If $v_1, dots, v_m$ are linearly independent, then there is $w$ such that $langle w, v_j rangle > 0$ for...
$begingroup$
Suppose $v_1, dots v_m$ is a linearly independent list in $V$. Show that there exists $w in V$ such that $langle w, v_j rangle > 0$ for all $j in {1, dots ,m}$.
I understand this question is saying given a linearly independent list, there is $w in V$ such that the vector $w$ is not orthogonal to any $v$ in that linearly independent set. I'm also confused as to why it is significant that the inner product be greater than zero and instead of just $neq 0$. Can someone give me a hint on how to do this problem?
I know that $langle v, v rangle >0$ for all $v$ not equal to zero, and since $v_1, dots v_m$ is linearly independent, then none of the $v_j$ will be zero, but it is impossible to have w equal to all $v_j$?
linear-algebra vector-spaces vectors inner-product-space
edited Dec 17 '18 at 0:47
Batominovski
33k33293
asked Jul 30 '14 at 21:55
SoapsSoaps
453315
$endgroup$
add a comment |
$begingroup$
Suppose $v_1, dots v_m$ is a linearly independent list in $V$. Show that there exists $w in V$ such that $langle w, v_j rangle > 0$ for all $j in {1, dots ,m}$.
I understand this question is saying given a linearly independent list, there is $w in V$ such that the vector $w$ is not orthogonal to any $v$ in that linearly independent set. I'm also confused as to why it is significant that the inner product be greater than zero and instead of just $neq 0$. Can someone give me a hint on how to do this problem?
I know that $langle v, v rangle >0$ for all $v$ not equal to zero, and since $v_1, dots v_m$ is linearly independent, then none of the $v_j$ will be zero, but it is impossible to have w equal to all $v_j$?
linear-algebra vector-spaces vectors inner-product-space
edited Dec 17 '18 at 0:47
Batominovski
33k33293
asked Jul 30 '14 at 21:55
SoapsSoaps
453315
$endgroup$
$begingroup$
Remember: it is possible to have $langle x,y rangle < 0$. For example, take $(1,0)$ and $(-1,0)$ in $Bbb R^2$. If you have an inner product space over $Bbb C$, then there are even more possibilities.
$endgroup$
– Omnomnomnom
Jul 30 '14 at 21:58
$begingroup$
Also, correction: "given a linearly independent list, there is w in V such that the vector w is not orthogonal to any v in that linearly independent set."
$endgroup$
– Omnomnomnom
Jul 30 '14 at 22:01
$begingroup$
The key is that we're not just finding one $w$ so that $(w,v_1) > 0$ and another so that $(w,v_2)$ and so on, we're finding a single $w$ for which $(w,v_1) > 0$ and $(w,v_2) > 0$ and so on.
$endgroup$
– Omnomnomnom
Jul 30 '14 at 22:04
$begingroup$
Are you given any information about $V$?
$endgroup$
– paw88789
Jul 30 '14 at 22:11
$begingroup$
Please try to write more descriptive titles in the future.
$endgroup$
– user147263
Jul 31 '14 at 1:55
add a comment |
$begingroup$
Suppose $v_1, dots v_m$ is a linearly independent list in $V$. Show that there exists $w in V$ such that $langle w, v_j rangle > 0$ for all $j in {1, dots ,m}$.
I understand this question is saying given a linearly independent list, there is $w in V$ such that the vector $w$ is not orthogonal to any $v$ in that linearly independent set. I'm also confused as to why it is significant that the inner product be greater than zero and instead of just $neq 0$. Can someone give me a hint on how to do this problem?
I know that $langle v, v rangle >0$ for all $v$ not equal to zero, and since $v_1, dots v_m$ is linearly independent, then none of the $v_j$ will be zero, but it is impossible to have w equal to all $v_j$?
linear-algebra vector-spaces vectors inner-product-space
edited Dec 17 '18 at 0:47
Batominovski
33k33293
asked Jul 30 '14 at 21:55
SoapsSoaps
453315
$endgroup$
Suppose $v_1, dots v_m$ is a linearly independent list in $V$. Show that there exists $w in V$ such that $langle w, v_j rangle > 0$ for all $j in {1, dots ,m}$.
