$Span(A)=S$; show that no proper subset of $A$ generates $S$ iff $A$ is linearly independent
$begingroup$
Let $Span(A)=S$.
Then, show that no proper subset of $A$ generates $S$ iff $A$ is linearly independent.
My approach:
$bbox[7px,border:1px solid green]{text{$(Rightarrow)$only-if part:}}$
Given: $Sp(A)=S$
No proper subset of $A$ generates $S$.
To Show: $A$ is linearly independent.
If possible, assume A is linearly dependent.
$Sp(A)=S$
$Rightarrow A$ generates $S$ but, $A$ is linearly dependent
$Rightarrow$ there exists a proper linearly independent subset $B$ of $A$, generating $A$ $_{...(1)}$
Since, $B$ generates $A$ and $A$ generates $S$, we have that $B$ generates $S$ $_{...(2)}$
, which is a contradiction.
Therefore, $A$ is linearly independent.
$bbox[7px,border:1px solid green]{text{$(Leftarrow)$if part:}}$
Given: $Sp(A)=S$
$A$ is linearly independent.
To Show: No proper subset of $A$ generates $S$.
Suppose, there exists a proper subset $B$ of $A$, generating $S$. Also, $Sp(A)=S$.
So, every element in $S$ can be written as a linear combination of vectors in $A$ as well as vectors in $B$. But, $B$ is also a subset of $A$. $_{...(3)}$
Doubts $(1),(2)$: If my approach is correct, I know intuitively that these conclusions are correct. But, I would like to know if there are theorems supporting this.
Doubt $(3)$: I am not sure how to proceed from there (I started with a proof by contradiction).
[EDIT: I made corrections in the statement before $(3)$, as pointed out(?)]
linear-algebra vector-spaces
asked Dec 23 '18 at 10:11
Za IraZa Ira
161115
$endgroup$
add a comment |
$begingroup$
Let $Span(A)=S$.
Then, show that no proper subset of $A$ generates $S$ iff $A$ is linearly independent.
My approach:
$bbox[7px,border:1px solid green]{text{$(Rightarrow)$only-if part:}}$
Given: $Sp(A)=S$
No proper subset of $A$ generates $S$.
To Show: $A$ is linearly independent.
If possible, assume A is linearly dependent.
$Sp(A)=S$
$Rightarrow A$ generates $S$ but, $A$ is linearly dependent
$Rightarrow$ there exists a proper linearly independent subset $B$ of $A$, generating $A$ $_{...(1)}$
Since, $B$ generates $A$ and $A$ generates $S$, we have that $B$ generates $S$ $_{...(2)}$
, which is a contradiction.
Therefore, $A$ is linearly independent.
$bbox[7px,border:1px solid green]{text{$(Leftarrow)$if part:}}$
Given: $Sp(A)=S$
$A$ is linearly independent.
To Show: No proper subset of $A$ generates $S$.
Suppose, there exists a proper subset $B$ of $A$, generating $S$. Also, $Sp(A)=S$.
So, every element in $S$ can be written as a linear combination of vectors in $A$ as well as vectors in $B$. But, $B$ is also a subset of $A$. $_{...(3)}$
Doubts $(1),(2)$: If my approach is correct, I know intuitively that these conclusions are correct. But, I would like to know if there are theorems supporting this.
Doubt $(3)$: I am not sure how to proceed from there (I started with a proof by contradiction).
[EDIT: I made corrections in the statement before $(3)$, as pointed out(?)]
linear-algebra vector-spaces
asked Dec 23 '18 at 10:11
Za IraZa Ira
161115
$endgroup$
add a comment |
$begingroup$
Let $Span(A)=S$.
Then, show that no proper subset of $A$ generates $S$ iff $A$ is linearly independent.
My approach:
$bbox[7px,border:1px solid green]{text{$(Rightarrow)$only-if part:}}$
Given: $Sp(A)=S$
No proper subset of $A$ generates $S$.
To Show: $A$ is linearly independent.
