Proof of $j=1$ where $v_j in span(v_1, …,v_{j-1})$
$begingroup$
In Linear Algebra Done Right, it presents the Linear Dependence Lemma which states that :
Suppose $v_1,...,v_m$ is a linearly dependent list in $V$. Then there exists $j in {1,2,...,m}$ such that the following hold: $v_j in span(v_1,...,v_{j-1})$. I follow the proof, but get confused on a special case where $j=1$. The book said choosing $j=1$ means that $v_1=0$, because if $j=1$ then the condition above is interpreted to mean that $v_1 in span()$.
I tried to follow the example in the proof:
Because the list $v_1,...,v_m$ is linearly dependent, there exist numbers $a_1,...,a_m in mathbb{F}$ , not all $0$ such that $$a_1v_1+...+a_mv_m = 0$$
Let $j$ be the first element of {1,...,m}, such that $a_j neq 0$. Then
$$a_1v_1 = -a_2v_2 -...-a_jv_j$$
$$v_1 = frac{-a_2}{a_1}v_2-...-frac{a_j}{a_1}v_j$$
Then does it mean $v_1$ is the span of $v_2,....v_j$?
linear-algebra
asked Dec 30 '18 at 22:38
JOHN JOHN
1518
$endgroup$
add a comment |
$begingroup$
In Linear Algebra Done Right, it presents the Linear Dependence Lemma which states that :
Suppose $v_1,...,v_m$ is a linearly dependent list in $V$. Then there exists $j in {1,2,...,m}$ such that the following hold: $v_j in span(v_1,...,v_{j-1})$. I follow the proof, but get confused on a special case where $j=1$. The book said choosing $j=1$ means that $v_1=0$, because if $j=1$ then the condition above is interpreted to mean that $v_1 in span()$.
I tried to follow the example in the proof:
Because the list $v_1,...,v_m$ is linearly dependent, there exist numbers $a_1,...,a_m in mathbb{F}$ , not all $0$ such that $$a_1v_1+...+a_mv_m = 0$$
Let $j$ be the first element of {1,...,m}, such that $a_j neq 0$. Then
$$a_1v_1 = -a_2v_2 -...-a_jv_j$$
$$v_1 = frac{-a_2}{a_1}v_2-...-frac{a_j}{a_1}v_j$$
Then does it mean $v_1$ is the span of $v_2,....v_j$?
linear-algebra
asked Dec 30 '18 at 22:38
JOHN JOHN
1518
$endgroup$
add a comment |
$begingroup$
In Linear Algebra Done Right, it presents the Linear Dependence Lemma which states that :
Suppose $v_1,...,v_m$ is a linearly dependent list in $V$. Then there exists $j in {1,2,...,m}$ such that the following hold: $v_j in span(v_1,...,v_{j-1})$. I follow the proof, but get confused on a special case where $j=1$. The book said choosing $j=1$ means that $v_1=0$, because if $j=1$ then the condition above is interpreted to mean that $v_1 in span()$.
I tried to follow the example in the proof:
Because the list $v_1,...,v_m$ is linearly dependent, there exist numbers $a_1,...,a_m in mathbb{F}$ , not all $0$ such that $$a_1v_1+...+a_mv_m = 0$$
Let $j$ be the first element of {1,...,m}, such that $a_j neq 0$. Then
$$a_1v_1 = -a_2v_2 -...-a_jv_j$$
$$v_1 = frac{-a_2}{a_1}v_2-...-frac{a_j}{a_1}v_j$$
Then does it mean $v_1$ is the span of $v_2,....v_j$?
linear-algebra
asked Dec 30 '18 at 22:38
JOHN JOHN
1518
$endgroup$
In Linear Algebra Done Right, it presents the Linear Dependence Lemma which states that :
Suppose $v_1,...,v_m$ is a linearly dependent list in $V$. Then there exists $j in {1,2,...,m}$ such that the following hold: $v_j in span(v_1,...,v_{j-1})$. I follow the proof, but get confused on a special case where $j=1$. The book said choosing $j=1$ means that $v_1=0$, because if $j=1$ then the condition above is interpreted to mean that $v_1 in span()$.
I tried to follow the example in the proof:
Because the list $v_1,...,v_m$ is linearly dependent, there exist numbers $a_1,...,a_m in mathbb{F}$ , not all $0$ such that $$a_1v_1+...+a_mv_m = 0$$
Let $j$ be the first element of {1,...,m}, such that $a_j neq 0$. Then
$$a_1v_1 = -a_2v_2 -...-a_jv_j$$
$$v_1 = frac{-a_2}{a_1}v_2-...-frac{a_j}{a_1}v_j$$
Then does it mean $v_1$ is the span of $v_2,....v_j$?
linear-algebra
linear-algebra
asked Dec 30 '18 at 22:38
JOHN JOHN
1518
asked Dec 30 '18 at 22:38
JOHN JOHN
1518
asked Dec 30 '18 at 22:38
JOHN JOHN
1518
asked Dec 30 '18 at 22:38
JOHN JOHN
1518
asked Dec 30 '18 at 22:38
JOHN JOHN
1518
1518
add a comment |
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
You've a mistake in the line after "Let $;j;$ be the first ..." . It must be;
$$a_jv_j=-a_1v_1-ldots-a_{j-1}v_{j-1}implies v_j=-frac{a_1}{a_j}v_1-ldots-frac{a_{j-1}}{a_j}v_{j-1}$$
and thus
$$v_jintext{Span},{v_1,...,v_{j-1}}$$
answered Dec 30 '18 at 22:46
DonAntonioDonAntonio
177k1492226
$endgroup$
add a comment |
$begingroup$
I have misunderstood the proof. It should be :
Let $j$ be the largest element of {1,...,m} such that $a_j neq 0$. It means that $a_{j+1} = 0, a_{j+2} = 0$, and so on.
