$A$, a linearly independent subset of a subspace $S$; $xnotin S$; show $Acup{x}$ is linearly independent
$begingroup$
Let $A$ be a linearly independent subset of a subspace $S$.
If $xnotin S$, show that $Acup{x}$ is linearly independent.
Theorem:
Let $B$ be linearly independent and $ynotin B$. Then, $Bcup{y}$ is linearly dependent iff $yin Span(B)$.
Using this theorem, I get that-
$A$ is linearly independent and $xnotin SRightarrow xnotin A$.
So, $Acup{x}$is linearly independent iff $xnotin Sp(A)$.
Therefore, it should be enough to show that-
$Asubseteq S$ (where, $A$ is linearly independent) and $xnotin S$
$Rightarrow xnotin Sp(A)$ $_{...(1)}$
$Rightarrow Acup{x}$ is linearly independent $_{...(2)}$
My question is, if my approach is correct how should I prove $(1)$ because $(2)$ obviously follows from there. Or is my approach incorrect?
[There are similar questions on the site, but I couldn't find one for my approach.]
linear-algebra vector-spaces
asked Dec 23 '18 at 4:57
Za IraZa Ira
161115
$endgroup$
|
show 2 more comments
$begingroup$
Let $A$ be a linearly independent subset of a subspace $S$.
If $xnotin S$, show that $Acup{x}$ is linearly independent.
Theorem:
Let $B$ be linearly independent and $ynotin B$. Then, $Bcup{y}$ is linearly dependent iff $yin Span(B)$.
Using this theorem, I get that-
$A$ is linearly independent and $xnotin SRightarrow xnotin A$.
So, $Acup{x}$is linearly independent iff $xnotin Sp(A)$.
Therefore, it should be enough to show that-
$Asubseteq S$ (where, $A$ is linearly independent) and $xnotin S$
$Rightarrow xnotin Sp(A)$ $_{...(1)}$
$Rightarrow Acup{x}$ is linearly independent $_{...(2)}$
My question is, if my approach is correct how should I prove $(1)$ because $(2)$ obviously follows from there. Or is my approach incorrect?
[There are similar questions on the site, but I couldn't find one for my approach.]
linear-algebra vector-spaces
asked Dec 23 '18 at 4:57
Za IraZa Ira
161115
$endgroup$
1
$begingroup$
According to your approach, take $A = S = leftlbrace left( 1, 0 right) rightrbrace$ in $mathbb{R}^2$. Now, take $x = left( 2, 0 right) notin S$. Is the set $S cup leftlbrace x rightrbrace$ linearly independent? Talking of ($1$) you mentioned, is $left( 2, 0 right) notin text{Span } left( S right)$?
$endgroup$
– Aniruddha Deshmukh
Dec 23 '18 at 5:04
$begingroup$
@AniruddhaDeshmukh word. Then, can you please explain where I am going wrong and what I should do.
$endgroup$
– Za Ira
Dec 23 '18 at 5:05
1
$begingroup$
Well, according to the statement you need to prove, it is already given to you that $x notin text{Span } left( S right)$. So, you need to start with it. If it is not in the span, what can you say about the dependence equation?
$endgroup$
– Aniruddha Deshmukh
Dec 23 '18 at 5:08
$begingroup$
@AniruddhaDeshmukh wait, so~ since $xnotin S$ and $A$ is a linearly independent subset of S, then $xnotin Span(A)$?
$endgroup$
– Za Ira
Dec 23 '18 at 5:17
1
$begingroup$
Yes. My bad! Well then, you just need to prove that $text{Span } left( S right) = S$ given that $S$ is a subspace. Then things will get easier.
$endgroup$
– Aniruddha Deshmukh
Dec 23 '18 at 5:46
|
show 2 more comments
$begingroup$
Let $A$ be a linearly independent subset of a subspace $S$.
If $xnotin S$, show that $Acup{x}$ is linearly independent.
Theorem:
Let $B$ be linearly independent and $ynotin B$. Then, $Bcup{y}$ is linearly dependent iff $yin Span(B)$.
Using this theorem, I get that-
$A$ is linearly independent and $xnotin SRightarrow xnotin A$.
So, $Acup{x}$is linearly independent iff $xnotin Sp(A)$.
Therefore, it should be enough to show that-
$Asubseteq S$ (where, $A$ is linearly independent) and $xnotin S$
$Rightarrow xnotin Sp(A)$ $_{...(1)}$
$Rightarrow Acup{x}$ is linearly independent $_{...(2)}$
My question is, if my approach is correct how should I prove $(1)$ because $(2)$ obviously follows from there. Or is my approach incorrect?
