Given that the vectors v 1 , v 2 , v 3 , v 4 are linearly independent.Are the following b 1 , b 2 , b 3 vectors linearly independent too? b 1 = 3 v 1 + 2 v 2 + v 3 + v 4 b 2 = 2 v 1 + 5 v 2 + 3 v 3 + 2 v 4 b 3 = 3 v 1 + 4 v 2 + 2 v 3 + 3 v 4 What I've done so far k 1 ( 3 v 1 + 2 v 2 + v 3 + v 4 ) + k 2 ( 2 v 1 + 5 v 2 + 3 v 3 + 2 v 4 ) + k 3 ( 3 v 1 + 4 v 2 + 2 v 3 + 3 v 4 ) = 0then I don't know what to do. I don't even know whether the 0 above is supposed to be a vector or a scalar.

Question

Given that the vectors       v          1        ,      v          2        ,      v          3        ,      v          4       are linearly independent.Are the following       b          1        ,      b          2        ,      b          3       vectors linearly independent too?      b          1        =  3      v          1        +  2      v          2        +      v          3        +      v          4            b          2        =  2      v          1        +  5      v          2        +  3      v          3        +  2      v          4            b          3        =  3      v          1        +  4      v          2        +  2      v          3        +  3      v          4      What I&#039;ve done so far      k          1        (  3      v          1        +  2      v          2        +      v          3        +      v          4        )  +      k          2        (  2      v          1        +  5      v          2        +  3      v          3        +  2      v          4        )  +      k          3        (  3      v          1        +  4      v          2        +  2      v          3        +  3      v          4        )  =  0then I don&#039;t know what to do. I don&#039;t even know whether the 0 above is supposed to be a vector or a scalar.

stangeix · Accepted Answer

Result: Let       α          i      j      ;   1  ≤  i  ,  j  ≤  n be scalars such that the system of equations:      ∑          j      =      1        n        x    j        α          i      j        =  0  ,    1  ≤  i  ≤  nhas only trivial solution. Let   {      v    1    ,      v    2    ,  …  ,      v    n    } be any linearly independent set of vectors. Then   {      ∑          i      =      1        n        α          i      1            v    i    ,      ∑          i      =      1        n        α          i      2            v    i    ,  …  ,      ∑          i      =      1        n        α          i      n            v    i    } is also linearly independent.Proof Suppose on the contrary that that   {      v    1    ,      v    2    ,  …  ,      v    n    } is linearly independent and   {      ∑          i      =      1        n        α          i      1            v    i    ,      ∑          i      =      1        n        α          i      2            v    i    ,  …  ,      ∑          i      =      1        n        α          i      n            v    i    } is linearly dependent. Then   ∃ scalars       c    1    ,      c    2    ,  …  ,      c    n   (not all zero) such that      c    1        ∑          i      =      1        n        α          i      1            v    i    +      c    2        ∑          i      =      1        n        α          i      2            v    i    +  ⋯  +      c    n        ∑          i      =      1        n        α          i      n            v    i    =      0    .Since   {      v    1    ,      v    2    ,  …  ,      v    n    } is linearly independent, we must have      ∑          j      =      1        n        c    j        α          i      j        =  0  ,    1  ≤  i  ≤  n  ,which is a contradiction.The similar technique can be applied if the &quot;sum&quot; set has cardinality lesser that n.

allucinemsj · Accepted Answer

Hint:To show they&#039;re independent, you have to show the matrix      [                                        b            1                                                b            2                                                b            3                                ]    =      [                            3                          2                          3                                      2                          5                          4                                      1                          3                          2                                      1                          2                          3                      ]  has maximal rank (3), which means thereis a subterminat of order 3 which is   ≠  0

Given that the vectors v_1,v_2,v_3,v_4 are linearly independent. Are the following b_1,b_2,b_3 vectors linearly independent too? b_1=3v_1+2v_2+v_3+v_4 b_2=2v_1+5v_2+3v_3+2v_4 b_3=3v_1+4v_2+2v_3+3v_4

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