Mixing
Unclear theorem about vector spaces
$begingroup$
Can anyone explain the following theorem? When I try to understand it, it seems like a contradiction of a basis.
Is the vector v mentioned in the theorem not a linear combination of all the vectors in S, given the way v is derived? If it is a linear combination it can't be a basis, right? There is some part I don't understand correctly, I suppose.
vector-spaces independence
asked Jan 19 at 15:46
ToTomToTom
505
$endgroup$
add a comment |
$begingroup$
Can anyone explain the following theorem? When I try to understand it, it seems like a contradiction of a basis.
Is the vector v mentioned in the theorem not a linear combination of all the vectors in S, given the way v is derived? If it is a linear combination it can't be a basis, right? There is some part I don't understand correctly, I suppose.
vector-spaces independence
asked Jan 19 at 15:46
ToTomToTom
505
$endgroup$
$begingroup$
The theorem states that you can replace any vector of your basis by a non-nul linear combination of the basis vectors.
$endgroup$
– Bermudes
Jan 19 at 15:52
add a comment |
$begingroup$
Can anyone explain the following theorem? When I try to understand it, it seems like a contradiction of a basis.
Is the vector v mentioned in the theorem not a linear combination of all the vectors in S, given the way v is derived? If it is a linear combination it can't be a basis, right? There is some part I don't understand correctly, I suppose.
vector-spaces independence
asked Jan 19 at 15:46
ToTomToTom
505
$endgroup$
Can anyone explain the following theorem? When I try to understand it, it seems like a contradiction of a basis.
Is the vector v mentioned in the theorem not a linear combination of all the vectors in S, given the way v is derived? If it is a linear combination it can't be a basis, right? There is some part I don't understand correctly, I suppose.
vector-spaces independence
vector-spaces independence
asked Jan 19 at 15:46
ToTomToTom
505
asked Jan 19 at 15:46
ToTomToTom
505
asked Jan 19 at 15:46
ToTomToTom
505
asked Jan 19 at 15:46
ToTomToTom
505
asked Jan 19 at 15:46
ToTomToTom
505
505
$begingroup$
The theorem states that you can replace any vector of your basis by a non-nul linear combination of the basis vectors.
$endgroup$
– Bermudes
Jan 19 at 15:52
add a comment |
$begingroup$
The theorem states that you can replace any vector of your basis by a non-nul linear combination of the basis vectors.
$endgroup$
– Bermudes
Jan 19 at 15:52
$begingroup$
The theorem states that you can replace any vector of your basis by a non-nul linear combination of the basis vectors.
$endgroup$
– Bermudes
Jan 19 at 15:52
$begingroup$
The theorem states that you can replace any vector of your basis by a non-nul linear combination of the basis vectors.
$endgroup$
– Bermudes
Jan 19 at 15:52
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
What this theorem is saying, essentially, is that, given one basis of a finite-dimensional space, you can swap out one of the basis vectors for another, provided the new vector is a linear combination of the old set: the new set will still be a basis. You'll notice this is a Lemma, not a Theorem: I expect the authors are moving towards some theorems involving change-of-basis, which is quite important in linear algebra. This particular Lemma is not that important by itself except as a stepping-stone to change-of-basis.
Remember what a basis is: a set of linearly independent vectors such that any arbitrary vector is a linear combination of vectors in the set.
answered Jan 19 at 15:52
Adrian KeisterAdrian Keister
5,26371933
$endgroup$
$begingroup$
Thank you, I would understand it if the theorem said v was a lin. comb. of V, not a lin. comb. of the basis. But if v is a lin. comb of the basis, the basis is not lin. indep. anymore, no?
$endgroup$
– ToTom
Jan 19 at 16:26
1
$begingroup$
It doesn't mean the basis is no longer linearly independent. It does mean the set ${v,S}$ is not linearly independent. But now, because you're subtracting one vector from $S$ and substituting $v$, you still have the same number of total vectors, so that $(Ssetminus{w_i})cup {v}$ is still linearly independent.
