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I'm having trouble proving this. My first instinct is to consider a zero vector (to somehow show linear independence of both), but I don't know I'm not sure. Any help is greatly appreciated :)

Let $\mathcal{B}$ be a basis for a vector space $V$. Prove or disprove that if $\vec{v_1},\ldots,\vec{v_n}\in V$ such that $\{[\vec{v_1}]_{\mathcal{B}},\ldots,[\vec{v_n}]_{\mathcal{B}}\}$ is a basis for $\mathbb{R}^n$, then $\{\vec{v_1},\ldots,\vec{v_n}\}$ is a basis for $V$.

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Suppose some linear combination $c_1 \vec{v_1}+ \ldots +c_n \vec{v_n}=\vec{0}$. Take coordinates of both sides:

$[c_1 \vec{v_1}+ \ldots +c_n \vec{v_n}]_\mathcal{B}=[\vec{0}]_\mathcal{B}$

Taking coordinates is a linear transformations therefore:

$c_1 [\vec{v_1}]_\mathcal B+ \ldots c_n [\vec {v_n}]_\mathcal B=0$.

It now follows that $c_1=c_2=\ldots=c_n=0$, because $\{[\vec{v_1}]_\mathcal B, \ldots, [\vec{v_n}]_\mathcal B \}$ is a basis for $\mathbb R^n$

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