Proof verification: Every set that contains a linearly dependent set is linearly dependent.












0












$begingroup$


I wrote a proof for a theorem and would appreciate if someone could check if my proof is sound. Thanks a lot.



$Theorem$:



Every set that contains a linearly dependent set is linearly dependent.



$Proof$:



Let $S={v_1, v_2,...,v_k}$ be a set of vectors. Let $L={v_1, v_2, ..., v_l} subset S$, $l lt k$, be a linearly dependent subset of $S$.



Since $L$ is linearly dependent, there must exist, without loss of generality, $v_1 in L$ such that



$$sum_{i=2}^l alpha_i v_i = v_1, alpha_i in mathbb{R}$$



Then we can write a sum



$$sum_{i=2}^l alpha_i v_i + sum_{i = l+1}^k 0 v_i= v_1 =$$
$$sum_{i=2}^k alpha_i v_i = v_1$$
where $alpha_i=0$ for $l+1 lt ilt k$.



Rearranging,
$$sum_{i=2}^k alpha_i v_i - v_1 = 0 iff sum_{i=1}^k alpha_i v_i = 0$$
with $alpha_i in mathbb{R}$ not all zero.
Thus, the set $S$ is linearly dependent. This completes the proof.






proof-verification vector-spaces







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$endgroup$












  • $begingroup$
    Perfect!$hspace{0pt}$
    $endgroup$
    – Theo Bendit
    Jan 8 at 7:15






  • 1




    $begingroup$
    @TheoBendit Sure about that?
    $endgroup$
    – José Carlos Santos
    Jan 8 at 7:20
















0












$begingroup$


I wrote a proof for a theorem and would appreciate if someone could check if my proof is sound. Thanks a lot.



$Theorem$:



Every set that contains a linearly dependent set is linearly dependent.



$Proof$:



Let $S={v_1, v_2,...,v_k}$ be a set of vectors. Let $L={v_1, v_2, ..., v_l} subset S$, $l lt k$, be a linearly dependent subset of $S$.



Since $L$ is linearly dependent, there must exist, without loss of generality, $v_1 in L$ such that



$$sum_{i=2}^l alpha_i v_i = v_1, alpha_i in mathbb{R}$$



Then we can write a sum



$$sum_{i=2}^l alpha_i v_i + sum_{i = l+1}^k 0 v_i= v_1 =$$
$$sum_{i=2}^k alpha_i v_i = v_1$$
where $alpha_i=0$ for $l+1 lt ilt k$.



Rearranging,
$$sum_{i=2}^k alpha_i v_i - v_1 = 0 iff sum_{i=1}^k alpha_i v_i = 0$$
with $alpha_i in mathbb{R}$ not all zero.
Thus, the set $S$ is linearly dependent. This completes the proof.






proof-verification vector-spaces







share|cite|improve this question









$endgroup$












  • $begingroup$
    Perfect!$hspace{0pt}$
    $endgroup$
    – Theo Bendit
    Jan 8 at 7:15






  • 1




    $begingroup$
    @TheoBendit Sure about that?
    $endgroup$
    – José Carlos Santos
    Jan 8 at 7:20














0












0








0





$begingroup$


I wrote a proof for a theorem and would appreciate if someone could check if my proof is sound. Thanks a lot.



$Theorem$:



Every set that contains a linearly dependent set is linearly dependent.



$Proof$:



Let $S={v_1, v_2,...,v_k}$ be a set of vectors. Let $L={v_1, v_2, ..., v_l} subset S$, $l lt k$, be a linearly dependent subset of $S$.



Since $L$ is linearly dependent, there must exist, without loss of generality, $v_1 in L$ such that



$$sum_{i=2}^l alpha_i v_i = v_1, alpha_i in mathbb{R}$$



Then we can write a sum



$$sum_{i=2}^l alpha_i v_i + sum_{i = l+1}^k 0 v_i= v_1 =$$
$$sum_{i=2}^k alpha_i v_i = v_1$$
where $alpha_i=0$ for $l+1 lt ilt k$.



Rearranging,
$$sum_{i=2}^k alpha_i v_i - v_1 = 0 iff sum_{i=1}^k alpha_i v_i = 0$$
with $alpha_i in mathbb{R}$ not all zero.
Thus, the set $S$ is linearly dependent. This completes the proof.






proof-verification vector-spaces







share|cite|improve this question









$endgroup$




I wrote a proof for a theorem and would appreciate if someone could check if my proof is sound. Thanks a lot.



