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I understand how to do this when I have values for the vectors, but what if there are no values? I also know that if the solution is trivial, it is independent. Basically, can I solve this with Gauss Jordan elimination when there are no values, and when one vector is not in the span?

Prove that a set {$v_1, v_2, v_3, v_4$} is linearly independent if {$v_2, v_3, v_4$} is linearly independent and $v_1$ is not in the Span {$v_2, v_3, v_4$).

Hermes Trismegistus
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3 Answers3

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You cannot use Gauss Jordan elimination, you will have to use the definition of linear independence: A set is linearly independent if it is the case that a linear combination of its vectors equals zero if and only if the coefficients of that linear combination are all zero.

Hint: Any linear combination of $\{v_2, v_3, v_4\}$ can be thought of as a linear combination of $\{v_1, v_2, v_3, v_4\}$ where the coefficient of $v_1$ is zero. For getting that $v_1$ is not in the span, prove this by contradiction, i.e., if $v_1$ is in the span then $\{v_1, v_2, v_3, v_4\}$ is not linearly independent.

Jim
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Let $a,b,c,d$ be the four vectors that are linearly independent. For the purpose of contradiction, assume that the $b,c,d$ are linearly independent and a is in the span of b,c,d. Further, assume that $a\neq 0$ else it doesn't make sense. Thus, there exists real scalars x,y,z not all 0 such that $a = xb + yc + zd $ $$\implies 1a - xb -yc -zd =0$$ which contradicts the definition of linear independence.

Inquest
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Hint: recall the definition of being linearly independent and the Span of vectors. i.e. if we know {v2, v3, v4} is lin. ind., what can we say about the coefficients ? And, if v1 is not in the Span{v2, v3, v4}, what can we say about the relationship between v1 and the remaining 3 vectors? (in terms of linear combinations) ?

Cecile
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  • So $c_1v_2 + c_2v_3 + c_3v_4$ should equal 0. If $v_1$ is not a part of this, but when you add $c_4v_1$ to the Span{$v_2 + v_3 + v_4$} it would still be 0? Is that something along the line of what you mean? – Hermes Trismegistus Feb 22 '13 at 18:36