I understand this question is saying given a linearly independent list, there is $w in V$ such that the vector $w$ is not orthogonal to any $v$ in that linearly independent set. I'm also confused as to why it is significant that the inner product be greater than zero and instead of just $neq 0$. Can someone give me a hint on how to do this problem?
I know that $langle v, v rangle >0$ for all $v$ not equal to zero, and since $v_1, dots v_m$ is linearly independent, then none of the $v_j$ will be zero, but it is impossible to have w equal to all $v_j$?
linear-algebra vector-spaces vectors inner-product-space
linear-algebra vector-spaces vectors inner-product-space
edited Dec 17 '18 at 0:47
Batominovski
33k33293
asked Jul 30 '14 at 21:55
SoapsSoaps
453315
edited Dec 17 '18 at 0:47
Batominovski
33k33293
asked Jul 30 '14 at 21:55
SoapsSoaps
453315
edited Dec 17 '18 at 0:47
Batominovski
33k33293
edited Dec 17 '18 at 0:47
Batominovski
33k33293
edited Dec 17 '18 at 0:47
Batominovski
33k33293
33k33293
asked Jul 30 '14 at 21:55
SoapsSoaps
453315
asked Jul 30 '14 at 21:55
SoapsSoaps
453315
asked Jul 30 '14 at 21:55
SoapsSoaps
453315
453315
$begingroup$
Remember: it is possible to have $langle x,y rangle < 0$. For example, take $(1,0)$ and $(-1,0)$ in $Bbb R^2$. If you have an inner product space over $Bbb C$, then there are even more possibilities.
$endgroup$
– Omnomnomnom
Jul 30 '14 at 21:58
$begingroup$
Also, correction: "given a linearly independent list, there is w in V such that the vector w is not orthogonal to any v in that linearly independent set."
$endgroup$
– Omnomnomnom
Jul 30 '14 at 22:01
$begingroup$
The key is that we're not just finding one $w$ so that $(w,v_1) > 0$ and another so that $(w,v_2)$ and so on, we're finding a single $w$ for which $(w,v_1) > 0$ and $(w,v_2) > 0$ and so on.
$endgroup$
– Omnomnomnom
Jul 30 '14 at 22:04
$begingroup$
Are you given any information about $V$?
$endgroup$
– paw88789
Jul 30 '14 at 22:11
$begingroup$
Please try to write more descriptive titles in the future.
$endgroup$
– user147263
Jul 31 '14 at 1:55
add a comment |
$begingroup$
Remember: it is possible to have $langle x,y rangle < 0$. For example, take $(1,0)$ and $(-1,0)$ in $Bbb R^2$. If you have an inner product space over $Bbb C$, then there are even more possibilities.
$endgroup$
– Omnomnomnom
Jul 30 '14 at 21:58
$begingroup$
Also, correction: "given a linearly independent list, there is w in V such that the vector w is not orthogonal to any v in that linearly independent set."
$endgroup$
– Omnomnomnom
Jul 30 '14 at 22:01
$begingroup$
The key is that we're not just finding one $w$ so that $(w,v_1) > 0$ and another so that $(w,v_2)$ and so on, we're finding a single $w$ for which $(w,v_1) > 0$ and $(w,v_2) > 0$ and so on.
$endgroup$
– Omnomnomnom
Jul 30 '14 at 22:04
$begingroup$
Are you given any information about $V$?
$endgroup$
– paw88789
Jul 30 '14 at 22:11
$begingroup$
Please try to write more descriptive titles in the future.
$endgroup$
– user147263
Jul 31 '14 at 1:55
$begingroup$
Remember: it is possible to have $langle x,y rangle < 0$. For example, take $(1,0)$ and $(-1,0)$ in $Bbb R^2$. If you have an inner product space over $Bbb C$, then there are even more possibilities.
