Prove $ \dim L(\{v_1+w,v_2+w,\dots,v_n+w\})\geq n-1,w\in V$

Question

Let $v_1,v_2,\dots,v_n\in V, $ linearly independent

Prove $ \dim L(\{v_1+w,v_2+w,\dots,v_n+w\})\geq n-1,w\in V$

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I tried

$\lambda_1(v_1+w)+\lambda_2(v_2+w)+\dots+\lambda_n(v_n+w)=0$

$\lambda_1v_1+\lambda_2v_2+\dots+\lambda_nv_n=-w(\lambda_1+\dots+\lambda_n)$

Answer 1

$v_i+w-(v_1+w)=v_i-v_1$ is an element of $L(v_1+w,...,v_n+w)$.

Suppose that $a_1(v_2-v_1)+...+a_{n-1}(v_n-v_1)=0$, it implies that

$a_1v_2+...+a_{n-1}v_n-(a_1+...+a_{n-1})v_1=0$, since $v_1,...,v_n$ are independentt, $a_i=0$.

Answer 2

Hint: A quick approach is to compare $\dim L(\{v_1+w,v_2+w,\dots,v_n+w\})$ with $\dim L(\{v_1+w,v_2+w,\dots,v_n+w,w\})$.

Perhaps your approach could be made to work, though.

Answer 3

Subtracting the first vector from the other $n-1$ we get $$L(v_1+w, v_2+w,\ldots, v_n+w) = L(v_1+w, v_2-v_1, \ldots, v_n-v_1).$$

It is easy to check that the vectors $v_2-v_1, \ldots, v_n-v_1$ are linearly independent so it follows that the dimension of the span is $\ge n-1$.