A singleton set with non-zero vector is Linearly Independent : LInear Algebra : B.Sc.Mathematics

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Theorem : 

    A singleton set with non-zero vector is Linearly Independent 

 Proof : 

        Suppose V ( F) is a vector space and S = {𝜶} is a subset of V where 𝜶 ≠ 0.

        Now we prove S is Linearly Independent.

         For a 𝞊 F such that a𝜶 = 0

                                     ⇒ a=0 or 𝜶 = 0 .

         Since 𝜶 ≠ 0 , we have a = 0.

           ∴ S is Linearly Independent.

                                              ***Hence the proof ***

     Explanation : 

         Linearly independent set means if the linear combination of vectors is zero then  all the scalars    

         included in that linear combination are all zero. 

         In the above theorem, S is a single point set containing 𝜶 ≠ 0and for that single point we need a               scalar  'a' with  a𝜶 =  0.

          Since 𝜶 ≠ 0, we have a = 0 . i,e, here the linear combination of single vector is zero which imply 

           the scalar included in that linear combination a = 0 .

          Thus , S is linearly independent.  

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*  Show that the set { ( 1 , 2 , 1 ) , ( 2 , 1 , 0 ) , ( 1 , -1 , 2  } forms a basis of V3 ( F )

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* linear sum of subspaces

* What is a vector space

* linear span of a set 

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 * Theorem on vector space  

 * Historical Introduction to Linear Algebra

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                                                     (ii) a 𝞊 F , 𝜶 𝞊 W ⇒ a𝜶 𝞊 W.

* Let V(F) be a vector space and let W ⊆ V. The necessary and sufficient conditions for W 

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