Mixing
Proof that any set linearly independent has at most $n$ elements (when the vector space has basis with n...
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My teacher gave us this proof today, but I don't know if I understood it entirely:
Theorem:
Suppose $V$ a vector space (finitely generated) over the reals.
$$B = {v_1,cdots, v_n}$$
where $B$ is a basis for $V$.
Suppose, also, a set $Ssubset V$ L.I., then $S$ is finite and has at most $n$ elements.
Proof:
Let $S = {y_1,cdots,y_m}$, with $m>n$. Suppose $S$ is L.I. Then:
$$alpha_1y_1+cdots+alpha_my_m = 0tag{1}$$ But since $S$ is greater
than $B$, and $B$ is a basis, $S$ can be spanned by $B$. So any
$y_iin S$ can be written as a linear combination of $B$, in this way:
$$\y_1 = beta_{11}v_1+cdots+beta_{n1}v_n\y_2 =
beta_{12}v_1+cdots+beta_{n2}v_n\cdots\y_m =
beta_{1m}v_1+cdots+beta_{nm}v_n$$ Substituting the system above in
$(1)$, we have:
$$alpha_1(beta_{11}v_1+cdots+beta_{n1}v_n)+cdots+alpha_m(beta_{1m}v_1+cdots+beta_{nm}v_n)
= 0implies\ (beta_{11}alpha_1+beta_{12}alpha_2+cdots+beta_{1m}alpha_m)v_1+cdots+(beta_{n1}alpha_1+beta_{n2}alpha_2+cdots+beta_{nm}alpha_m)v_m
= 0$$ But since $B$ is L.I., then the linear combination above implies the coefficients are all $0$. Therefore we end up with the system:
$$begin{cases}beta_{11}alpha_1+beta_{12}alpha_2+cdots+beta_{1m}alpha_m = 0\
cdots \
beta_{n1}alpha_1+beta_{n2}alpha_2+cdots+beta_{nm}alpha_m = 0end{cases}$$
that has $n$ equations, and $m$ variables, therefore is compatible and
indeterminated, that implies it has infinitely many solutions. One of
them is not trivial, so $S$ is L.D.
I think wikipedia has a similar proof, but there, it proves directly that any basis has the same number of elements (could you tell me why a vector $v_i$ suddenly appears in the middle of the proof?. In this proof, the conclusion is that $S$ has at most $n$ elements.
Is my proof correct? I understood it correctly? (I didn't copy, this is my understanding of the proof).
linear-algebra vector-spaces proof-verification
asked Dec 8 '14 at 23:51
Guerlando OCsGuerlando OCs
11321856
$endgroup$
add a comment |
$begingroup$
My teacher gave us this proof today, but I don't know if I understood it entirely:
Theorem:
Suppose $V$ a vector space (finitely generated) over the reals.
$$B = {v_1,cdots, v_n}$$
where $B$ is a basis for $V$.
Suppose, also, a set $Ssubset V$ L.I., then $S$ is finite and has at most $n$ elements.
Proof:
Let $S = {y_1,cdots,y_m}$, with $m>n$. Suppose $S$ is L.I. Then:
$$alpha_1y_1+cdots+alpha_my_m = 0tag{1}$$ But since $S$ is greater
than $B$, and $B$ is a basis, $S$ can be spanned by $B$. So any
$y_iin S$ can be written as a linear combination of $B$, in this way:
$$\y_1 = beta_{11}v_1+cdots+beta_{n1}v_n\y_2 =
beta_{12}v_1+cdots+beta_{n2}v_n\cdots\y_m =
beta_{1m}v_1+cdots+beta_{nm}v_n$$ Substituting the system above in
$(1)$, we have:
$$alpha_1(beta_{11}v_1+cdots+beta_{n1}v_n)+cdots+alpha_m(beta_{1m}v_1+cdots+beta_{nm}v_n)
= 0implies\ (beta_{11}alpha_1+beta_{12}alpha_2+cdots+beta_{1m}alpha_m)v_1+cdots+(beta_{n1}alpha_1+beta_{n2}alpha_2+cdots+beta_{nm}alpha_m)v_m
= 0$$ But since $B$ is L.I., then the linear combination above implies the coefficients are all $0$. Therefore we end up with the system:
$$begin{cases}beta_{11}alpha_1+beta_{12}alpha_2+cdots+beta_{1m}alpha_m = 0\
cdots \
beta_{n1}alpha_1+beta_{n2}alpha_2+cdots+beta_{nm}alpha_m = 0end{cases}$$
that has $n$ equations, and $m$ variables, therefore is compatible and
indeterminated, that implies it has infinitely many solutions. One of
them is not trivial, so $S$ is L.D.