If possible, assume A is linearly dependent.
$Sp(A)=S$
$Rightarrow A$ generates $S$ but, $A$ is linearly dependent
$Rightarrow$ there exists a proper linearly independent subset $B$ of $A$, generating $A$ $_{...(1)}$
Since, $B$ generates $A$ and $A$ generates $S$, we have that $B$ generates $S$ $_{...(2)}$
, which is a contradiction.
Therefore, $A$ is linearly independent.
$bbox[7px,border:1px solid green]{text{$(Leftarrow)$if part:}}$
Given: $Sp(A)=S$
$A$ is linearly independent.
To Show: No proper subset of $A$ generates $S$.
Suppose, there exists a proper subset $B$ of $A$, generating $S$. Also, $Sp(A)=S$.
So, every element in $S$ can be written as a linear combination of vectors in $A$ as well as vectors in $B$. But, $B$ is also a subset of $A$. $_{...(3)}$
Doubts $(1),(2)$: If my approach is correct, I know intuitively that these conclusions are correct. But, I would like to know if there are theorems supporting this.
Doubt $(3)$: I am not sure how to proceed from there (I started with a proof by contradiction).
[EDIT: I made corrections in the statement before $(3)$, as pointed out(?)]
linear-algebra vector-spaces
asked Dec 23 '18 at 10:11
Za IraZa Ira
161115
$endgroup$
Let $Span(A)=S$.
Then, show that no proper subset of $A$ generates $S$ iff $A$ is linearly independent.
My approach:
$bbox[7px,border:1px solid green]{text{$(Rightarrow)$only-if part:}}$
Given: $Sp(A)=S$
No proper subset of $A$ generates $S$.
To Show: $A$ is linearly independent.
If possible, assume A is linearly dependent.
$Sp(A)=S$
$Rightarrow A$ generates $S$ but, $A$ is linearly dependent
$Rightarrow$ there exists a proper linearly independent subset $B$ of $A$, generating $A$ $_{...(1)}$
Since, $B$ generates $A$ and $A$ generates $S$, we have that $B$ generates $S$ $_{...(2)}$
, which is a contradiction.
Therefore, $A$ is linearly independent.
$bbox[7px,border:1px solid green]{text{$(Leftarrow)$if part:}}$
Given: $Sp(A)=S$
$A$ is linearly independent.
To Show: No proper subset of $A$ generates $S$.
Suppose, there exists a proper subset $B$ of $A$, generating $S$. Also, $Sp(A)=S$.
So, every element in $S$ can be written as a linear combination of vectors in $A$ as well as vectors in $B$. But, $B$ is also a subset of $A$. $_{...(3)}$
Doubts $(1),(2)$: If my approach is correct, I know intuitively that these conclusions are correct. But, I would like to know if there are theorems supporting this.
Doubt $(3)$: I am not sure how to proceed from there (I started with a proof by contradiction).
[EDIT: I made corrections in the statement before $(3)$, as pointed out(?)]
linear-algebra vector-spaces
linear-algebra vector-spaces
asked Dec 23 '18 at 10:11
Za IraZa Ira
161115
asked Dec 23 '18 at 10:11
Za IraZa Ira
161115
edited Dec 24 '18 at 1:30
Za Ira
asked Dec 23 '18 at 10:11
Za IraZa Ira
161115
asked Dec 23 '18 at 10:11
Za IraZa Ira
161115
asked Dec 23 '18 at 10:11
Za IraZa Ira
161115
161115
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There are a few mistakes in your writing, though you have the correct idea in your mind. For example, the statement $S$ can written as a linear combination of ... makes no sense.
Here is the correct way to write the proof: suppose no proper subset of $A$ generates $S$. If $A$ is not linearly indpendent then there exists an equation of the form $sum c_ix_i=0$ with $x_i in A$ for all $i$ and not all of the coefficients are $0$. If $c_{i_0} neq 0$ we can rewrite the equation in the form $x_{i_0}=sum_{ineq i_0}b_ix_i$. Now let $A'=span(Asetminus {x_{i_0}})$. Then $A'$ is a proper subset of $A$ which generates $S$.