$Theorem$:



Every set that contains a linearly dependent set is linearly dependent.



$Proof$:



Let $S={v_1, v_2,...,v_k}$ be a set of vectors. Let $L={v_1, v_2, ..., v_l} subset S$, $l lt k$, be a linearly dependent subset of $S$.



Since $L$ is linearly dependent, there must exist, without loss of generality, $v_1 in L$ such that



$$sum_{i=2}^l alpha_i v_i = v_1, alpha_i in mathbb{R}$$



Then we can write a sum



$$sum_{i=2}^l alpha_i v_i + sum_{i = l+1}^k 0 v_i= v_1 =$$
$$sum_{i=2}^k alpha_i v_i = v_1$$
where $alpha_i=0$ for $l+1 lt ilt k$.



Rearranging,
$$sum_{i=2}^k alpha_i v_i - v_1 = 0 iff sum_{i=1}^k alpha_i v_i = 0$$
with $alpha_i in mathbb{R}$ not all zero.
Thus, the set $S$ is linearly dependent. This completes the proof.






proof-verification vector-spaces




proof-verification vector-spaces






share|cite|improve this question













share|cite|improve this question











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share|cite|improve this question










asked Jan 8 at 7:14









Ulrich Paul WohakUlrich Paul Wohak

52




52












  • $begingroup$
    Perfect!$hspace{0pt}$
    $endgroup$
    – Theo Bendit
    Jan 8 at 7:15






  • 1




    $begingroup$
    @TheoBendit Sure about that?
    $endgroup$
    – José Carlos Santos
    Jan 8 at 7:20


















  • $begingroup$
    Perfect!$hspace{0pt}$
    $endgroup$
    – Theo Bendit
    Jan 8 at 7:15






  • 1




    $begingroup$
    @TheoBendit Sure about that?
    $endgroup$
    – José Carlos Santos
    Jan 8 at 7:20
















$begingroup$
Perfect!$hspace{0pt}$
$endgroup$
– Theo Bendit
Jan 8 at 7:15




$begingroup$
Perfect!$hspace{0pt}$
$endgroup$
– Theo Bendit
Jan 8 at 7:15




1




1




$begingroup$
@TheoBendit Sure about that?
$endgroup$
– José Carlos Santos
Jan 8 at 7:20




$begingroup$
@TheoBendit Sure about that?
$endgroup$
– José Carlos Santos
Jan 8 at 7:20










2 Answers
2






active

oldest

votes


















2












$begingroup$

It is not correct. You actually did not use the fact that $L$ is linearly dependent at all. In the second paragraph of you proof, you should have written that, since $L$ is linearly dependente, there are numbers $alpha_1,ldots,alpha_linmathbb R$ not all of which are $0$ such that $sum_{j=1}^lalpha_jv_j=0$.



Besides, you assumed that $S$ is finite, but the statement that you are trying to prove doesn't contain that assumption.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you Jose! Assuming that $S$ is finite, and starting with: Since $L$ is linearly dependent, there must exist $alpha_i in mathbb{R}$ not all zero such that $sum_{i=1}^l alpha_i v_i = 0$, it follows that, without loss of generality, $v_1 = sum_{i=2}^l alpha_i v_i$. If I then continue with the rest of my proof, would that work?
    $endgroup$
    – Ulrich Paul Wohak
    Jan 8 at 7:27












  • $begingroup$
    Yes, it would be fine.
    $endgroup$
    – José Carlos Santos
    Jan 8 at 7:31










  • $begingroup$
    Muchas gracias! :)
    $endgroup$
    – Ulrich Paul Wohak
    Jan 8 at 7:32



















0












$begingroup$

Suppose on contrary let us consider a set S which is linearly independent and a subset T which is linearly dependent. Now we claim that every suset of linearly independent set is linearly independent
Now if it is singleton then we have done because every single vector is linearly independent And if it is not singleton then we have,suppose it is linearly dependent then there exit some scalars atleast one of them is non zero which is not possible. This shows that every suset of LI set is LI Since we have a subset T which is linearly dependent of a set S which is linearly independent then by above result S have to be linearly dependent which contradicts our assumption. Hence,every suset of LD is LD






share|cite|improve this answer









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    2 Answers
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    2 Answers
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    $begingroup$

    It is not correct. You actually did not use the fact that $L$ is linearly dependent at all. In the second paragraph of you proof, you should have written that, since $L$ is linearly dependente, there are numbers $alpha_1,ldots,alpha_linmathbb R$ not all of which are $0$ such that $sum_{j=1}^lalpha_jv_j=0$.