I think wikipedia has a similar proof, but there, it proves directly that any basis has the same number of elements (could you tell me why a vector $v_i$ suddenly appears in the middle of the proof?. In this proof, the conclusion is that $S$ has at most $n$ elements.
Is my proof correct? I understood it correctly? (I didn't copy, this is my understanding of the proof).
linear-algebra vector-spaces proof-verification
asked Dec 8 '14 at 23:51
Guerlando OCsGuerlando OCs
11321856
$endgroup$
$begingroup$
Your understanding seems to be correct. In (1) you wanted to say that for $S$ linearly independent, $alpha_1y_1+cdots+alpha_my_m = 0$ if and only if $alpha_{i}=0$, $forall i in lbrace 1, dots, m rbrace$. Then your proof essentially shows that there exists a set of $alpha_{i}$'s that are not all zero and still satisfy (1). Hence, S is not L.I. Note that it is important for the argument that $Rightarrow$ in the middle of your proof is actually an equivalence $Leftrightarrow$.
$endgroup$
– megas
Dec 9 '14 at 0:12
add a comment |
$begingroup$
My teacher gave us this proof today, but I don't know if I understood it entirely:
Theorem:
Suppose $V$ a vector space (finitely generated) over the reals.
$$B = {v_1,cdots, v_n}$$
where $B$ is a basis for $V$.
Suppose, also, a set $Ssubset V$ L.I., then $S$ is finite and has at most $n$ elements.
Proof:
Let $S = {y_1,cdots,y_m}$, with $m>n$. Suppose $S$ is L.I. Then:
$$alpha_1y_1+cdots+alpha_my_m = 0tag{1}$$ But since $S$ is greater
than $B$, and $B$ is a basis, $S$ can be spanned by $B$. So any
$y_iin S$ can be written as a linear combination of $B$, in this way:
$$\y_1 = beta_{11}v_1+cdots+beta_{n1}v_n\y_2 =
beta_{12}v_1+cdots+beta_{n2}v_n\cdots\y_m =
beta_{1m}v_1+cdots+beta_{nm}v_n$$ Substituting the system above in
$(1)$, we have:
$$alpha_1(beta_{11}v_1+cdots+beta_{n1}v_n)+cdots+alpha_m(beta_{1m}v_1+cdots+beta_{nm}v_n)
= 0implies\ (beta_{11}alpha_1+beta_{12}alpha_2+cdots+beta_{1m}alpha_m)v_1+cdots+(beta_{n1}alpha_1+beta_{n2}alpha_2+cdots+beta_{nm}alpha_m)v_m
= 0$$ But since $B$ is L.I., then the linear combination above implies the coefficients are all $0$. Therefore we end up with the system:
$$begin{cases}beta_{11}alpha_1+beta_{12}alpha_2+cdots+beta_{1m}alpha_m = 0\
cdots \
beta_{n1}alpha_1+beta_{n2}alpha_2+cdots+beta_{nm}alpha_m = 0end{cases}$$
that has $n$ equations, and $m$ variables, therefore is compatible and
indeterminated, that implies it has infinitely many solutions. One of
them is not trivial, so $S$ is L.D.
I think wikipedia has a similar proof, but there, it proves directly that any basis has the same number of elements (could you tell me why a vector $v_i$ suddenly appears in the middle of the proof?. In this proof, the conclusion is that $S$ has at most $n$ elements.
Is my proof correct? I understood it correctly? (I didn't copy, this is my understanding of the proof).
linear-algebra vector-spaces proof-verification
asked Dec 8 '14 at 23:51
Guerlando OCsGuerlando OCs
11321856
$endgroup$
My teacher gave us this proof today, but I don't know if I understood it entirely:
Theorem:
Suppose $V$ a vector space (finitely generated) over the reals.
$$B = {v_1,cdots, v_n}$$
where $B$ is a basis for $V$.
Suppose, also, a set $Ssubset V$ L.I., then $S$ is finite and has at most $n$ elements.