$endgroup$
– Omnomnomnom
Jul 30 '14 at 21:58
$begingroup$
Remember: it is possible to have $langle x,y rangle < 0$. For example, take $(1,0)$ and $(-1,0)$ in $Bbb R^2$. If you have an inner product space over $Bbb C$, then there are even more possibilities.
$endgroup$
– Omnomnomnom
Jul 30 '14 at 21:58
$begingroup$
Also, correction: "given a linearly independent list, there is w in V such that the vector w is not orthogonal to any v in that linearly independent set."
$endgroup$
– Omnomnomnom
Jul 30 '14 at 22:01
$begingroup$
Also, correction: "given a linearly independent list, there is w in V such that the vector w is not orthogonal to any v in that linearly independent set."
$endgroup$
– Omnomnomnom
Jul 30 '14 at 22:01
$begingroup$
The key is that we're not just finding one $w$ so that $(w,v_1) > 0$ and another so that $(w,v_2)$ and so on, we're finding a single $w$ for which $(w,v_1) > 0$ and $(w,v_2) > 0$ and so on.
$endgroup$
– Omnomnomnom
Jul 30 '14 at 22:04
$begingroup$
The key is that we're not just finding one $w$ so that $(w,v_1) > 0$ and another so that $(w,v_2)$ and so on, we're finding a single $w$ for which $(w,v_1) > 0$ and $(w,v_2) > 0$ and so on.
$endgroup$
– Omnomnomnom
Jul 30 '14 at 22:04
$begingroup$
Are you given any information about $V$?
$endgroup$
– paw88789
Jul 30 '14 at 22:11
$begingroup$
Are you given any information about $V$?
$endgroup$
– paw88789
Jul 30 '14 at 22:11
$begingroup$
Please try to write more descriptive titles in the future.
$endgroup$
– user147263
Jul 31 '14 at 1:55
$begingroup$
Please try to write more descriptive titles in the future.
$endgroup$
– user147263
Jul 31 '14 at 1:55
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Consider a linear functional $phi$ from $span(v_1,…,v_m)$ to scalar field $mathbb F$ (I assume $mathbb R$ or $mathbb C$) given by formula $$phi(a_1v_1+cdots+a_mv_m)=a_1+cdots+a_m.$$ Clearly $phi(v_j)=1$ for $j in {1,…,m}$. By the Riesz Representation Theorem there is a vector $w in span(v_1,…,v_m) subset V$ such that $phi(v)=langle v,w rangle$ for every $v in span(v_1,…,v_m)$. Thus $langle v_j,w rangle = phi(v_j) = 1 >0$ for all $j in {1,…,m}$.
answered Dec 16 '18 at 23:31
Paweł PodPaweł Pod
112
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add a comment |
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Hint: suppose that $w_{n-1}$ is such that $langle w, v_j rangle > 0$ for $j = 1,dots,n-1$. Let $S_{n-1}$ be the span of $v_1,dots,v_{n-1}$. Suppose furthermore (without loss of generality) that $w in S_{n-1}$.
Let $v^perp$ be the component of $v_n$ perpendicular to $S_{n-1}$. Let $w_{n} = w_{n-1} + a v^perp$ for some constant $a>0$. We may select $a>0$ so that $w_n$ satisfies $langle w, v_j rangle > 0$ for $j = 1,dots,n$. Note that $w_n in S_n$, the span of $v_1,dots,v_n$.