$endgroup$
– Adrian Keister
Jan 19 at 16:30
$begingroup$
Thanks, now I got it. I didn't think that the linear combination of v was intended to be the sum of the original basis. Now it makes sense.
$endgroup$
– ToTom
Jan 19 at 16:36
$begingroup$
You're very welcome!
$endgroup$
– Adrian Keister
Jan 19 at 16:38
add a comment |
$begingroup$
The key part of the lemma is that the $i$th component of $v$ must be nonzero. This ensures that $v$ is outside the span of the other $w$s. Since the $w$s form a basis, it follows that $v$ is linearly independent from the other $w$s, and therefore replacing $w_i$ by $v$ yields a basis. The $i$ in the equation $gamma_i neq 0$ is the specific $i$ mentioned earlier in the lemma.
answered Jan 19 at 16:49
SimonSimon
763513
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
What this theorem is saying, essentially, is that, given one basis of a finite-dimensional space, you can swap out one of the basis vectors for another, provided the new vector is a linear combination of the old set: the new set will still be a basis. You'll notice this is a Lemma, not a Theorem: I expect the authors are moving towards some theorems involving change-of-basis, which is quite important in linear algebra. This particular Lemma is not that important by itself except as a stepping-stone to change-of-basis.
Remember what a basis is: a set of linearly independent vectors such that any arbitrary vector is a linear combination of vectors in the set.
answered Jan 19 at 15:52
Adrian KeisterAdrian Keister
5,26371933
$endgroup$
$begingroup$
Thank you, I would understand it if the theorem said v was a lin. comb. of V, not a lin. comb. of the basis. But if v is a lin. comb of the basis, the basis is not lin. indep. anymore, no?
$endgroup$
– ToTom
Jan 19 at 16:26
1
$begingroup$
It doesn't mean the basis is no longer linearly independent. It does mean the set ${v,S}$ is not linearly independent. But now, because you're subtracting one vector from $S$ and substituting $v$, you still have the same number of total vectors, so that $(Ssetminus{w_i})cup {v}$ is still linearly independent.
$endgroup$
– Adrian Keister
Jan 19 at 16:30
$begingroup$
Thanks, now I got it. I didn't think that the linear combination of v was intended to be the sum of the original basis. Now it makes sense.
$endgroup$
– ToTom
Jan 19 at 16:36
$begingroup$
You're very welcome!
$endgroup$
– Adrian Keister
Jan 19 at 16:38
add a comment |
$begingroup$
What this theorem is saying, essentially, is that, given one basis of a finite-dimensional space, you can swap out one of the basis vectors for another, provided the new vector is a linear combination of the old set: the new set will still be a basis. You'll notice this is a Lemma, not a Theorem: I expect the authors are moving towards some theorems involving change-of-basis, which is quite important in linear algebra. This particular Lemma is not that important by itself except as a stepping-stone to change-of-basis.
Remember what a basis is: a set of linearly independent vectors such that any arbitrary vector is a linear combination of vectors in the set.
answered Jan 19 at 15:52
Adrian KeisterAdrian Keister
5,26371933
$endgroup$
$begingroup$
Thank you, I would understand it if the theorem said v was a lin. comb. of V, not a lin. comb. of the basis. But if v is a lin. comb of the basis, the basis is not lin. indep. anymore, no?
$endgroup$
– ToTom
Jan 19 at 16:26
1
$begingroup$
It doesn't mean the basis is no longer linearly independent. It does mean the set ${v,S}$ is not linearly independent. But now, because you're subtracting one vector from $S$ and substituting $v$, you still have the same number of total vectors, so that $(Ssetminus{w_i})cup {v}$ is still linearly independent.
$endgroup$
– Adrian Keister
Jan 19 at 16:30
$begingroup$
Thanks, now I got it. I didn't think that the linear combination of v was intended to be the sum of the original basis. Now it makes sense.