[There are similar questions on the site, but I couldn't find one for my approach.]
linear-algebra vector-spaces
asked Dec 23 '18 at 4:57
Za IraZa Ira
161115
$endgroup$
Let $A$ be a linearly independent subset of a subspace $S$.
If $xnotin S$, show that $Acup{x}$ is linearly independent.
Theorem:
Let $B$ be linearly independent and $ynotin B$. Then, $Bcup{y}$ is linearly dependent iff $yin Span(B)$.
Using this theorem, I get that-
$A$ is linearly independent and $xnotin SRightarrow xnotin A$.
So, $Acup{x}$is linearly independent iff $xnotin Sp(A)$.
Therefore, it should be enough to show that-
$Asubseteq S$ (where, $A$ is linearly independent) and $xnotin S$
$Rightarrow xnotin Sp(A)$ $_{...(1)}$
$Rightarrow Acup{x}$ is linearly independent $_{...(2)}$
My question is, if my approach is correct how should I prove $(1)$ because $(2)$ obviously follows from there. Or is my approach incorrect?
[There are similar questions on the site, but I couldn't find one for my approach.]
linear-algebra vector-spaces
linear-algebra vector-spaces
asked Dec 23 '18 at 4:57
Za IraZa Ira
161115
asked Dec 23 '18 at 4:57
Za IraZa Ira
161115
asked Dec 23 '18 at 4:57
Za IraZa Ira
161115
asked Dec 23 '18 at 4:57
Za IraZa Ira
161115
asked Dec 23 '18 at 4:57
Za IraZa Ira
161115
161115
1
$begingroup$
According to your approach, take $A = S = leftlbrace left( 1, 0 right) rightrbrace$ in $mathbb{R}^2$. Now, take $x = left( 2, 0 right) notin S$. Is the set $S cup leftlbrace x rightrbrace$ linearly independent? Talking of ($1$) you mentioned, is $left( 2, 0 right) notin text{Span } left( S right)$?
$endgroup$
– Aniruddha Deshmukh
Dec 23 '18 at 5:04
$begingroup$
@AniruddhaDeshmukh word. Then, can you please explain where I am going wrong and what I should do.
$endgroup$
– Za Ira
Dec 23 '18 at 5:05
1
$begingroup$
Well, according to the statement you need to prove, it is already given to you that $x notin text{Span } left( S right)$. So, you need to start with it. If it is not in the span, what can you say about the dependence equation?
$endgroup$
– Aniruddha Deshmukh
Dec 23 '18 at 5:08
$begingroup$
@AniruddhaDeshmukh wait, so~ since $xnotin S$ and $A$ is a linearly independent subset of S, then $xnotin Span(A)$?
$endgroup$
– Za Ira
Dec 23 '18 at 5:17
1
$begingroup$
Yes. My bad! Well then, you just need to prove that $text{Span } left( S right) = S$ given that $S$ is a subspace. Then things will get easier.
$endgroup$
– Aniruddha Deshmukh
Dec 23 '18 at 5:46
|
show 2 more comments
1
$begingroup$
According to your approach, take $A = S = leftlbrace left( 1, 0 right) rightrbrace$ in $mathbb{R}^2$. Now, take $x = left( 2, 0 right) notin S$. Is the set $S cup leftlbrace x rightrbrace$ linearly independent? Talking of ($1$) you mentioned, is $left( 2, 0 right) notin text{Span } left( S right)$?
$endgroup$
– Aniruddha Deshmukh
Dec 23 '18 at 5:04
$begingroup$
@AniruddhaDeshmukh word. Then, can you please explain where I am going wrong and what I should do.
$endgroup$
– Za Ira
Dec 23 '18 at 5:05
1
$begingroup$
Well, according to the statement you need to prove, it is already given to you that $x notin text{Span } left( S right)$. So, you need to start with it. If it is not in the span, what can you say about the dependence equation?
$endgroup$
– Aniruddha Deshmukh
Dec 23 '18 at 5:08
$begingroup$
@AniruddhaDeshmukh wait, so~ since $xnotin S$ and $A$ is a linearly independent subset of S, then $xnotin Span(A)$?
$endgroup$
– Za Ira
Dec 23 '18 at 5:17
1
$begingroup$
Yes. My bad! Well then, you just need to prove that $text{Span } left( S right) = S$ given that $S$ is a subspace. Then things will get easier.
$endgroup$
– Aniruddha Deshmukh
Dec 23 '18 at 5:46
1
1
$begingroup$
According to your approach, take $A = S = leftlbrace left( 1, 0 right) rightrbrace$ in $mathbb{R}^2$. Now, take $x = left( 2, 0 right) notin S$. Is the set $S cup leftlbrace x rightrbrace$ linearly independent? Talking of ($1$) you mentioned, is $left( 2, 0 right) notin text{Span } left( S right)$?
