Proving linear independence and that $dim V$ is greater than or equals 3.












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Let $V= operatorname{span}{v_1,v_2,v_3,v_4}$ be a vector space such that $v_i$ are unit vectors for all $i$ and $v_i.v_j<0$ if $i$ does not equals to $j$.



(i) Show that no two vectors among {$v_1,v_2,v_3,v_4$} are linearly dependent.



(ii) Prove that $dim V$ is greater than or equals to 3.



My approach:
(i) If all the vectors are linearly dependent, this would mean that $v_i=av_j$.
However, since |$v_i$| = 1, it does not = a|$v_j$| unless $a = 1$. Thus, it can be proven that no two vectors in the set are linearly dependent.



Please correct me if my approach is incorrect, and may I ask how do I approach the second part of the question?






linear-algebra matrices eigenvalues-eigenvectors independence







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    In 1) you only get $|a|=1$ not $a=1$. So $a =pm 1$, If $a=-1$ you have to use the fact that $v_3.v_1<0$ and $v_3.v_2<0$ to get a contradiction.
    – Kavi Rama Murthy
    Nov 27 '18 at 6:33


















1














Let $V= operatorname{span}{v_1,v_2,v_3,v_4}$ be a vector space such that $v_i$ are unit vectors for all $i$ and $v_i.v_j<0$ if $i$ does not equals to $j$.



(i) Show that no two vectors among {$v_1,v_2,v_3,v_4$} are linearly dependent.



(ii) Prove that $dim V$ is greater than or equals to 3.



My approach:
(i) If all the vectors are linearly dependent, this would mean that $v_i=av_j$.
However, since |$v_i$| = 1, it does not = a|$v_j$| unless $a = 1$. Thus, it can be proven that no two vectors in the set are linearly dependent.



Please correct me if my approach is incorrect, and may I ask how do I approach the second part of the question?






linear-algebra matrices eigenvalues-eigenvectors independence







share|cite|improve this question




















  • 1




    In 1) you only get $|a|=1$ not $a=1$. So $a =pm 1$, If $a=-1$ you have to use the fact that $v_3.v_1<0$ and $v_3.v_2<0$ to get a contradiction.
    – Kavi Rama Murthy
    Nov 27 '18 at 6:33
















1












1








1







Let $V= operatorname{span}{v_1,v_2,v_3,v_4}$ be a vector space such that $v_i$ are unit vectors for all $i$ and $v_i.v_j<0$ if $i$ does not equals to $j$.



(i) Show that no two vectors among {$v_1,v_2,v_3,v_4$} are linearly dependent.



(ii) Prove that $dim V$ is greater than or equals to 3.



My approach:
(i) If all the vectors are linearly dependent, this would mean that $v_i=av_j$.
However, since |$v_i$| = 1, it does not = a|$v_j$| unless $a = 1$. Thus, it can be proven that no two vectors in the set are linearly dependent.



Please correct me if my approach is incorrect, and may I ask how do I approach the second part of the question?






linear-algebra matrices eigenvalues-eigenvectors independence







share|cite|improve this question















Let $V= operatorname{span}{v_1,v_2,v_3,v_4}$ be a vector space such that $v_i$ are unit vectors for all $i$ and $v_i.v_j<0$ if $i$ does not equals to $j$.



(i) Show that no two vectors among {$v_1,v_2,v_3,v_4$} are linearly dependent.



(ii) Prove that $dim V$ is greater than or equals to 3.



My approach:
(i) If all the vectors are linearly dependent, this would mean that $v_i=av_j$.
However, since |$v_i$| = 1, it does not = a|$v_j$| unless $a = 1$. Thus, it can be proven that no two vectors in the set are linearly dependent.