$endgroup$
– ToTom
Jan 19 at 16:36
$begingroup$
You're very welcome!
$endgroup$
– Adrian Keister
Jan 19 at 16:38
add a comment |
$begingroup$
What this theorem is saying, essentially, is that, given one basis of a finite-dimensional space, you can swap out one of the basis vectors for another, provided the new vector is a linear combination of the old set: the new set will still be a basis. You'll notice this is a Lemma, not a Theorem: I expect the authors are moving towards some theorems involving change-of-basis, which is quite important in linear algebra. This particular Lemma is not that important by itself except as a stepping-stone to change-of-basis.
Remember what a basis is: a set of linearly independent vectors such that any arbitrary vector is a linear combination of vectors in the set.
answered Jan 19 at 15:52
Adrian KeisterAdrian Keister
5,26371933
$endgroup$
What this theorem is saying, essentially, is that, given one basis of a finite-dimensional space, you can swap out one of the basis vectors for another, provided the new vector is a linear combination of the old set: the new set will still be a basis. You'll notice this is a Lemma, not a Theorem: I expect the authors are moving towards some theorems involving change-of-basis, which is quite important in linear algebra. This particular Lemma is not that important by itself except as a stepping-stone to change-of-basis.
Remember what a basis is: a set of linearly independent vectors such that any arbitrary vector is a linear combination of vectors in the set.
answered Jan 19 at 15:52
Adrian KeisterAdrian Keister
5,26371933
answered Jan 19 at 15:52
Adrian KeisterAdrian Keister
5,26371933
answered Jan 19 at 15:52
Adrian KeisterAdrian Keister
5,26371933
answered Jan 19 at 15:52
Adrian KeisterAdrian Keister
5,26371933
5,26371933
$begingroup$
Thank you, I would understand it if the theorem said v was a lin. comb. of V, not a lin. comb. of the basis. But if v is a lin. comb of the basis, the basis is not lin. indep. anymore, no?
$endgroup$
– ToTom
Jan 19 at 16:26
1
$begingroup$
It doesn't mean the basis is no longer linearly independent. It does mean the set ${v,S}$ is not linearly independent. But now, because you're subtracting one vector from $S$ and substituting $v$, you still have the same number of total vectors, so that $(Ssetminus{w_i})cup {v}$ is still linearly independent.
$endgroup$
– Adrian Keister
Jan 19 at 16:30
$begingroup$
Thanks, now I got it. I didn't think that the linear combination of v was intended to be the sum of the original basis. Now it makes sense.
$endgroup$
– ToTom
Jan 19 at 16:36
$begingroup$
You're very welcome!
$endgroup$
– Adrian Keister
Jan 19 at 16:38
add a comment |
$begingroup$
Thank you, I would understand it if the theorem said v was a lin. comb. of V, not a lin. comb. of the basis. But if v is a lin. comb of the basis, the basis is not lin. indep. anymore, no?
$endgroup$
– ToTom
Jan 19 at 16:26
1
$begingroup$
It doesn't mean the basis is no longer linearly independent. It does mean the set ${v,S}$ is not linearly independent. But now, because you're subtracting one vector from $S$ and substituting $v$, you still have the same number of total vectors, so that $(Ssetminus{w_i})cup {v}$ is still linearly independent.
$endgroup$
– Adrian Keister
Jan 19 at 16:30
$begingroup$
Thanks, now I got it. I didn't think that the linear combination of v was intended to be the sum of the original basis. Now it makes sense.
$endgroup$
– ToTom
Jan 19 at 16:36
$begingroup$
You're very welcome!
$endgroup$
– Adrian Keister
Jan 19 at 16:38
$begingroup$
Thank you, I would understand it if the theorem said v was a lin. comb. of V, not a lin. comb. of the basis. But if v is a lin. comb of the basis, the basis is not lin. indep. anymore, no?