$endgroup$
– Aniruddha Deshmukh
Dec 23 '18 at 5:04
$begingroup$
According to your approach, take $A = S = leftlbrace left( 1, 0 right) rightrbrace$ in $mathbb{R}^2$. Now, take $x = left( 2, 0 right) notin S$. Is the set $S cup leftlbrace x rightrbrace$ linearly independent? Talking of ($1$) you mentioned, is $left( 2, 0 right) notin text{Span } left( S right)$?
$endgroup$
– Aniruddha Deshmukh
Dec 23 '18 at 5:04
$begingroup$
@AniruddhaDeshmukh word. Then, can you please explain where I am going wrong and what I should do.
$endgroup$
– Za Ira
Dec 23 '18 at 5:05
$begingroup$
@AniruddhaDeshmukh word. Then, can you please explain where I am going wrong and what I should do.
$endgroup$
– Za Ira
Dec 23 '18 at 5:05
1
1
$begingroup$
Well, according to the statement you need to prove, it is already given to you that $x notin text{Span } left( S right)$. So, you need to start with it. If it is not in the span, what can you say about the dependence equation?
$endgroup$
– Aniruddha Deshmukh
Dec 23 '18 at 5:08
$begingroup$
Well, according to the statement you need to prove, it is already given to you that $x notin text{Span } left( S right)$. So, you need to start with it. If it is not in the span, what can you say about the dependence equation?
$endgroup$
– Aniruddha Deshmukh
Dec 23 '18 at 5:08
$begingroup$
@AniruddhaDeshmukh wait, so~ since $xnotin S$ and $A$ is a linearly independent subset of S, then $xnotin Span(A)$?
$endgroup$
– Za Ira
Dec 23 '18 at 5:17
$begingroup$
@AniruddhaDeshmukh wait, so~ since $xnotin S$ and $A$ is a linearly independent subset of S, then $xnotin Span(A)$?
$endgroup$
– Za Ira
Dec 23 '18 at 5:17
1
1
$begingroup$
Yes. My bad! Well then, you just need to prove that $text{Span } left( S right) = S$ given that $S$ is a subspace. Then things will get easier.
$endgroup$
– Aniruddha Deshmukh
Dec 23 '18 at 5:46
$begingroup$
Yes. My bad! Well then, you just need to prove that $text{Span } left( S right) = S$ given that $S$ is a subspace. Then things will get easier.
$endgroup$
– Aniruddha Deshmukh
Dec 23 '18 at 5:46
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
I think you overdid it a little bit. With the theorem we are pretty much done, since $xnotin Simplies xnotinoperatorname{span}A$.
answered Dec 23 '18 at 5:04
Chris CusterChris Custer
13.6k3827
$endgroup$
$begingroup$
As @AniruddhaDeshmukh pointed out with an example, the corollary I mentioned is wrong somewhere. Hence, please explain how $xnotin span(A)Rightarrow Acup{x}$ is linearly independent.
$endgroup$
– Za Ira
Dec 23 '18 at 5:30
1
$begingroup$
Suppose we have a linear combination of the elements of $A$ and $x$ equal to zero. Then it has to be the trivial one.
$endgroup$
– Chris Custer
Dec 23 '18 at 5:36
add a comment |
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1 Answer
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$begingroup$
I think you overdid it a little bit. With the theorem we are pretty much done, since $xnotin Simplies xnotinoperatorname{span}A$.
answered Dec 23 '18 at 5:04
Chris CusterChris Custer
13.6k3827
$endgroup$
$begingroup$
As @AniruddhaDeshmukh pointed out with an example, the corollary I mentioned is wrong somewhere. Hence, please explain how $xnotin span(A)Rightarrow Acup{x}$ is linearly independent.
$endgroup$
– Za Ira
Dec 23 '18 at 5:30
1
$begingroup$
Suppose we have a linear combination of the elements of $A$ and $x$ equal to zero. Then it has to be the trivial one.
$endgroup$
– Chris Custer
Dec 23 '18 at 5:36
add a comment |
$begingroup$
I think you overdid it a little bit. With the theorem we are pretty much done, since $xnotin Simplies xnotinoperatorname{span}A$.
answered Dec 23 '18 at 5:04
Chris CusterChris Custer
13.6k3827
$endgroup$
$begingroup$
As @AniruddhaDeshmukh pointed out with an example, the corollary I mentioned is wrong somewhere. Hence, please explain how $xnotin span(A)Rightarrow Acup{x}$ is linearly independent.