In particular, for $i leq n-1$, we have
$$
langle w_n, v_i rangle = langle w_{n-1}, v_i rangle + alangle v^perp, v_i rangle =
langle w_{n-1}, v_i rangle
$$
and then
$$
langle w_n, v_nrangle = langle w_{n-1}, v_n rangle + alangle v^perp, v_n rangle
= langle w_{n-1}, v_n rangle + a|v^perp|^2 \
geq
a |v_n|^2-|w_{n-1}| |v_n| =
(a |v_n|-|w_{n-1}|) |v_n|
$$
So, in particular, it suffices to set $a > frac{|w_{n-1}|}{|v_n|}$.
answered Jul 30 '14 at 22:49
OmnomnomnomOmnomnomnom
128k791183
$endgroup$
$begingroup$
It's not quite true that $langle w_{n-1} + v^perp, v_n rangle > 0$ (consider $w_1 = v_1 = (1,0)$, $v_2 = (-1,1)$). However, one can take $w_n := w_{n-1} + av^perp$ for some $a gg 0$
$endgroup$
– zcn
Jul 30 '14 at 23:06
$begingroup$
@zcn thanks for the catch; I was thinking about $langle w_{n-1} + v^perp, v^perp rangle$.
$endgroup$
– Omnomnomnom
Jul 30 '14 at 23:11
$begingroup$
@Soaps feel free to ask for clarification if one of the steps is unclear.
$endgroup$
– Omnomnomnom
Jul 30 '14 at 23:12
$begingroup$
I just don't understand in general I'm sorry, do you think you could explain it in English maybe? Is $w_{n-1}$ different than $w_n$ in this case, I just don't really understand what your approach was?
$endgroup$
– Soaps
Jul 31 '14 at 2:17
$begingroup$
@Soaps the idea here is that I'm using an inductive approach. If you can find a $w$ for $n-1$ vectors, then you can use this process to find a $w$ for $n$ vectors.
$endgroup$
– Omnomnomnom
Jul 31 '14 at 2:23
|
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Consider a linear functional $phi$ from $span(v_1,…,v_m)$ to scalar field $mathbb F$ (I assume $mathbb R$ or $mathbb C$) given by formula $$phi(a_1v_1+cdots+a_mv_m)=a_1+cdots+a_m.$$ Clearly $phi(v_j)=1$ for $j in {1,…,m}$. By the Riesz Representation Theorem there is a vector $w in span(v_1,…,v_m) subset V$ such that $phi(v)=langle v,w rangle$ for every $v in span(v_1,…,v_m)$. Thus $langle v_j,w rangle = phi(v_j) = 1 >0$ for all $j in {1,…,m}$.
answered Dec 16 '18 at 23:31
Paweł PodPaweł Pod
112
$endgroup$
add a comment |
$begingroup$
Consider a linear functional $phi$ from $span(v_1,…,v_m)$ to scalar field $mathbb F$ (I assume $mathbb R$ or $mathbb C$) given by formula $$phi(a_1v_1+cdots+a_mv_m)=a_1+cdots+a_m.$$ Clearly $phi(v_j)=1$ for $j in {1,…,m}$. By the Riesz Representation Theorem there is a vector $w in span(v_1,…,v_m) subset V$ such that $phi(v)=langle v,w rangle$ for every $v in span(v_1,…,v_m)$. Thus $langle v_j,w rangle = phi(v_j) = 1 >0$ for all $j in {1,…,m}$.
answered Dec 16 '18 at 23:31
Paweł PodPaweł Pod
112
$endgroup$
add a comment |
$begingroup$
Consider a linear functional $phi$ from $span(v_1,…,v_m)$ to scalar field $mathbb F$ (I assume $mathbb R$ or $mathbb C$) given by formula $$phi(a_1v_1+cdots+a_mv_m)=a_1+cdots+a_m.$$ Clearly $phi(v_j)=1$ for $j in {1,…,m}$. By the Riesz Representation Theorem there is a vector $w in span(v_1,…,v_m) subset V$ such that $phi(v)=langle v,w rangle$ for every $v in span(v_1,…,v_m)$. Thus $langle v_j,w rangle = phi(v_j) = 1 >0$ for all $j in {1,…,m}$.
