
Mar 30 at 18:17
1
$begingroup$
Yes, $K_1 neq 0$ is enough since $a, b neq 0$.
$endgroup$
– Michael Biro
Mar 30 at 18:22
add a comment |
$begingroup$
Yes, for two non-zero vectors, being dependent means that each is a multiple of the other, and uniqueness of multiplication means that there is only one multiple that works. You do want to be a little careful in not assuming that your $K_1 neq 0$ in the operation $-fracK_2K_1$ (but you can easily verify that separately).
In other words, if $a = w_1 b$ and $a = w_2b$ for $w_1 neq w_2$ then
$$b = fracw_1 - w_2w_1 - w_2 b = fracw_1b - w_2bw_1-w_2 = fraca - aw_1 - w_2 = 0$$
answered Mar 30 at 17:57
Michael BiroMichael Biro
11.7k21831
$endgroup$
$begingroup$
so I have shown that the solution is unique if i'm strictly working under the assumption that $K_1$ is non-zero, right?
$endgroup$
– topologicalmagician
Mar 30 at 18:17
1
$begingroup$
Yes, $K_1 neq 0$ is enough since $a, b neq 0$.
$endgroup$
– Michael Biro
Mar 30 at 18:22
add a comment |
$begingroup$
Yes, for two non-zero vectors, being dependent means that each is a multiple of the other, and uniqueness of multiplication means that there is only one multiple that works. You do want to be a little careful in not assuming that your $K_1 neq 0$ in the operation $-fracK_2K_1$ (but you can easily verify that separately).
In other words, if $a = w_1 b$ and $a = w_2b$ for $w_1 neq w_2$ then
$$b = fracw_1 - w_2w_1 - w_2 b = fracw_1b - w_2bw_1-w_2 = fraca - aw_1 - w_2 = 0$$
answered Mar 30 at 17:57
Michael BiroMichael Biro
11.7k21831
$endgroup$
Yes, for two non-zero vectors, being dependent means that each is a multiple of the other, and uniqueness of multiplication means that there is only one multiple that works. You do want to be a little careful in not assuming that your $K_1 neq 0$ in the operation $-fracK_2K_1$ (but you can easily verify that separately).
In other words, if $a = w_1 b$ and $a = w_2b$ for $w_1 neq w_2$ then
$$b = fracw_1 - w_2w_1 - w_2 b = fracw_1b - w_2bw_1-w_2 = fraca - aw_1 - w_2 = 0$$
answered Mar 30 at 17:57
Michael BiroMichael Biro
11.7k21831
answered Mar 30 at 17:57
Michael BiroMichael Biro
11.7k21831
answered Mar 30 at 17:57
Michael BiroMichael Biro
11.7k21831
answered Mar 30 at 17:57
Michael BiroMichael Biro
11.7k21831
11.7k21831
$begingroup$
so I have shown that the solution is unique if i'm strictly working under the assumption that $K_1$ is non-zero, right?
$endgroup$
– topologicalmagician
Mar 30 at 18:17
1
$begingroup$
Yes, $K_1 neq 0$ is enough since $a, b neq 0$.
$endgroup$
– Michael Biro
Mar 30 at 18:22
add a comment |
$begingroup$
so I have shown that the solution is unique if i'm strictly working under the assumption that $K_1$ is non-zero, right?
$endgroup$
– topologicalmagician
Mar 30 at 18:17
1
$begingroup$
Yes, $K_1 neq 0$ is enough since $a, b neq 0$.
$endgroup$
– Michael Biro
Mar 30 at 18:22
$begingroup$
so I have shown that the solution is unique if i'm strictly working under the assumption that $K_1$ is non-zero, right?
$endgroup$
– topologicalmagician
Mar 30 at 18:17
$begingroup$
so I have shown that the solution is unique if i'm strictly working under the assumption that $K_1$ is non-zero, right?
$endgroup$
– topologicalmagician
Mar 30 at 18:17
1
1
$begingroup$
Yes, $K_1 neq 0$ is enough since $a, b neq 0$.
$endgroup$
– Michael Biro
Mar 30 at 18:22
$begingroup$
Yes, $K_1 neq 0$ is enough since $a, b neq 0$.
$endgroup$
– Michael Biro
Mar 30 at 18:22
add a comment |
$begingroup$
No, because the other solution is always $$K_1 = K_2 = 0.$$
(Yes, there must be at least one other, which you selected.)
By the way, at the end you expressed unambiguosly the vector $a$, NOT the pair of scalars $K_1, K_2.$
answered Mar 30 at 18:00
MarianDMarianD
2,2611618
$endgroup$
$begingroup$
But have I shown it is unique if i'm strictly working under the assumption that $K_1 neq 0$?
$endgroup$
– topologicalmagician
Mar 30 at 18:12
$begingroup$
Unique must be the pair of coefficients $K_1, K_2$ to vectors $a, b$ be linearly independent. As they are NOT independent, you had a chance to choose a nonzero pair of coefficients (from many such pairs - coefficients $c.K_1, c.K_2$ for $c ne 0$ are examples of other such pairs).
$endgroup$
– MarianD
Mar 30 at 21:20
add a comment |
$begingroup$
No, because the other solution is always $$K_1 = K_2 = 0.$$
(Yes, there must be at least one other, which you selected.)
By the way, at the end you expressed unambiguosly the vector $a$, NOT the pair of scalars $K_1, K_2.$
answered Mar 30 at 18:00
MarianDMarianD
2,2611618
$endgroup$
$begingroup$
But have I shown it is unique if i'm strictly working under the assumption that $K_1 neq 0$?