$endgroup$
– Za Ira
Dec 23 '18 at 5:30
1
$begingroup$
Suppose we have a linear combination of the elements of $A$ and $x$ equal to zero. Then it has to be the trivial one.
$endgroup$
– Chris Custer
Dec 23 '18 at 5:36
add a comment |
$begingroup$
I think you overdid it a little bit. With the theorem we are pretty much done, since $xnotin Simplies xnotinoperatorname{span}A$.
answered Dec 23 '18 at 5:04
Chris CusterChris Custer
13.6k3827
$endgroup$
I think you overdid it a little bit. With the theorem we are pretty much done, since $xnotin Simplies xnotinoperatorname{span}A$.
answered Dec 23 '18 at 5:04
Chris CusterChris Custer
13.6k3827
answered Dec 23 '18 at 5:04
Chris CusterChris Custer
13.6k3827
answered Dec 23 '18 at 5:04
Chris CusterChris Custer
13.6k3827
answered Dec 23 '18 at 5:04
Chris CusterChris Custer
13.6k3827
13.6k3827
$begingroup$
As @AniruddhaDeshmukh pointed out with an example, the corollary I mentioned is wrong somewhere. Hence, please explain how $xnotin span(A)Rightarrow Acup{x}$ is linearly independent.
$endgroup$
– Za Ira
Dec 23 '18 at 5:30
1
$begingroup$
Suppose we have a linear combination of the elements of $A$ and $x$ equal to zero. Then it has to be the trivial one.
$endgroup$
– Chris Custer
Dec 23 '18 at 5:36
add a comment |
$begingroup$
As @AniruddhaDeshmukh pointed out with an example, the corollary I mentioned is wrong somewhere. Hence, please explain how $xnotin span(A)Rightarrow Acup{x}$ is linearly independent.
$endgroup$
– Za Ira
Dec 23 '18 at 5:30
1
$begingroup$
Suppose we have a linear combination of the elements of $A$ and $x$ equal to zero. Then it has to be the trivial one.
$endgroup$
– Chris Custer
Dec 23 '18 at 5:36
$begingroup$
As @AniruddhaDeshmukh pointed out with an example, the corollary I mentioned is wrong somewhere. Hence, please explain how $xnotin span(A)Rightarrow Acup{x}$ is linearly independent.
$endgroup$
– Za Ira
Dec 23 '18 at 5:30
$begingroup$
As @AniruddhaDeshmukh pointed out with an example, the corollary I mentioned is wrong somewhere. Hence, please explain how $xnotin span(A)Rightarrow Acup{x}$ is linearly independent.
$endgroup$
– Za Ira
Dec 23 '18 at 5:30
1
1
$begingroup$
Suppose we have a linear combination of the elements of $A$ and $x$ equal to zero. Then it has to be the trivial one.
$endgroup$
– Chris Custer
Dec 23 '18 at 5:36
$begingroup$
Suppose we have a linear combination of the elements of $A$ and $x$ equal to zero. Then it has to be the trivial one.
$endgroup$
– Chris Custer
Dec 23 '18 at 5:36
add a comment |
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1
$begingroup$
According to your approach, take $A = S = leftlbrace left( 1, 0 right) rightrbrace$ in $mathbb{R}^2$. Now, take $x = left( 2, 0 right) notin S$. Is the set $S cup leftlbrace x rightrbrace$ linearly independent? Talking of ($1$) you mentioned, is $left( 2, 0 right) notin text{Span } left( S right)$?
$endgroup$
– Aniruddha Deshmukh
Dec 23 '18 at 5:04
$begingroup$
@AniruddhaDeshmukh word. Then, can you please explain where I am going wrong and what I should do.
$endgroup$
– Za Ira
Dec 23 '18 at 5:05
1
$begingroup$
Well, according to the statement you need to prove, it is already given to you that $x notin text{Span } left( S right)$. So, you need to start with it. If it is not in the span, what can you say about the dependence equation?
$endgroup$
– Aniruddha Deshmukh
Dec 23 '18 at 5:08
$begingroup$
@AniruddhaDeshmukh wait, so~ since $xnotin S$ and $A$ is a linearly independent subset of S, then $xnotin Span(A)$?
$endgroup$
– Za Ira
Dec 23 '18 at 5:17
1
$begingroup$
Yes. My bad! Well then, you just need to prove that $text{Span } left( S right) = S$ given that $S$ is a subspace. Then things will get easier.
$endgroup$
– Aniruddha Deshmukh
Dec 23 '18 at 5:46