Conversely, suppose $A$ is linearly independent and let $A'$
be a proper subset of $A$. Then there is a vector $x in Asetminus A'$. This vector cannot be written as a linear combination of elements of $A'$ because $A$ is linearly independent. Hence $A'$ does not span $S$.
answered Dec 23 '18 at 12:01
Kavi Rama MurthyKavi Rama Murthy
64.7k42766
$endgroup$
$begingroup$
What do you mean by $span(A/x_{i_{o}})$? Assuming $x_i$ s are elements of $A$, is $A'=span(A/c_{i_{o}})$ what you meant? Also, how do we get that $A'$ generates $S$ from that statement, I am sorry but I am unable to understand.
$endgroup$
– Za Ira
Dec 23 '18 at 14:00
1
$begingroup$
@ZaIra span of a set is the set of all linear combinations of elements of that set. Every element of $S$ is a linear combination of elements of $A$. If one of these elements is $x_{i_0}$ then use the fact that $x_{i_0}$ is itself a linear combination of other elements of $A$ to conclude that $Asetminus x_{i_0}$ spans $S$.
$endgroup$
– Kavi Rama Murthy
Dec 23 '18 at 23:19
$begingroup$
oh i realize you used $setminus$ and not $/$. Thank-you!
$endgroup$
– Za Ira
Dec 24 '18 at 1:32
add a comment |
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$begingroup$
There are a few mistakes in your writing, though you have the correct idea in your mind. For example, the statement $S$ can written as a linear combination of ... makes no sense.
Here is the correct way to write the proof: suppose no proper subset of $A$ generates $S$. If $A$ is not linearly indpendent then there exists an equation of the form $sum c_ix_i=0$ with $x_i in A$ for all $i$ and not all of the coefficients are $0$. If $c_{i_0} neq 0$ we can rewrite the equation in the form $x_{i_0}=sum_{ineq i_0}b_ix_i$. Now let $A'=span(Asetminus {x_{i_0}})$. Then $A'$ is a proper subset of $A$ which generates $S$.
Conversely, suppose $A$ is linearly independent and let $A'$
be a proper subset of $A$. Then there is a vector $x in Asetminus A'$. This vector cannot be written as a linear combination of elements of $A'$ because $A$ is linearly independent. Hence $A'$ does not span $S$.
answered Dec 23 '18 at 12:01
Kavi Rama MurthyKavi Rama Murthy
64.7k42766
$endgroup$
$begingroup$
What do you mean by $span(A/x_{i_{o}})$? Assuming $x_i$ s are elements of $A$, is $A'=span(A/c_{i_{o}})$ what you meant? Also, how do we get that $A'$ generates $S$ from that statement, I am sorry but I am unable to understand.
$endgroup$
– Za Ira
Dec 23 '18 at 14:00
1
$begingroup$
@ZaIra span of a set is the set of all linear combinations of elements of that set. Every element of $S$ is a linear combination of elements of $A$. If one of these elements is $x_{i_0}$ then use the fact that $x_{i_0}$ is itself a linear combination of other elements of $A$ to conclude that $Asetminus x_{i_0}$ spans $S$.
$endgroup$
– Kavi Rama Murthy
Dec 23 '18 at 23:19
$begingroup$
oh i realize you used $setminus$ and not $/$. Thank-you!
$endgroup$
– Za Ira
Dec 24 '18 at 1:32
add a comment |
$begingroup$
There are a few mistakes in your writing, though you have the correct idea in your mind. For example, the statement $S$ can written as a linear combination of ... makes no sense.