    Besides, you assumed that $S$ is finite, but the statement that you are trying to prove doesn't contain that assumption.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you Jose! Assuming that $S$ is finite, and starting with: Since $L$ is linearly dependent, there must exist $alpha_i in mathbb{R}$ not all zero such that $sum_{i=1}^l alpha_i v_i = 0$, it follows that, without loss of generality, $v_1 = sum_{i=2}^l alpha_i v_i$. If I then continue with the rest of my proof, would that work?
      $endgroup$
      – Ulrich Paul Wohak
      Jan 8 at 7:27












    • $begingroup$
      Yes, it would be fine.
      $endgroup$
      – José Carlos Santos
      Jan 8 at 7:31










    • $begingroup$
      Muchas gracias! :)
      $endgroup$
      – Ulrich Paul Wohak
      Jan 8 at 7:32
















    2












    $begingroup$

    It is not correct. You actually did not use the fact that $L$ is linearly dependent at all. In the second paragraph of you proof, you should have written that, since $L$ is linearly dependente, there are numbers $alpha_1,ldots,alpha_linmathbb R$ not all of which are $0$ such that $sum_{j=1}^lalpha_jv_j=0$.



    Besides, you assumed that $S$ is finite, but the statement that you are trying to prove doesn't contain that assumption.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you Jose! Assuming that $S$ is finite, and starting with: Since $L$ is linearly dependent, there must exist $alpha_i in mathbb{R}$ not all zero such that $sum_{i=1}^l alpha_i v_i = 0$, it follows that, without loss of generality, $v_1 = sum_{i=2}^l alpha_i v_i$. If I then continue with the rest of my proof, would that work?
      $endgroup$
      – Ulrich Paul Wohak
      Jan 8 at 7:27












    • $begingroup$
      Yes, it would be fine.
      $endgroup$
      – José Carlos Santos
      Jan 8 at 7:31










    • $begingroup$
      Muchas gracias! :)
      $endgroup$
      – Ulrich Paul Wohak
      Jan 8 at 7:32














    2












    2








    2





    $begingroup$

    It is not correct. You actually did not use the fact that $L$ is linearly dependent at all. In the second paragraph of you proof, you should have written that, since $L$ is linearly dependente, there are numbers $alpha_1,ldots,alpha_linmathbb R$ not all of which are $0$ such that $sum_{j=1}^lalpha_jv_j=0$.



    Besides, you assumed that $S$ is finite, but the statement that you are trying to prove doesn't contain that assumption.






    share|cite|improve this answer









    $endgroup$



    It is not correct. You actually did not use the fact that $L$ is linearly dependent at all. In the second paragraph of you proof, you should have written that, since $L$ is linearly dependente, there are numbers $alpha_1,ldots,alpha_linmathbb R$ not all of which are $0$ such that $sum_{j=1}^lalpha_jv_j=0$.



    Besides, you assumed that $S$ is finite, but the statement that you are trying to prove doesn't contain that assumption.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 8 at 7:20









    José Carlos SantosJosé Carlos Santos

    171k23132240




    171k23132240












    • $begingroup$
      Thank you Jose! Assuming that $S$ is finite, and starting with: Since $L$ is linearly dependent, there must exist $alpha_i in mathbb{R}$ not all zero such that $sum_{i=1}^l alpha_i v_i = 0$, it follows that, without loss of generality, $v_1 = sum_{i=2}^l alpha_i v_i$. If I then continue with the rest of my proof, would that work?
      $endgroup$
      – Ulrich Paul Wohak
      Jan 8 at 7:27












    • $begingroup$
      Yes, it would be fine.
      $endgroup$
      – José Carlos Santos
      Jan 8 at 7:31










    • $begingroup$
      Muchas gracias! :)
      $endgroup$
      – Ulrich Paul Wohak
      Jan 8 at 7:32


