Therefore, $v_j in span(v1,...,v_{j-1})$ actually means if the list is linearly independent, one of the vectors is in the span of the PREVIOUS ones.
So, in the case of $j=1$, $v_1 in span()$
More reference here
answered Dec 30 '18 at 23:02
JOHN JOHN
1518
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
You've a mistake in the line after "Let $;j;$ be the first ..." . It must be;
$$a_jv_j=-a_1v_1-ldots-a_{j-1}v_{j-1}implies v_j=-frac{a_1}{a_j}v_1-ldots-frac{a_{j-1}}{a_j}v_{j-1}$$
and thus
$$v_jintext{Span},{v_1,...,v_{j-1}}$$
answered Dec 30 '18 at 22:46
DonAntonioDonAntonio
177k1492226
$endgroup$
add a comment |
$begingroup$
You've a mistake in the line after "Let $;j;$ be the first ..." . It must be;
$$a_jv_j=-a_1v_1-ldots-a_{j-1}v_{j-1}implies v_j=-frac{a_1}{a_j}v_1-ldots-frac{a_{j-1}}{a_j}v_{j-1}$$
and thus
$$v_jintext{Span},{v_1,...,v_{j-1}}$$
answered Dec 30 '18 at 22:46
DonAntonioDonAntonio
177k1492226
$endgroup$
add a comment |
$begingroup$
You've a mistake in the line after "Let $;j;$ be the first ..." . It must be;
$$a_jv_j=-a_1v_1-ldots-a_{j-1}v_{j-1}implies v_j=-frac{a_1}{a_j}v_1-ldots-frac{a_{j-1}}{a_j}v_{j-1}$$
and thus
$$v_jintext{Span},{v_1,...,v_{j-1}}$$
answered Dec 30 '18 at 22:46
DonAntonioDonAntonio
177k1492226
$endgroup$
You've a mistake in the line after "Let $;j;$ be the first ..." . It must be;
$$a_jv_j=-a_1v_1-ldots-a_{j-1}v_{j-1}implies v_j=-frac{a_1}{a_j}v_1-ldots-frac{a_{j-1}}{a_j}v_{j-1}$$
and thus
$$v_jintext{Span},{v_1,...,v_{j-1}}$$
answered Dec 30 '18 at 22:46
DonAntonioDonAntonio
177k1492226
answered Dec 30 '18 at 22:46
DonAntonioDonAntonio
177k1492226
answered Dec 30 '18 at 22:46
DonAntonioDonAntonio
177k1492226
answered Dec 30 '18 at 22:46
DonAntonioDonAntonio
177k1492226
177k1492226
add a comment |
add a comment |
$begingroup$
I have misunderstood the proof. It should be :
Let $j$ be the largest element of {1,...,m} such that $a_j neq 0$. It means that $a_{j+1} = 0, a_{j+2} = 0$, and so on.
Therefore, $v_j in span(v1,...,v_{j-1})$ actually means if the list is linearly independent, one of the vectors is in the span of the PREVIOUS ones.
So, in the case of $j=1$, $v_1 in span()$
More reference here
answered Dec 30 '18 at 23:02
JOHN JOHN
1518
$endgroup$
add a comment |
$begingroup$
I have misunderstood the proof. It should be :
Let $j$ be the largest element of {1,...,m} such that $a_j neq 0$. It means that $a_{j+1} = 0, a_{j+2} = 0$, and so on.
Therefore, $v_j in span(v1,...,v_{j-1})$ actually means if the list is linearly independent, one of the vectors is in the span of the PREVIOUS ones.
So, in the case of $j=1$, $v_1 in span()$
More reference here
answered Dec 30 '18 at 23:02
JOHN JOHN
1518
$endgroup$
add a comment |
$begingroup$
I have misunderstood the proof. It should be :
Let $j$ be the largest element of {1,...,m} such that $a_j neq 0$. It means that $a_{j+1} = 0, a_{j+2} = 0$, and so on.
Therefore, $v_j in span(v1,...,v_{j-1})$ actually means if the list is linearly independent, one of the vectors is in the span of the PREVIOUS ones.
So, in the case of $j=1$, $v_1 in span()$
More reference here
answered Dec 30 '18 at 23:02
JOHN JOHN
1518
$endgroup$
I have misunderstood the proof. It should be :
Let $j$ be the largest element of {1,...,m} such that $a_j neq 0$. It means that $a_{j+1} = 0, a_{j+2} = 0$, and so on.
Therefore, $v_j in span(v1,...,v_{j-1})$ actually means if the list is linearly independent, one of the vectors is in the span of the PREVIOUS ones.
So, in the case of $j=1$, $v_1 in span()$
More reference here
answered Dec 30 '18 at 23:02
JOHN JOHN
1518
answered Dec 30 '18 at 23:02
JOHN JOHN
1518
answered Dec 30 '18 at 23:02
JOHN JOHN
1518
answered Dec 30 '18 at 23:02
JOHN JOHN
1518
1518
add a comment |
add a comment |
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