answered Dec 16 '18 at 23:31
Paweł PodPaweł Pod
112
$endgroup$
Consider a linear functional $phi$ from $span(v_1,…,v_m)$ to scalar field $mathbb F$ (I assume $mathbb R$ or $mathbb C$) given by formula $$phi(a_1v_1+cdots+a_mv_m)=a_1+cdots+a_m.$$ Clearly $phi(v_j)=1$ for $j in {1,…,m}$. By the Riesz Representation Theorem there is a vector $w in span(v_1,…,v_m) subset V$ such that $phi(v)=langle v,w rangle$ for every $v in span(v_1,…,v_m)$. Thus $langle v_j,w rangle = phi(v_j) = 1 >0$ for all $j in {1,…,m}$.
answered Dec 16 '18 at 23:31
Paweł PodPaweł Pod
112
edited Dec 16 '18 at 23:45
answered Dec 16 '18 at 23:31
Paweł PodPaweł Pod
112
answered Dec 16 '18 at 23:31
Paweł PodPaweł Pod
112
answered Dec 16 '18 at 23:31
Paweł PodPaweł Pod
112
112
add a comment |
add a comment |
$begingroup$
Hint: suppose that $w_{n-1}$ is such that $langle w, v_j rangle > 0$ for $j = 1,dots,n-1$. Let $S_{n-1}$ be the span of $v_1,dots,v_{n-1}$. Suppose furthermore (without loss of generality) that $w in S_{n-1}$.
Let $v^perp$ be the component of $v_n$ perpendicular to $S_{n-1}$. Let $w_{n} = w_{n-1} + a v^perp$ for some constant $a>0$. We may select $a>0$ so that $w_n$ satisfies $langle w, v_j rangle > 0$ for $j = 1,dots,n$. Note that $w_n in S_n$, the span of $v_1,dots,v_n$.
In particular, for $i leq n-1$, we have
$$
langle w_n, v_i rangle = langle w_{n-1}, v_i rangle + alangle v^perp, v_i rangle =
langle w_{n-1}, v_i rangle
$$
and then
$$
langle w_n, v_nrangle = langle w_{n-1}, v_n rangle + alangle v^perp, v_n rangle
= langle w_{n-1}, v_n rangle + a|v^perp|^2 \
geq
a |v_n|^2-|w_{n-1}| |v_n| =
(a |v_n|-|w_{n-1}|) |v_n|
$$
So, in particular, it suffices to set $a > frac{|w_{n-1}|}{|v_n|}$.
answered Jul 30 '14 at 22:49
OmnomnomnomOmnomnomnom
128k791183
$endgroup$
$begingroup$
It's not quite true that $langle w_{n-1} + v^perp, v_n rangle > 0$ (consider $w_1 = v_1 = (1,0)$, $v_2 = (-1,1)$). However, one can take $w_n := w_{n-1} + av^perp$ for some $a gg 0$
$endgroup$
– zcn
Jul 30 '14 at 23:06
$begingroup$
@zcn thanks for the catch; I was thinking about $langle w_{n-1} + v^perp, v^perp rangle$.
$endgroup$
– Omnomnomnom
Jul 30 '14 at 23:11
$begingroup$
@Soaps feel free to ask for clarification if one of the steps is unclear.
$endgroup$
– Omnomnomnom
Jul 30 '14 at 23:12
$begingroup$
I just don't understand in general I'm sorry, do you think you could explain it in English maybe? Is $w_{n-1}$ different than $w_n$ in this case, I just don't really understand what your approach was?
$endgroup$
– Soaps
Jul 31 '14 at 2:17
$begingroup$
@Soaps the idea here is that I'm using an inductive approach. If you can find a $w$ for $n-1$ vectors, then you can use this process to find a $w$ for $n$ vectors.
$endgroup$
– Omnomnomnom
Jul 31 '14 at 2:23
|
show 1 more comment
$begingroup$
Hint: suppose that $w_{n-1}$ is such that $langle w, v_j rangle > 0$ for $j = 1,dots,n-1$. Let $S_{n-1}$ be the span of $v_1,dots,v_{n-1}$. Suppose furthermore (without loss of generality) that $w in S_{n-1}$.
