Floating local kernels and positive interior blocks

If $a_i(v,v)=0$, each nonnegative term ${\rho }_i{\int }_K\mathrm{\mid }\mathrm{\nabla }v{\mathrm{\mid }}^2$ is zero. Since $\mathrm{\nabla }v$ is constant on each triangle, it vanishes on every triangle, so $v$ is constant there. Continuity across every shared edge makes neighboring constants equal; connectedness of the triangle adjacency graph makes $v$ one constant on all of ${\mathrm{\Omega }}_i$. Conversely, a constant has zero gradient, hence lies in the kernel. This proves the floating-block nullspace claim.

A vector in the interior block's kernel defines a local $P_1$ function with zero boundary trace and zero energy. Step 1.1 makes it constant and the zero boundary trace makes that constant zero. Thus the interior quadratic form is strictly positive on every nonzero vector, so its symmetric matrix $K_{II}^{(i)}$ is positive definite. If there are no interior vertices, the unique map on the zero vector space is invertible.

If the global energy vanishes, step 1.1 applied triangle by triangle gives one constant on every subdomain. Interface continuity makes adjacent subdomain constants equal; connectedness propagates one constant through $\mathrm{\Omega }$. The nonempty Dirichlet segment forces it to be zero. Hence only the zero global vector has zero quadratic energy, proving positive definiteness.