Here is the correct way to write the proof: suppose no proper subset of $A$ generates $S$. If $A$ is not linearly indpendent then there exists an equation of the form $sum c_ix_i=0$ with $x_i in A$ for all $i$ and not all of the coefficients are $0$. If $c_{i_0} neq 0$ we can rewrite the equation in the form $x_{i_0}=sum_{ineq i_0}b_ix_i$. Now let $A'=span(Asetminus {x_{i_0}})$. Then $A'$ is a proper subset of $A$ which generates $S$.
Conversely, suppose $A$ is linearly independent and let $A'$
be a proper subset of $A$. Then there is a vector $x in Asetminus A'$. This vector cannot be written as a linear combination of elements of $A'$ because $A$ is linearly independent. Hence $A'$ does not span $S$.
answered Dec 23 '18 at 12:01
Kavi Rama MurthyKavi Rama Murthy
64.7k42766
$endgroup$
$begingroup$
What do you mean by $span(A/x_{i_{o}})$? Assuming $x_i$ s are elements of $A$, is $A'=span(A/c_{i_{o}})$ what you meant? Also, how do we get that $A'$ generates $S$ from that statement, I am sorry but I am unable to understand.
$endgroup$
– Za Ira
Dec 23 '18 at 14:00
1
$begingroup$
@ZaIra span of a set is the set of all linear combinations of elements of that set. Every element of $S$ is a linear combination of elements of $A$. If one of these elements is $x_{i_0}$ then use the fact that $x_{i_0}$ is itself a linear combination of other elements of $A$ to conclude that $Asetminus x_{i_0}$ spans $S$.
$endgroup$
– Kavi Rama Murthy
Dec 23 '18 at 23:19
$begingroup$
oh i realize you used $setminus$ and not $/$. Thank-you!
$endgroup$
– Za Ira
Dec 24 '18 at 1:32
add a comment |
$begingroup$
There are a few mistakes in your writing, though you have the correct idea in your mind. For example, the statement $S$ can written as a linear combination of ... makes no sense.
Here is the correct way to write the proof: suppose no proper subset of $A$ generates $S$. If $A$ is not linearly indpendent then there exists an equation of the form $sum c_ix_i=0$ with $x_i in A$ for all $i$ and not all of the coefficients are $0$. If $c_{i_0} neq 0$ we can rewrite the equation in the form $x_{i_0}=sum_{ineq i_0}b_ix_i$. Now let $A'=span(Asetminus {x_{i_0}})$. Then $A'$ is a proper subset of $A$ which generates $S$.
Conversely, suppose $A$ is linearly independent and let $A'$
be a proper subset of $A$. Then there is a vector $x in Asetminus A'$. This vector cannot be written as a linear combination of elements of $A'$ because $A$ is linearly independent. Hence $A'$ does not span $S$.
answered Dec 23 '18 at 12:01
Kavi Rama MurthyKavi Rama Murthy
64.7k42766
$endgroup$
There are a few mistakes in your writing, though you have the correct idea in your mind. For example, the statement $S$ can written as a linear combination of ... makes no sense.
Here is the correct way to write the proof: suppose no proper subset of $A$ generates $S$. If $A$ is not linearly indpendent then there exists an equation of the form $sum c_ix_i=0$ with $x_i in A$ for all $i$ and not all of the coefficients are $0$. If $c_{i_0} neq 0$ we can rewrite the equation in the form $x_{i_0}=sum_{ineq i_0}b_ix_i$. Now let $A'=span(Asetminus {x_{i_0}})$. Then $A'$ is a proper subset of $A$ which generates $S$.
Conversely, suppose $A$ is linearly independent and let $A'$
be a proper subset of $A$. Then there is a vector $x in Asetminus A'$. This vector cannot be written as a linear combination of elements of $A'$ because $A$ is linearly independent. Hence $A'$ does not span $S$.
answered Dec 23 '18 at 12:01
Kavi Rama MurthyKavi Rama Murthy
64.7k42766
answered Dec 23 '18 at 12:01
Kavi Rama MurthyKavi Rama Murthy
64.7k42766
answered Dec 23 '18 at 12:01
Kavi Rama MurthyKavi Rama Murthy
64.7k42766
answered Dec 23 '18 at 12:01
Kavi Rama MurthyKavi Rama Murthy
64.7k42766
64.7k42766
$begingroup$
What do you mean by $span(A/x_{i_{o}})$? Assuming $x_i$ s are elements of $A$, is $A'=span(A/c_{i_{o}})$ what you meant? Also, how do we get that $A'$ generates $S$ from that statement, I am sorry but I am unable to understand.
