Suppose v1,v2,........,vm is a linearly independent in V and w∈V. Show that v1,v2,........,vm is linearly independent if and only if w∉span(v1,v2,........,vm )
Suppose that "w\\in \\text{Span}(v_1,...,v_m)", then we have "w=\\sum \\alpha_i v_i" and thus "\\sum \\alpha_iv_i-w=0" is a non-trivial (as at least the coefficient before "w" is "-1\\neq 0" ) linear combination of "(v_1,...,v_m,w)" that gives zero and thus "(v_1,...,v_m,w)" is not linearly independent.
Now let's suppose that "(v_1,...,v_m,w)" is linearly dependent, then there exists a non-trivial linear combination such that "\\sum \\alpha_i v_i +bw=0". We can not have "b=0" as in this case we would have "\\sum \\alpha_i v_i=0" for sum non-trivial combination "(\\alpha_i)". Therefore we have "w=\\sum(-\\alpha_i\/b)v_i \\in \\text{Span}(v_1,...,v_m)".
"\\neg((v_1,...,v_m,w) \\text{ are linearly independent}) \\leftrightarrow \\neg (w\\notin \\text{Span}(v_1,...,v_m))" and therefore we have (by contraposition)
"((v_1,...,v_m,w) \\text{ are linearly independent}) \\leftrightarrow (w\\notin \\text{Span}(v_1,...,v_m))"
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