If $v_1, dots, v_m$ are linearly independent, then there is $w$ such that $langle w, v_j rangle > 0$ for...












2












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















2












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














2












2








2


1



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 0:47









Batominovski

33k33293




33k33293










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


















  • $begingroup$
    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










2 Answers
2






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1












$begingroup$

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 $j in {1,…,m}$.






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$endgroup$





















    0












    $begingroup$

    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 $a > frac{|w_{n-1}|}{|v_n|}$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      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
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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

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    1












    $begingroup$

    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 $j in {1,…,m}$.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      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 $j in {1,…,m}$.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        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 $j in {1,…,m}$.






        share|cite|improve this answer











        $endgroup$



        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 $j in {1,…,m}$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 16 '18 at 23:45

























        answered Dec 16 '18 at 23:31









        Paweł PodPaweł Pod

        112




        112























            0












            $begingroup$

            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 $a > frac{|w_{n-1}|}{|v_n|}$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              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
















            0












            $begingroup$

            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 $a > frac{|w_{n-1}|}{|v_n|}$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              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














            0












            0








            0





            $begingroup$

            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 $a > frac{|w_{n-1}|}{|v_n|}$.






            share|cite|improve this answer











            $endgroup$



            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 $a > frac{|w_{n-1}|}{|v_n|}$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jul 31 '14 at 2:35

























            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


















            • $begingroup$
              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


















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