Proof:
Let $S = {y_1,cdots,y_m}$, with $m>n$. Suppose $S$ is L.I. Then:
$$alpha_1y_1+cdots+alpha_my_m = 0tag{1}$$ But since $S$ is greater
than $B$, and $B$ is a basis, $S$ can be spanned by $B$. So any
$y_iin S$ can be written as a linear combination of $B$, in this way:
$$\y_1 = beta_{11}v_1+cdots+beta_{n1}v_n\y_2 =
beta_{12}v_1+cdots+beta_{n2}v_n\cdots\y_m =
beta_{1m}v_1+cdots+beta_{nm}v_n$$ Substituting the system above in
$(1)$, we have:
$$alpha_1(beta_{11}v_1+cdots+beta_{n1}v_n)+cdots+alpha_m(beta_{1m}v_1+cdots+beta_{nm}v_n)
= 0implies\ (beta_{11}alpha_1+beta_{12}alpha_2+cdots+beta_{1m}alpha_m)v_1+cdots+(beta_{n1}alpha_1+beta_{n2}alpha_2+cdots+beta_{nm}alpha_m)v_m
= 0$$ But since $B$ is L.I., then the linear combination above implies the coefficients are all $0$. Therefore we end up with the system:
$$begin{cases}beta_{11}alpha_1+beta_{12}alpha_2+cdots+beta_{1m}alpha_m = 0\
cdots \
beta_{n1}alpha_1+beta_{n2}alpha_2+cdots+beta_{nm}alpha_m = 0end{cases}$$
that has $n$ equations, and $m$ variables, therefore is compatible and
indeterminated, that implies it has infinitely many solutions. One of
them is not trivial, so $S$ is L.D.
I think wikipedia has a similar proof, but there, it proves directly that any basis has the same number of elements (could you tell me why a vector $v_i$ suddenly appears in the middle of the proof?. In this proof, the conclusion is that $S$ has at most $n$ elements.
Is my proof correct? I understood it correctly? (I didn't copy, this is my understanding of the proof).
linear-algebra vector-spaces proof-verification
linear-algebra vector-spaces proof-verification
asked Dec 8 '14 at 23:51
Guerlando OCsGuerlando OCs
11321856
asked Dec 8 '14 at 23:51
Guerlando OCsGuerlando OCs
11321856
edited Dec 9 '14 at 0:20
Guerlando OCs
asked Dec 8 '14 at 23:51
Guerlando OCsGuerlando OCs
11321856
asked Dec 8 '14 at 23:51
Guerlando OCsGuerlando OCs
11321856
asked Dec 8 '14 at 23:51
Guerlando OCsGuerlando OCs
11321856
11321856
$begingroup$
Your understanding seems to be correct. In (1) you wanted to say that for $S$ linearly independent, $alpha_1y_1+cdots+alpha_my_m = 0$ if and only if $alpha_{i}=0$, $forall i in lbrace 1, dots, m rbrace$. Then your proof essentially shows that there exists a set of $alpha_{i}$'s that are not all zero and still satisfy (1). Hence, S is not L.I. Note that it is important for the argument that $Rightarrow$ in the middle of your proof is actually an equivalence $Leftrightarrow$.
$endgroup$
– megas
Dec 9 '14 at 0:12
add a comment |
$begingroup$
Your understanding seems to be correct. In (1) you wanted to say that for $S$ linearly independent, $alpha_1y_1+cdots+alpha_my_m = 0$ if and only if $alpha_{i}=0$, $forall i in lbrace 1, dots, m rbrace$. Then your proof essentially shows that there exists a set of $alpha_{i}$'s that are not all zero and still satisfy (1). Hence, S is not L.I. Note that it is important for the argument that $Rightarrow$ in the middle of your proof is actually an equivalence $Leftrightarrow$.
$endgroup$
– megas
Dec 9 '14 at 0:12
$begingroup$
Your understanding seems to be correct. In (1) you wanted to say that for $S$ linearly independent, $alpha_1y_1+cdots+alpha_my_m = 0$ if and only if $alpha_{i}=0$, $forall i in lbrace 1, dots, m rbrace$. Then your proof essentially shows that there exists a set of $alpha_{i}$'s that are not all zero and still satisfy (1). Hence, S is not L.I. Note that it is important for the argument that $Rightarrow$ in the middle of your proof is actually an equivalence $Leftrightarrow$.
$endgroup$
– megas
Dec 9 '14 at 0:12
$begingroup$
Your understanding seems to be correct. In (1) you wanted to say that for $S$ linearly independent, $alpha_1y_1+cdots+alpha_my_m = 0$ if and only if $alpha_{i}=0$, $forall i in lbrace 1, dots, m rbrace$. Then your proof essentially shows that there exists a set of $alpha_{i}$'s that are not all zero and still satisfy (1). Hence, S is not L.I. Note that it is important for the argument that $Rightarrow$ in the middle of your proof is actually an equivalence $Leftrightarrow$.