Let $v^perp$ be the component of $v_n$ perpendicular to $S_{n-1}$. Let $w_{n} = w_{n-1} + a v^perp$ for some constant $a>0$. We may select $a>0$ so that $w_n$ satisfies $langle w, v_j rangle > 0$ for $j = 1,dots,n$. Note that $w_n in S_n$, the span of $v_1,dots,v_n$.
In particular, for $i leq n-1$, we have
$$
langle w_n, v_i rangle = langle w_{n-1}, v_i rangle + alangle v^perp, v_i rangle =
langle w_{n-1}, v_i rangle
$$
and then
$$
langle w_n, v_nrangle = langle w_{n-1}, v_n rangle + alangle v^perp, v_n rangle
= langle w_{n-1}, v_n rangle + a|v^perp|^2 \
geq
a |v_n|^2-|w_{n-1}| |v_n| =
(a |v_n|-|w_{n-1}|) |v_n|
$$
So, in particular, it suffices to set $a > frac{|w_{n-1}|}{|v_n|}$.
answered Jul 30 '14 at 22:49
OmnomnomnomOmnomnomnom
128k791183
$endgroup$
$begingroup$
It's not quite true that $langle w_{n-1} + v^perp, v_n rangle > 0$ (consider $w_1 = v_1 = (1,0)$, $v_2 = (-1,1)$). However, one can take $w_n := w_{n-1} + av^perp$ for some $a gg 0$
$endgroup$
– zcn
Jul 30 '14 at 23:06
$begingroup$
@zcn thanks for the catch; I was thinking about $langle w_{n-1} + v^perp, v^perp rangle$.
$endgroup$
– Omnomnomnom
Jul 30 '14 at 23:11
$begingroup$
@Soaps feel free to ask for clarification if one of the steps is unclear.
$endgroup$
– Omnomnomnom
Jul 30 '14 at 23:12
$begingroup$
I just don't understand in general I'm sorry, do you think you could explain it in English maybe? Is $w_{n-1}$ different than $w_n$ in this case, I just don't really understand what your approach was?
$endgroup$
– Soaps
Jul 31 '14 at 2:17
$begingroup$
@Soaps the idea here is that I'm using an inductive approach. If you can find a $w$ for $n-1$ vectors, then you can use this process to find a $w$ for $n$ vectors.
$endgroup$
– Omnomnomnom
Jul 31 '14 at 2:23
|
show 1 more comment
$begingroup$
Hint: suppose that $w_{n-1}$ is such that $langle w, v_j rangle > 0$ for $j = 1,dots,n-1$. Let $S_{n-1}$ be the span of $v_1,dots,v_{n-1}$. Suppose furthermore (without loss of generality) that $w in S_{n-1}$.
Let $v^perp$ be the component of $v_n$ perpendicular to $S_{n-1}$. Let $w_{n} = w_{n-1} + a v^perp$ for some constant $a>0$. We may select $a>0$ so that $w_n$ satisfies $langle w, v_j rangle > 0$ for $j = 1,dots,n$. Note that $w_n in S_n$, the span of $v_1,dots,v_n$.
In particular, for $i leq n-1$, we have
$$
langle w_n, v_i rangle = langle w_{n-1}, v_i rangle + alangle v^perp, v_i rangle =
langle w_{n-1}, v_i rangle
$$
and then
$$
langle w_n, v_nrangle = langle w_{n-1}, v_n rangle + alangle v^perp, v_n rangle
= langle w_{n-1}, v_n rangle + a|v^perp|^2 \
geq
a |v_n|^2-|w_{n-1}| |v_n| =
(a |v_n|-|w_{n-1}|) |v_n|
$$
So, in particular, it suffices to set $a > frac{|w_{n-1}|}{|v_n|}$.
answered Jul 30 '14 at 22:49
OmnomnomnomOmnomnomnom
128k791183
$endgroup$
Hint: suppose that $w_{n-1}$ is such that $langle w, v_j rangle > 0$ for $j = 1,dots,n-1$. Let $S_{n-1}$ be the span of $v_1,dots,v_{n-1}$. Suppose furthermore (without loss of generality) that $w in S_{n-1}$.
