Let $d\in\mathbb N$. $(K,\mathcal P,\mathcal N)$ is called finite element $:\Leftrightarrow$
- $K\subseteq\mathbb R^d$ is compact with piecewise smooth boundary
- $\mathcal P\subseteq\mathbb R^K$ is a finite dimensional $\mathbb R$-vector space
- $\mathcal N$ is a basis of $\mathcal P^\ast$
Finite elements $(K,\mathcal P,\mathcal N)$ and $(\tilde K,\tilde{\mathcal P},\tilde{\mathcal N})$ are called affine equivalent $:\Leftrightarrow$
- $F^\ast\tilde{\mathcal P}=\mathcal P$
- $F_\ast\mathcal N=\tilde{\mathcal N}$
for some nonsingular affine transformation $F:K\to\tilde K$ with $$F^\ast(\tilde f):=\tilde f\circ F$$ and $$(F_\ast N)(\tilde f):=N(F^\ast(\tilde f))\;\;\;\text{for all }N\in\mathcal N$$ for all $\tilde f\in\mathbb R^{\tilde K}$.
Now, if $K\subseteq\mathbb R^2$ is a nondegenerate triangle with vertices $p^1,p^2$ and $p^3$, then a finite element $(K,\mathcal P,\mathcal N)$ is called linear Lagrange element $:\Leftrightarrow$
- $\mathcal P$ is the set of polynomials $K\to\mathbb R$ of degree at most $1$
- $\mathcal N=(N_1,N_2,N_3)$ with $$N_i(f):=f(p^i)\;\;\;\text{for all }f\in\mathcal P$$
Now, I want to show that all linear Lagrange elements are affine equivalent.
I've found a proof which states the following: Let $(K,\mathcal P,\mathcal N)$ be a linear Lagrange element and $(\phi_1,\phi_2,\phi_3)$ be the basis of $\mathcal P$ dual to $\mathcal N$, i.e. $$N_i(\phi_j)=\delta_{ij}\;\;\;\text{for all }i,j\in\left\{1,2,3\right\}\;.$$ If $p\in K$, then it's easy to see that $(\phi_1(p),\phi_2(p),\phi_3(p))$ are the barycentric coordinates of $p$ with respect to $(\phi_1,\phi_2,\phi_3)$, i.e. $\lambda_i:=\phi_i(p)\ge0$ for all $i\in\left\{1,2,3\right\}$, $$\sum_{i=1}^3\lambda_i=1$$ and $$p=\sum_{i=1}^3\lambda_ip^i\;.$$ Now, they conclude the proof with the remark that $$K\to\left\{\lambda\in[0,1]^3:\sum_{i=1}^3\lambda_i=1\right\}\;,\;\;\;p\mapsto(\phi_1(p),\phi_2(p),\phi_3(p))^T\tag1$$ is invertible. Why does that prove the claim?