- #1
*melinda*
- 86
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How to prove stuff about linear algebra?
Question:
Suppose [itex](v_1, v_2, ..., v_n)[/itex] is linearly independent in [itex]V[/itex] and [itex]w\in V[/itex].
Prove that if [itex](v_1 +w, v_2 +w, ..., v_n +w)[/itex] is linearly dependent, then [itex]w\in span(v_1, ...,v_n)[/itex].
To prove this I tried...
If [itex](v_1, v_2, ..., v_n)[/itex] is linearly independent then [itex]a_1 v_1 + ...+a_n v_n =0[/itex] for all [itex](a_1 , ..., a_n )=0[/itex].
then,
[itex]a_1 (v_1 +w)+a_2 (v_2 +w)+...+a_n (v_n +w)=0[/itex]
is not linearly independent, but can be rewritten as,
[itex]a_1 v_1 + ...+a_n v_n +(\sum a_i )w=0[/itex]
so,
[itex]a_1 v_1 + ...+a_n v_n = -(\sum a_i )w[/itex].
Since [itex]w[/itex] is a linear combination of vectors in [itex]V[/itex], [itex]w\in span(V)[/itex].
Did I do this right?
Is there a better way of doing this?
Any input is much appreciated!
Question:
Suppose [itex](v_1, v_2, ..., v_n)[/itex] is linearly independent in [itex]V[/itex] and [itex]w\in V[/itex].
Prove that if [itex](v_1 +w, v_2 +w, ..., v_n +w)[/itex] is linearly dependent, then [itex]w\in span(v_1, ...,v_n)[/itex].
To prove this I tried...
If [itex](v_1, v_2, ..., v_n)[/itex] is linearly independent then [itex]a_1 v_1 + ...+a_n v_n =0[/itex] for all [itex](a_1 , ..., a_n )=0[/itex].
then,
[itex]a_1 (v_1 +w)+a_2 (v_2 +w)+...+a_n (v_n +w)=0[/itex]
is not linearly independent, but can be rewritten as,
[itex]a_1 v_1 + ...+a_n v_n +(\sum a_i )w=0[/itex]
so,
[itex]a_1 v_1 + ...+a_n v_n = -(\sum a_i )w[/itex].
Since [itex]w[/itex] is a linear combination of vectors in [itex]V[/itex], [itex]w\in span(V)[/itex].
Did I do this right?
Is there a better way of doing this?
Any input is much appreciated!