Let $v^perp$ be the component of $v_n$ perpendicular to $S_{n-1}$. Let $w_{n} = w_{n-1} + a v^perp$ for some constant $a>0$. We may select $a>0$ so that $w_n$ satisfies $langle w, v_j rangle > 0$ for $j = 1,dots,n$. Note that $w_n in S_n$, the span of $v_1,dots,v_n$.
In particular, for $i leq n-1$, we have
$$
langle w_n, v_i rangle = langle w_{n-1}, v_i rangle + alangle v^perp, v_i rangle =
langle w_{n-1}, v_i rangle
$$
and then
$$
langle w_n, v_nrangle = langle w_{n-1}, v_n rangle + alangle v^perp, v_n rangle
= langle w_{n-1}, v_n rangle + a|v^perp|^2 \
geq
a |v_n|^2-|w_{n-1}| |v_n| =
(a |v_n|-|w_{n-1}|) |v_n|
$$
So, in particular, it suffices to set $a > frac{|w_{n-1}|}{|v_n|}$.
answered Jul 30 '14 at 22:49
OmnomnomnomOmnomnomnom
128k791183
edited Jul 31 '14 at 2:35
answered Jul 30 '14 at 22:49
OmnomnomnomOmnomnomnom
128k791183
answered Jul 30 '14 at 22:49
OmnomnomnomOmnomnomnom
128k791183
answered Jul 30 '14 at 22:49
OmnomnomnomOmnomnomnom
128k791183
128k791183
$begingroup$
It's not quite true that $langle w_{n-1} + v^perp, v_n rangle > 0$ (consider $w_1 = v_1 = (1,0)$, $v_2 = (-1,1)$). However, one can take $w_n := w_{n-1} + av^perp$ for some $a gg 0$
$endgroup$
– zcn
Jul 30 '14 at 23:06
$begingroup$
@zcn thanks for the catch; I was thinking about $langle w_{n-1} + v^perp, v^perp rangle$.
$endgroup$
– Omnomnomnom
Jul 30 '14 at 23:11
$begingroup$
@Soaps feel free to ask for clarification if one of the steps is unclear.
$endgroup$
– Omnomnomnom
Jul 30 '14 at 23:12
$begingroup$
I just don't understand in general I'm sorry, do you think you could explain it in English maybe? Is $w_{n-1}$ different than $w_n$ in this case, I just don't really understand what your approach was?
$endgroup$
– Soaps
Jul 31 '14 at 2:17
$begingroup$
@Soaps the idea here is that I'm using an inductive approach. If you can find a $w$ for $n-1$ vectors, then you can use this process to find a $w$ for $n$ vectors.
$endgroup$
– Omnomnomnom
Jul 31 '14 at 2:23
|
show 1 more comment
$begingroup$
It's not quite true that $langle w_{n-1} + v^perp, v_n rangle > 0$ (consider $w_1 = v_1 = (1,0)$, $v_2 = (-1,1)$). However, one can take $w_n := w_{n-1} + av^perp$ for some $a gg 0$
$endgroup$
– zcn
Jul 30 '14 at 23:06
$begingroup$
@zcn thanks for the catch; I was thinking about $langle w_{n-1} + v^perp, v^perp rangle$.
$endgroup$
– Omnomnomnom
Jul 30 '14 at 23:11
$begingroup$
@Soaps feel free to ask for clarification if one of the steps is unclear.
$endgroup$
– Omnomnomnom
Jul 30 '14 at 23:12
$begingroup$
I just don't understand in general I'm sorry, do you think you could explain it in English maybe? Is $w_{n-1}$ different than $w_n$ in this case, I just don't really understand what your approach was?