Please correct me if my approach is incorrect, and may I ask how do I approach the second part of the question?






linear-algebra matrices eigenvalues-eigenvectors independence




linear-algebra matrices eigenvalues-eigenvectors independence






share|cite|improve this question















share|cite|improve this question













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edited Nov 27 '18 at 10:17









Bernard

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118k639112










asked Nov 27 '18 at 6:20









Cheryl

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  • 1




    In 1) you only get $|a|=1$ not $a=1$. So $a =pm 1$, If $a=-1$ you have to use the fact that $v_3.v_1<0$ and $v_3.v_2<0$ to get a contradiction.
    – Kavi Rama Murthy
    Nov 27 '18 at 6:33
















  • 1




    In 1) you only get $|a|=1$ not $a=1$. So $a =pm 1$, If $a=-1$ you have to use the fact that $v_3.v_1<0$ and $v_3.v_2<0$ to get a contradiction.
    – Kavi Rama Murthy
    Nov 27 '18 at 6:33










1




1




In 1) you only get $|a|=1$ not $a=1$. So $a =pm 1$, If $a=-1$ you have to use the fact that $v_3.v_1<0$ and $v_3.v_2<0$ to get a contradiction.
– Kavi Rama Murthy
Nov 27 '18 at 6:33






In 1) you only get $|a|=1$ not $a=1$. So $a =pm 1$, If $a=-1$ you have to use the fact that $v_3.v_1<0$ and $v_3.v_2<0$ to get a contradiction.
– Kavi Rama Murthy
Nov 27 '18 at 6:33












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(i) Assume that exists $i,jin{1,2,3,4}$ such that $v_{i}=av_{j}$. Considering the norm, we have $1=||v_{i}||=||av_{j}||=|a|,||v_{j}||=|a| Rightarrow a=-1,$, since $a=v_{i}cdot v_{j}<0$. However, we have $v_{i}cdot v_{k}=a v_{j}cdot v_{k}=- v_{j}cdot v_{k} > 0$ for $kneq i,j$. This is absurd.



(ii) Thanks to (i) $,operatorname{dim} V >1$, so it is sufficient to prove that $operatorname{dim} V neq 2$.

We assume for absurd that $operatorname{dim} V = 2$.
Let $w= v_{2} - frac{(v_{2}cdot v_{1})}{||v_{1}||}, v_{1} in V$ such that $wcdot v_{1}=0$. Then $operatorname{span}{v_{1},v_{2},v_{3},v_{4}}=operatorname{span}{v_{1},w}$, and we can write
$$v_{3}=a_{1}v_{1}+a_{2}w; ,; v_{4}=b_{1}v_{1}+b_{2}w ;;text{with}; a_{i},b_{i}<0$$
but $0>v_{3}cdot v_{4}=a_{1}b_{1}+a_{2}b_{2}>0$.

Follow that $operatorname{dim}Vgeq 3$.






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    (i) Assume that exists $i,jin{1,2,3,4}$ such that $v_{i}=av_{j}$. Considering the norm, we have $1=||v_{i}||=||av_{j}||=|a|,||v_{j}||=|a| Rightarrow a=-1,$, since $a=v_{i}cdot v_{j}<0$. However, we have $v_{i}cdot v_{k}=a v_{j}cdot v_{k}=- v_{j}cdot v_{k} > 0$ for $kneq i,j$. This is absurd.



    (ii) Thanks to (i) $,operatorname{dim} V >1$, so it is sufficient to prove that $operatorname{dim} V neq 2$.

    We assume for absurd that $operatorname{dim} V = 2$.
    Let $w= v_{2} - frac{(v_{2}cdot v_{1})}{||v_{1}||}, v_{1} in V$ such that $wcdot v_{1}=0$. Then $operatorname{span}{v_{1},v_{2},v_{3},v_{4}}=operatorname{span}{v_{1},w}$, and we can write
    $$v_{3}=a_{1}v_{1}+a_{2}w; ,; v_{4}=b_{1}v_{1}+b_{2}w ;;text{with}; a_{i},b_{i}<0$$
    but $0>v_{3}cdot v_{4}=a_{1}b_{1}+a_{2}b_{2}>0$.