    • $begingroup$
      Thank you Jose! Assuming that $S$ is finite, and starting with: Since $L$ is linearly dependent, there must exist $alpha_i in mathbb{R}$ not all zero such that $sum_{i=1}^l alpha_i v_i = 0$, it follows that, without loss of generality, $v_1 = sum_{i=2}^l alpha_i v_i$. If I then continue with the rest of my proof, would that work?
      $endgroup$
      – Ulrich Paul Wohak
      Jan 8 at 7:27












    • $begingroup$
      Yes, it would be fine.
      $endgroup$
      – José Carlos Santos
      Jan 8 at 7:31










    • $begingroup$
      Muchas gracias! :)
      $endgroup$
      – Ulrich Paul Wohak
      Jan 8 at 7:32
















    $begingroup$
    Thank you Jose! Assuming that $S$ is finite, and starting with: Since $L$ is linearly dependent, there must exist $alpha_i in mathbb{R}$ not all zero such that $sum_{i=1}^l alpha_i v_i = 0$, it follows that, without loss of generality, $v_1 = sum_{i=2}^l alpha_i v_i$. If I then continue with the rest of my proof, would that work?
    $endgroup$
    – Ulrich Paul Wohak
    Jan 8 at 7:27






    $begingroup$
    Thank you Jose! Assuming that $S$ is finite, and starting with: Since $L$ is linearly dependent, there must exist $alpha_i in mathbb{R}$ not all zero such that $sum_{i=1}^l alpha_i v_i = 0$, it follows that, without loss of generality, $v_1 = sum_{i=2}^l alpha_i v_i$. If I then continue with the rest of my proof, would that work?
    $endgroup$
    – Ulrich Paul Wohak
    Jan 8 at 7:27














    $begingroup$
    Yes, it would be fine.
    $endgroup$
    – José Carlos Santos
    Jan 8 at 7:31




    $begingroup$
    Yes, it would be fine.
    $endgroup$
    – José Carlos Santos
    Jan 8 at 7:31












    $begingroup$
    Muchas gracias! :)
    $endgroup$
    – Ulrich Paul Wohak
    Jan 8 at 7:32




    $begingroup$
    Muchas gracias! :)
    $endgroup$
    – Ulrich Paul Wohak
    Jan 8 at 7:32











    0












    $begingroup$

    Suppose on contrary let us consider a set S which is linearly independent and a subset T which is linearly dependent. Now we claim that every suset of linearly independent set is linearly independent
    Now if it is singleton then we have done because every single vector is linearly independent And if it is not singleton then we have,suppose it is linearly dependent then there exit some scalars atleast one of them is non zero which is not possible. This shows that every suset of LI set is LI Since we have a subset T which is linearly dependent of a set S which is linearly independent then by above result S have to be linearly dependent which contradicts our assumption. Hence,every suset of LD is LD






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Suppose on contrary let us consider a set S which is linearly independent and a subset T which is linearly dependent. Now we claim that every suset of linearly independent set is linearly independent
      Now if it is singleton then we have done because every single vector is linearly independent And if it is not singleton then we have,suppose it is linearly dependent then there exit some scalars atleast one of them is non zero which is not possible. This shows that every suset of LI set is LI Since we have a subset T which is linearly dependent of a set S which is linearly independent then by above result S have to be linearly dependent which contradicts our assumption. Hence,every suset of LD is LD






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Suppose on contrary let us consider a set S which is linearly independent and a subset T which is linearly dependent. Now we claim that every suset of linearly independent set is linearly independent
        Now if it is singleton then we have done because every single vector is linearly independent And if it is not singleton then we have,suppose it is linearly dependent then there exit some scalars atleast one of them is non zero which is not possible. This shows that every suset of LI set is LI Since we have a subset T which is linearly dependent of a set S which is linearly independent then by above result S have to be linearly dependent which contradicts our assumption. Hence,every suset of LD is LD






        share|cite|improve this answer









        $endgroup$



        Suppose on contrary let us consider a set S which is linearly independent and a subset T which is linearly dependent. Now we claim that every suset of linearly independent set is linearly independent
        Now if it is singleton then we have done because every single vector is linearly independent And if it is not singleton then we have,suppose it is linearly dependent then there exit some scalars atleast one of them is non zero which is not possible. This shows that every suset of LI set is LI Since we have a subset T which is linearly dependent of a set S which is linearly independent then by above result S have to be linearly dependent which contradicts our assumption. Hence,every suset of LD is LD







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 8 at 9:00









        Abhishek TripathiAbhishek Tripathi

        1




        1






























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