= Unisolvent finite element
{title2=$p\longmapsto(\ell_1(p),\ldots,\ell_n(p))\text{ is bijective}$}
A <finite element> with local shape space $P$ and $n=\dim P$ <finite element degrees of freedom> $\ell_1,\ldots,\ell_n$ is unisolvent when each possible list of values determines exactly one member of $P$. By the <rank-nullity theorem>, it suffices to show that $\ell_i(p)=0$ for every $i$ implies $p=0$. Equivalently, one may construct a cardinal <basis> $N_1,\ldots,N_n$ with $\ell_i(N_j)=\delta_{ij}$. Such a <basis> is <linearly independent> by applying each $\ell_i$, and its size makes it span $P$; then $p=\sum_i\ell_i(p)N_i$.
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