    Follow that $operatorname{dim}Vgeq 3$.






    share|cite|improve this answer




























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      (i) Assume that exists $i,jin{1,2,3,4}$ such that $v_{i}=av_{j}$. Considering the norm, we have $1=||v_{i}||=||av_{j}||=|a|,||v_{j}||=|a| Rightarrow a=-1,$, since $a=v_{i}cdot v_{j}<0$. However, we have $v_{i}cdot v_{k}=a v_{j}cdot v_{k}=- v_{j}cdot v_{k} > 0$ for $kneq i,j$. This is absurd.



      (ii) Thanks to (i) $,operatorname{dim} V >1$, so it is sufficient to prove that $operatorname{dim} V neq 2$.

      We assume for absurd that $operatorname{dim} V = 2$.
      Let $w= v_{2} - frac{(v_{2}cdot v_{1})}{||v_{1}||}, v_{1} in V$ such that $wcdot v_{1}=0$. Then $operatorname{span}{v_{1},v_{2},v_{3},v_{4}}=operatorname{span}{v_{1},w}$, and we can write
      $$v_{3}=a_{1}v_{1}+a_{2}w; ,; v_{4}=b_{1}v_{1}+b_{2}w ;;text{with}; a_{i},b_{i}<0$$
      but $0>v_{3}cdot v_{4}=a_{1}b_{1}+a_{2}b_{2}>0$.

      Follow that $operatorname{dim}Vgeq 3$.






      share|cite|improve this answer


























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        (i) Assume that exists $i,jin{1,2,3,4}$ such that $v_{i}=av_{j}$. Considering the norm, we have $1=||v_{i}||=||av_{j}||=|a|,||v_{j}||=|a| Rightarrow a=-1,$, since $a=v_{i}cdot v_{j}<0$. However, we have $v_{i}cdot v_{k}=a v_{j}cdot v_{k}=- v_{j}cdot v_{k} > 0$ for $kneq i,j$. This is absurd.



        (ii) Thanks to (i) $,operatorname{dim} V >1$, so it is sufficient to prove that $operatorname{dim} V neq 2$.

        We assume for absurd that $operatorname{dim} V = 2$.
        Let $w= v_{2} - frac{(v_{2}cdot v_{1})}{||v_{1}||}, v_{1} in V$ such that $wcdot v_{1}=0$. Then $operatorname{span}{v_{1},v_{2},v_{3},v_{4}}=operatorname{span}{v_{1},w}$, and we can write
        $$v_{3}=a_{1}v_{1}+a_{2}w; ,; v_{4}=b_{1}v_{1}+b_{2}w ;;text{with}; a_{i},b_{i}<0$$
        but $0>v_{3}cdot v_{4}=a_{1}b_{1}+a_{2}b_{2}>0$.

        Follow that $operatorname{dim}Vgeq 3$.






        share|cite|improve this answer














        (i) Assume that exists $i,jin{1,2,3,4}$ such that $v_{i}=av_{j}$. Considering the norm, we have $1=||v_{i}||=||av_{j}||=|a|,||v_{j}||=|a| Rightarrow a=-1,$, since $a=v_{i}cdot v_{j}<0$. However, we have $v_{i}cdot v_{k}=a v_{j}cdot v_{k}=- v_{j}cdot v_{k} > 0$ for $kneq i,j$. This is absurd.



        (ii) Thanks to (i) $,operatorname{dim} V >1$, so it is sufficient to prove that $operatorname{dim} V neq 2$.

        We assume for absurd that $operatorname{dim} V = 2$.
        Let $w= v_{2} - frac{(v_{2}cdot v_{1})}{||v_{1}||}, v_{1} in V$ such that $wcdot v_{1}=0$. Then $operatorname{span}{v_{1},v_{2},v_{3},v_{4}}=operatorname{span}{v_{1},w}$, and we can write
        $$v_{3}=a_{1}v_{1}+a_{2}w; ,; v_{4}=b_{1}v_{1}+b_{2}w ;;text{with}; a_{i},b_{i}<0$$
        but $0>v_{3}cdot v_{4}=a_{1}b_{1}+a_{2}b_{2}>0$.

        Follow that $operatorname{dim}Vgeq 3$.







        share|cite|improve this answer














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        edited Nov 27 '18 at 10:23

























        answered Nov 27 '18 at 10:08









        Andrea Cavaliere

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