$A$, a linearly independent subset of a subspace $S$; $xnotin S$; show $Acup{x}$ is linearly independent












0












$begingroup$



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







share|cite|improve this question









$endgroup$








  • 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
















0












$begingroup$



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







share|cite|improve this question









$endgroup$








  • 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














0












0








0





$begingroup$



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







share|cite|improve this question









$endgroup$





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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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














  • 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










1 Answer
1






active

oldest

votes


















2












$begingroup$

I think you overdid it a little bit. With the theorem we are pretty much done, since $xnotin Simplies xnotinoperatorname{span}A$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

I think you overdid it a little bit. With the theorem we are pretty much done, since $xnotin Simplies xnotinoperatorname{span}A$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















2












$begingroup$

I think you overdid it a little bit. With the theorem we are pretty much done, since $xnotin Simplies xnotinoperatorname{span}A$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














2












2








2





$begingroup$

I think you overdid it a little bit. With the theorem we are pretty much done, since $xnotin Simplies xnotinoperatorname{span}A$.






share|cite|improve this answer









$endgroup$



I think you overdid it a little bit. With the theorem we are pretty much done, since $xnotin Simplies xnotinoperatorname{span}A$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 23 '18 at 5:04









Chris CusterChris Custer

14.4k3827




14.4k3827












  • $begingroup$
    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






  • 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


















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