$endgroup$
– megas
Dec 9 '14 at 0:12
add a comment |
1 Answer
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It seems nothing wrong in your understanding. Probably what makes you confused is that the proof on wikipedia is a generalization of what you learnt in your class: your dimension theorem is a version only on finite dimensional vector spaces and that on wikipedia is the full version on both finite or infinite dimensional vector spaces.
In general, in a vector space $V$, the span of a set $S subseteq V$ is define as all finite linear combinations of elements in $S$. A set $S$ is linearly independent if, a finite linear combination of elements in $S$ is the zero vector implies the coefficients are all zero. The standard way of representing a finite combination is by $sum_{i in I}$ where $|I| < infty$ as in the proof from wikipedia.
answered Dec 9 '14 at 2:11
EmpiricistEmpiricist
6,83011433
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It seems nothing wrong in your understanding. Probably what makes you confused is that the proof on wikipedia is a generalization of what you learnt in your class: your dimension theorem is a version only on finite dimensional vector spaces and that on wikipedia is the full version on both finite or infinite dimensional vector spaces.
In general, in a vector space $V$, the span of a set $S subseteq V$ is define as all finite linear combinations of elements in $S$. A set $S$ is linearly independent if, a finite linear combination of elements in $S$ is the zero vector implies the coefficients are all zero. The standard way of representing a finite combination is by $sum_{i in I}$ where $|I| < infty$ as in the proof from wikipedia.
answered Dec 9 '14 at 2:11
EmpiricistEmpiricist
6,83011433
$endgroup$
add a comment |
$begingroup$
It seems nothing wrong in your understanding. Probably what makes you confused is that the proof on wikipedia is a generalization of what you learnt in your class: your dimension theorem is a version only on finite dimensional vector spaces and that on wikipedia is the full version on both finite or infinite dimensional vector spaces.
In general, in a vector space $V$, the span of a set $S subseteq V$ is define as all finite linear combinations of elements in $S$. A set $S$ is linearly independent if, a finite linear combination of elements in $S$ is the zero vector implies the coefficients are all zero. The standard way of representing a finite combination is by $sum_{i in I}$ where $|I| < infty$ as in the proof from wikipedia.
answered Dec 9 '14 at 2:11
EmpiricistEmpiricist
6,83011433
$endgroup$
add a comment |
$begingroup$
It seems nothing wrong in your understanding. Probably what makes you confused is that the proof on wikipedia is a generalization of what you learnt in your class: your dimension theorem is a version only on finite dimensional vector spaces and that on wikipedia is the full version on both finite or infinite dimensional vector spaces.
In general, in a vector space $V$, the span of a set $S subseteq V$ is define as all finite linear combinations of elements in $S$. A set $S$ is linearly independent if, a finite linear combination of elements in $S$ is the zero vector implies the coefficients are all zero. The standard way of representing a finite combination is by $sum_{i in I}$ where $|I| < infty$ as in the proof from wikipedia.
answered Dec 9 '14 at 2:11
EmpiricistEmpiricist
6,83011433
$endgroup$
It seems nothing wrong in your understanding. Probably what makes you confused is that the proof on wikipedia is a generalization of what you learnt in your class: your dimension theorem is a version only on finite dimensional vector spaces and that on wikipedia is the full version on both finite or infinite dimensional vector spaces.
In general, in a vector space $V$, the span of a set $S subseteq V$ is define as all finite linear combinations of elements in $S$. A set $S$ is linearly independent if, a finite linear combination of elements in $S$ is the zero vector implies the coefficients are all zero. The standard way of representing a finite combination is by $sum_{i in I}$ where $|I| < infty$ as in the proof from wikipedia.
answered Dec 9 '14 at 2:11
EmpiricistEmpiricist
6,83011433
edited Dec 9 '14 at 3:09
answered Dec 9 '14 at 2:11
EmpiricistEmpiricist
6,83011433
answered Dec 9 '14 at 2:11
EmpiricistEmpiricist
6,83011433
answered Dec 9 '14 at 2:11
EmpiricistEmpiricist
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6,83011433
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$begingroup$
Your understanding seems to be correct. In (1) you wanted to say that for $S$ linearly independent, $alpha_1y_1+cdots+alpha_my_m = 0$ if and only if $alpha_{i}=0$, $forall i in lbrace 1, dots, m rbrace$. Then your proof essentially shows that there exists a set of $alpha_{i}$'s that are not all zero and still satisfy (1). Hence, S is not L.I. Note that it is important for the argument that $Rightarrow$ in the middle of your proof is actually an equivalence $Leftrightarrow$.
$endgroup$
– megas
Dec 9 '14 at 0:12