$endgroup$
– topologicalmagician
Mar 30 at 18:12
$begingroup$
Unique must be the pair of coefficients $K_1, K_2$ to vectors $a, b$ be linearly independent. As they are NOT independent, you had a chance to choose a nonzero pair of coefficients (from many such pairs - coefficients $c.K_1, c.K_2$ for $c ne 0$ are examples of other such pairs).
$endgroup$
– MarianD
Mar 30 at 21:20
add a comment |
$begingroup$
No, because the other solution is always $$K_1 = K_2 = 0.$$
(Yes, there must be at least one other, which you selected.)
By the way, at the end you expressed unambiguosly the vector $a$, NOT the pair of scalars $K_1, K_2.$
answered Mar 30 at 18:00
MarianDMarianD
2,2611618
$endgroup$
No, because the other solution is always $$K_1 = K_2 = 0.$$
(Yes, there must be at least one other, which you selected.)
By the way, at the end you expressed unambiguosly the vector $a$, NOT the pair of scalars $K_1, K_2.$
answered Mar 30 at 18:00
MarianDMarianD
2,2611618
edited Mar 30 at 18:08
answered Mar 30 at 18:00
MarianDMarianD
2,2611618
answered Mar 30 at 18:00
MarianDMarianD
2,2611618
answered Mar 30 at 18:00
MarianDMarianD
2,2611618
2,2611618
$begingroup$
But have I shown it is unique if i'm strictly working under the assumption that $K_1 neq 0$?
$endgroup$
– topologicalmagician
Mar 30 at 18:12
$begingroup$
Unique must be the pair of coefficients $K_1, K_2$ to vectors $a, b$ be linearly independent. As they are NOT independent, you had a chance to choose a nonzero pair of coefficients (from many such pairs - coefficients $c.K_1, c.K_2$ for $c ne 0$ are examples of other such pairs).
$endgroup$
– MarianD
Mar 30 at 21:20
add a comment |
$begingroup$
But have I shown it is unique if i'm strictly working under the assumption that $K_1 neq 0$?
$endgroup$
– topologicalmagician
Mar 30 at 18:12
$begingroup$
Unique must be the pair of coefficients $K_1, K_2$ to vectors $a, b$ be linearly independent. As they are NOT independent, you had a chance to choose a nonzero pair of coefficients (from many such pairs - coefficients $c.K_1, c.K_2$ for $c ne 0$ are examples of other such pairs).
$endgroup$
– MarianD
Mar 30 at 21:20
$begingroup$
But have I shown it is unique if i'm strictly working under the assumption that $K_1 neq 0$?
$endgroup$
– topologicalmagician
Mar 30 at 18:12
$begingroup$
But have I shown it is unique if i'm strictly working under the assumption that $K_1 neq 0$?
$endgroup$
– topologicalmagician
Mar 30 at 18:12
$begingroup$
Unique must be the pair of coefficients $K_1, K_2$ to vectors $a, b$ be linearly independent. As they are NOT independent, you had a chance to choose a nonzero pair of coefficients (from many such pairs - coefficients $c.K_1, c.K_2$ for $c ne 0$ are examples of other such pairs).
$endgroup$
– MarianD
Mar 30 at 21:20
$begingroup$
Unique must be the pair of coefficients $K_1, K_2$ to vectors $a, b$ be linearly independent. As they are NOT independent, you had a chance to choose a nonzero pair of coefficients (from many such pairs - coefficients $c.K_1, c.K_2$ for $c ne 0$ are examples of other such pairs).
$endgroup$
– MarianD
Mar 30 at 21:20
add a comment |
$begingroup$
It looks alright except that you may be dividing by zero.
Note that if $K_1$ is zero then $K_2b = 0$ too, which implies $K_2 = 0$, in contradiction. From here your proof is fine.
answered Mar 30 at 17:57
MariahMariah
2,1471718
$endgroup$
add a comment |
$begingroup$
It looks alright except that you may be dividing by zero.
Note that if $K_1$ is zero then $K_2b = 0$ too, which implies $K_2 = 0$, in contradiction. From here your proof is fine.
answered Mar 30 at 17:57
MariahMariah
2,1471718
$endgroup$
add a comment |
$begingroup$
It looks alright except that you may be dividing by zero.
Note that if $K_1$ is zero then $K_2b = 0$ too, which implies $K_2 = 0$, in contradiction. From here your proof is fine.
answered Mar 30 at 17:57
MariahMariah
2,1471718
$endgroup$
It looks alright except that you may be dividing by zero.
Note that if $K_1$ is zero then $K_2b = 0$ too, which implies $K_2 = 0$, in contradiction. From here your proof is fine.
answered Mar 30 at 17:57
MariahMariah
2,1471718
answered Mar 30 at 17:57
MariahMariah
2,1471718
answered Mar 30 at 17:57
MariahMariah
2,1471718
answered Mar 30 at 17:57
MariahMariah
2,1471718
2,1471718
add a comment |
add a comment |
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$begingroup$
Are you sure you have the hypothesis right? If $a neq 0$ and $b = 0$, then you can't find the $w$ you're looking for.
$endgroup$
– Michael Biro
Mar 30 at 17:47
$begingroup$
@MichaelBiro fixed it
$endgroup$
– topologicalmagician
Mar 30 at 17:48