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Show that if each of the vectors $\left\{v_1, v_2, . . . , v_n\right\}$ is a linear combination of the vectors $\left\{w_1, w_2, . . . , w_n\right\}$, then $\left\{v_1, v_2, . . . , v_n\right\}$ is linearly dependent.

What I attempted:

I formed the dependency equation $a_1v_1+a_2v_2+...+ a_nv_n = 0$. For $\left\{v_1, v_2, . . . , v_n\right\}$ to be linearly dependent, then it must be that $a_i \neq 0$ $\forall i$. But I can't seem to show just by using $\left\{v_1, v_2, . . . , v_n\right\}$ is a linear combination of $\left\{w_1, w_2, . . . , w_n\right\}$. Any hint you guys can provide?

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    $\begingroup$ this is false, let $v_i=w_i=e_i$ be the standard basis vectors in $\Bbb R^n$. $\endgroup$ – Adam Hughes Apr 18 '16 at 2:58
  • $\begingroup$ Yes, @llawliet_78 are you sure that the number of $v$s is supposed to be equal to the number of $w$s? $\endgroup$ – Aidan Sims Apr 18 '16 at 3:30

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