$endgroup$
– Soaps
Jul 31 '14 at 2:17
$begingroup$
@Soaps the idea here is that I'm using an inductive approach. If you can find a $w$ for $n-1$ vectors, then you can use this process to find a $w$ for $n$ vectors.
$endgroup$
– Omnomnomnom
Jul 31 '14 at 2:23
$begingroup$
It's not quite true that $langle w_{n-1} + v^perp, v_n rangle > 0$ (consider $w_1 = v_1 = (1,0)$, $v_2 = (-1,1)$). However, one can take $w_n := w_{n-1} + av^perp$ for some $a gg 0$
$endgroup$
– zcn
Jul 30 '14 at 23:06
$begingroup$
It's not quite true that $langle w_{n-1} + v^perp, v_n rangle > 0$ (consider $w_1 = v_1 = (1,0)$, $v_2 = (-1,1)$). However, one can take $w_n := w_{n-1} + av^perp$ for some $a gg 0$
$endgroup$
– zcn
Jul 30 '14 at 23:06
$begingroup$
@zcn thanks for the catch; I was thinking about $langle w_{n-1} + v^perp, v^perp rangle$.
$endgroup$
– Omnomnomnom
Jul 30 '14 at 23:11
$begingroup$
@zcn thanks for the catch; I was thinking about $langle w_{n-1} + v^perp, v^perp rangle$.
$endgroup$
– Omnomnomnom
Jul 30 '14 at 23:11
$begingroup$
@Soaps feel free to ask for clarification if one of the steps is unclear.
$endgroup$
– Omnomnomnom
Jul 30 '14 at 23:12
$begingroup$
@Soaps feel free to ask for clarification if one of the steps is unclear.
$endgroup$
– Omnomnomnom
Jul 30 '14 at 23:12
$begingroup$
I just don't understand in general I'm sorry, do you think you could explain it in English maybe? Is $w_{n-1}$ different than $w_n$ in this case, I just don't really understand what your approach was?
$endgroup$
– Soaps
Jul 31 '14 at 2:17
$begingroup$
I just don't understand in general I'm sorry, do you think you could explain it in English maybe? Is $w_{n-1}$ different than $w_n$ in this case, I just don't really understand what your approach was?
$endgroup$
– Soaps
Jul 31 '14 at 2:17
$begingroup$
@Soaps the idea here is that I'm using an inductive approach. If you can find a $w$ for $n-1$ vectors, then you can use this process to find a $w$ for $n$ vectors.
$endgroup$
– Omnomnomnom
Jul 31 '14 at 2:23
$begingroup$
@Soaps the idea here is that I'm using an inductive approach. If you can find a $w$ for $n-1$ vectors, then you can use this process to find a $w$ for $n$ vectors.
$endgroup$
– Omnomnomnom
Jul 31 '14 at 2:23
|
show 1 more comment
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$begingroup$
Remember: it is possible to have $langle x,y rangle < 0$. For example, take $(1,0)$ and $(-1,0)$ in $Bbb R^2$. If you have an inner product space over $Bbb C$, then there are even more possibilities.
$endgroup$
– Omnomnomnom
Jul 30 '14 at 21:58
$begingroup$
Also, correction: "given a linearly independent list, there is w in V such that the vector w is not orthogonal to any v in that linearly independent set."
$endgroup$
– Omnomnomnom
Jul 30 '14 at 22:01
$begingroup$
The key is that we're not just finding one $w$ so that $(w,v_1) > 0$ and another so that $(w,v_2)$ and so on, we're finding a single $w$ for which $(w,v_1) > 0$ and $(w,v_2) > 0$ and so on.
$endgroup$
– Omnomnomnom
Jul 30 '14 at 22:04
$begingroup$
Are you given any information about $V$?
$endgroup$
– paw88789
Jul 30 '14 at 22:11
$begingroup$
Please try to write more descriptive titles in the future.
$endgroup$
– user147263
Jul 31 '14 at 1:55