Dimension of the vector space spanned by ${e,f,g,..}$ equals the maximal number of linearly independent...











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I am reading Gelfand's Lectures on Linear Algebra.



He defines a vector space to have dimension $n$ if there exists a linearly independent set of $n$ vectors, and if every set of $n+1$ vectors is linearly dependent.



He asks to prove that the dimension of the vector space generated by ${e,f,g,..}$ equals the maximal number of linearly independent vectors in ${e,f,g...}$.



My approach:



Let $N$ be the maximal number of linearly independent vectors in ${e,f,g..}$.



Then since ${e,f,g..}$ contains $N$ linearly independent vectors its dimension must be greater than or equal to $N$.



Suppose it were strictly greater than $N$. Then there must exist a linearly independent set of $N+1$ vectors, contradicting the fact that the maximal number of linearly independent vectors among ${e,f,g..}$ was $N$.



Is this argument correct?










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    down vote

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    I am reading Gelfand's Lectures on Linear Algebra.



    He defines a vector space to have dimension $n$ if there exists a linearly independent set of $n$ vectors, and if every set of $n+1$ vectors is linearly dependent.



    He asks to prove that the dimension of the vector space generated by ${e,f,g,..}$ equals the maximal number of linearly independent vectors in ${e,f,g...}$.



    My approach:



    Let $N$ be the maximal number of linearly independent vectors in ${e,f,g..}$.



    Then since ${e,f,g..}$ contains $N$ linearly independent vectors its dimension must be greater than or equal to $N$.



    Suppose it were strictly greater than $N$. Then there must exist a linearly independent set of $N+1$ vectors, contradicting the fact that the maximal number of linearly independent vectors among ${e,f,g..}$ was $N$.



    Is this argument correct?










    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I am reading Gelfand's Lectures on Linear Algebra.



      He defines a vector space to have dimension $n$ if there exists a linearly independent set of $n$ vectors, and if every set of $n+1$ vectors is linearly dependent.



      He asks to prove that the dimension of the vector space generated by ${e,f,g,..}$ equals the maximal number of linearly independent vectors in ${e,f,g...}$.



      My approach:



      Let $N$ be the maximal number of linearly independent vectors in ${e,f,g..}$.



      Then since ${e,f,g..}$ contains $N$ linearly independent vectors its dimension must be greater than or equal to $N$.



      Suppose it were strictly greater than $N$. Then there must exist a linearly independent set of $N+1$ vectors, contradicting the fact that the maximal number of linearly independent vectors among ${e,f,g..}$ was $N$.



      Is this argument correct?










      share|cite|improve this question













      I am reading Gelfand's Lectures on Linear Algebra.



      He defines a vector space to have dimension $n$ if there exists a linearly independent set of $n$ vectors, and if every set of $n+1$ vectors is linearly dependent.



      He asks to prove that the dimension of the vector space generated by ${e,f,g,..}$ equals the maximal number of linearly independent vectors in ${e,f,g...}$.



      My approach:



      Let $N$ be the maximal number of linearly independent vectors in ${e,f,g..}$.



      Then since ${e,f,g..}$ contains $N$ linearly independent vectors its dimension must be greater than or equal to $N$.



      Suppose it were strictly greater than $N$. Then there must exist a linearly independent set of $N+1$ vectors, contradicting the fact that the maximal number of linearly independent vectors among ${e,f,g..}$ was $N$.



      Is this argument correct?







      linear-algebra






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      asked Nov 18 at 22:23









      trynalearn

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