$endgroup$
– Za Ira
Dec 23 '18 at 14:00
1
$begingroup$
@ZaIra span of a set is the set of all linear combinations of elements of that set. Every element of $S$ is a linear combination of elements of $A$. If one of these elements is $x_{i_0}$ then use the fact that $x_{i_0}$ is itself a linear combination of other elements of $A$ to conclude that $Asetminus x_{i_0}$ spans $S$.
$endgroup$
– Kavi Rama Murthy
Dec 23 '18 at 23:19
$begingroup$
oh i realize you used $setminus$ and not $/$. Thank-you!
$endgroup$
– Za Ira
Dec 24 '18 at 1:32
add a comment |
$begingroup$
What do you mean by $span(A/x_{i_{o}})$? Assuming $x_i$ s are elements of $A$, is $A'=span(A/c_{i_{o}})$ what you meant? Also, how do we get that $A'$ generates $S$ from that statement, I am sorry but I am unable to understand.
$endgroup$
– Za Ira
Dec 23 '18 at 14:00
1
$begingroup$
@ZaIra span of a set is the set of all linear combinations of elements of that set. Every element of $S$ is a linear combination of elements of $A$. If one of these elements is $x_{i_0}$ then use the fact that $x_{i_0}$ is itself a linear combination of other elements of $A$ to conclude that $Asetminus x_{i_0}$ spans $S$.
$endgroup$
– Kavi Rama Murthy
Dec 23 '18 at 23:19
$begingroup$
oh i realize you used $setminus$ and not $/$. Thank-you!
$endgroup$
– Za Ira
Dec 24 '18 at 1:32
$begingroup$
What do you mean by $span(A/x_{i_{o}})$? Assuming $x_i$ s are elements of $A$, is $A'=span(A/c_{i_{o}})$ what you meant? Also, how do we get that $A'$ generates $S$ from that statement, I am sorry but I am unable to understand.
$endgroup$
– Za Ira
Dec 23 '18 at 14:00
$begingroup$
What do you mean by $span(A/x_{i_{o}})$? Assuming $x_i$ s are elements of $A$, is $A'=span(A/c_{i_{o}})$ what you meant? Also, how do we get that $A'$ generates $S$ from that statement, I am sorry but I am unable to understand.
$endgroup$
– Za Ira
Dec 23 '18 at 14:00
1
1
$begingroup$
@ZaIra span of a set is the set of all linear combinations of elements of that set. Every element of $S$ is a linear combination of elements of $A$. If one of these elements is $x_{i_0}$ then use the fact that $x_{i_0}$ is itself a linear combination of other elements of $A$ to conclude that $Asetminus x_{i_0}$ spans $S$.
$endgroup$
– Kavi Rama Murthy
Dec 23 '18 at 23:19
$begingroup$
@ZaIra span of a set is the set of all linear combinations of elements of that set. Every element of $S$ is a linear combination of elements of $A$. If one of these elements is $x_{i_0}$ then use the fact that $x_{i_0}$ is itself a linear combination of other elements of $A$ to conclude that $Asetminus x_{i_0}$ spans $S$.
$endgroup$
– Kavi Rama Murthy
Dec 23 '18 at 23:19
$begingroup$
oh i realize you used $setminus$ and not $/$. Thank-you!
$endgroup$
– Za Ira
Dec 24 '18 at 1:32
$begingroup$
oh i realize you used $setminus$ and not $/$. Thank-you!
$endgroup$
– Za Ira
Dec 24 '18 at 1:32
add a comment |
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