$endgroup$
– ToTom
Jan 19 at 16:26
$begingroup$
Thank you, I would understand it if the theorem said v was a lin. comb. of V, not a lin. comb. of the basis. But if v is a lin. comb of the basis, the basis is not lin. indep. anymore, no?
$endgroup$
– ToTom
Jan 19 at 16:26
1
1
$begingroup$
It doesn't mean the basis is no longer linearly independent. It does mean the set ${v,S}$ is not linearly independent. But now, because you're subtracting one vector from $S$ and substituting $v$, you still have the same number of total vectors, so that $(Ssetminus{w_i})cup {v}$ is still linearly independent.
$endgroup$
– Adrian Keister
Jan 19 at 16:30
$begingroup$
It doesn't mean the basis is no longer linearly independent. It does mean the set ${v,S}$ is not linearly independent. But now, because you're subtracting one vector from $S$ and substituting $v$, you still have the same number of total vectors, so that $(Ssetminus{w_i})cup {v}$ is still linearly independent.
$endgroup$
– Adrian Keister
Jan 19 at 16:30
$begingroup$
Thanks, now I got it. I didn't think that the linear combination of v was intended to be the sum of the original basis. Now it makes sense.
$endgroup$
– ToTom
Jan 19 at 16:36
$begingroup$
Thanks, now I got it. I didn't think that the linear combination of v was intended to be the sum of the original basis. Now it makes sense.
$endgroup$
– ToTom
Jan 19 at 16:36
$begingroup$
You're very welcome!
$endgroup$
– Adrian Keister
Jan 19 at 16:38
$begingroup$
You're very welcome!
$endgroup$
– Adrian Keister
Jan 19 at 16:38
add a comment |
$begingroup$
The key part of the lemma is that the $i$th component of $v$ must be nonzero. This ensures that $v$ is outside the span of the other $w$s. Since the $w$s form a basis, it follows that $v$ is linearly independent from the other $w$s, and therefore replacing $w_i$ by $v$ yields a basis. The $i$ in the equation $gamma_i neq 0$ is the specific $i$ mentioned earlier in the lemma.
answered Jan 19 at 16:49
SimonSimon
763513
$endgroup$
add a comment |
$begingroup$
The key part of the lemma is that the $i$th component of $v$ must be nonzero. This ensures that $v$ is outside the span of the other $w$s. Since the $w$s form a basis, it follows that $v$ is linearly independent from the other $w$s, and therefore replacing $w_i$ by $v$ yields a basis. The $i$ in the equation $gamma_i neq 0$ is the specific $i$ mentioned earlier in the lemma.
answered Jan 19 at 16:49
SimonSimon
763513
$endgroup$
add a comment |
$begingroup$
The key part of the lemma is that the $i$th component of $v$ must be nonzero. This ensures that $v$ is outside the span of the other $w$s. Since the $w$s form a basis, it follows that $v$ is linearly independent from the other $w$s, and therefore replacing $w_i$ by $v$ yields a basis. The $i$ in the equation $gamma_i neq 0$ is the specific $i$ mentioned earlier in the lemma.
answered Jan 19 at 16:49
SimonSimon
763513
$endgroup$
The key part of the lemma is that the $i$th component of $v$ must be nonzero. This ensures that $v$ is outside the span of the other $w$s. Since the $w$s form a basis, it follows that $v$ is linearly independent from the other $w$s, and therefore replacing $w_i$ by $v$ yields a basis. The $i$ in the equation $gamma_i neq 0$ is the specific $i$ mentioned earlier in the lemma.
answered Jan 19 at 16:49
SimonSimon
763513
answered Jan 19 at 16:49
SimonSimon
763513
answered Jan 19 at 16:49
SimonSimon
763513
answered Jan 19 at 16:49
SimonSimon
763513
763513
add a comment |
add a comment |
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$begingroup$
The theorem states that you can replace any vector of your basis by a non-nul linear combination of the basis vectors.
$endgroup$
– Bermudes
Jan 19 at 15:52