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A finite element degree of freedom is one of the linear functionals chosen to specify a local shape function in a finite-dimensional vector space $P$. Examples are point evaluation functionals and edge or cell integrals. The finite element is an unisolvent finite element when the complete list gives a bijection from $P$ to the corresponding list of real values.

Label the polygon vertices of the nondegenerate triangle $K$ as $a_1,a_2,a_3$, and put $m_{ij}=(a_i+a_j)/2$ for its three edge midpoints. The six monomials $1,x_1,x_2,x_1^2,x_1{x}_2,x_2^2$ form a basis of ${\mathcal{P}}_2(K)$, so its dimension is six. Nondegeneracy matters: $K$ has nonempty interior, and a polynomial vanishing on that interior is the zero polynomial.

Let ${\lambda }_1,{\lambda }_2,{\lambda }_3$ be the affine functions of the barycentric coordinate system, characterized by ${\lambda }_i(a_j)={\delta }_{ij}$ and ${\sum }_i{\lambda }_i=1$. Define six quadratic polynomialsAt $a_k$, $N_i(a_k)={\delta }_{ik}$ and all $N_{ij}(a_k)$ vanish. At $m_{kl}$, exactly ${\lambda }_k$ and ${\lambda }_l$ equal $1/2$, while the third is zero. Thus every $N_i$ vanishes at all three edge midpoints, and $N_{ij}(m_{kl})$ is one for the matching unordered pair and zero otherwise.

These six polynomials have the cardinal property for the six point evaluation functionals. Evaluating any vanishing linear combination at the nodes forces each of its coefficients to vanish. They are therefore six linearly independent members of a six-dimensional vector space, hence a basis.

In particular, every $p\in {\mathcal{P}}_2(K)$ has the unique expansionThis proves both existence and uniqueness for arbitrary six prescribed nodal values, which is precisely an unisolvent finite element. It also identifies the finite element degrees of freedom and shape functions of the quadratic Lagrange triangle explicitly.

The quadratic Lagrange triangle uses degree-at-most-two polynomials on a nondegenerate triangle, with point evaluation at its three polygon vertices and three edge midpoints as the six finite element degrees of freedom. If ${\lambda }_i$ are the affine functions of its barycentric coordinate system, the cardinal shape functions areAt a polygon vertex, one coordinate in the barycentric coordinate system is one and the others vanish; at an edge midpoint, two are $1/2$ and the third is zero. These values prove the cardinal property. Since $\dim {\mathcal{P}}_2(K)=6$, this is an unisolvent finite element. Matching the two endpoint values and midpoint value on a shared edge of a polygon matches the whole quadratic restriction, producing a continuous global conforming finite element space.