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<p>Because $\vect{u}_i$ was assumed to be nozero, <acroref type="theorem" acro="PIP" /> says $\innerproduct{\vect{u}_i}{\vect{u}_i}$ is nonzero and thus $\alpha_i$ must be zero. So we conclude that $\alpha_i=0$ for all $1\leq i\leq n$ in any relation of linear dependence on $S$. But this says that $S$ is a linearly independent set since the only way to form a relation of linear dependence is the trivial way (<acroref type="definition" acro="LICV" />). Boom!</p>
+<p>Because $\vect{u}_i$ was assumed to be nonzero, <acroref type="theorem" acro="PIP" /> says $\innerproduct{\vect{u}_i}{\vect{u}_i}$ is nonzero and thus $\alpha_i$ must be zero. So we conclude that $\alpha_i=0$ for all $1\leq i\leq n$ in any relation of linear dependence on $S$. But this says that $S$ is a linearly independent set since the only way to form a relation of linear dependence is the trivial way (<acroref type="definition" acro="LICV" />). Boom!</p>