 Solution

Let $x_1,\dots ,x_N$ be the interior nodes, and let ${\varphi }_1,\dots ,{\varphi }_N$ be the nodal basis of the common conforming finite element space, with ${\varphi }_j(x_i)={\delta }_{ij}$ and zero boundary values. These piecewise-linear hat functions give the expansionsTesting the first equation with ${\varphi }_i$ and then the second equation with ${\varphi }_i$ gives the linear systemHere $S$ is the diffusion stiffness matrix and $M$ the mass matrix. Their entries are symmetric in $i,j$, so both are symmetric matrices.

For $c\in {\mathbb{R}}^N$, put $v_c={\sum }_j{c}_j{\varphi }_j$. The two quadratic forms areIf $c\ne 0$, the linear independence of the basis ensures $v_c\ne 0$, so $c^{\mathsf{T}}Mc>0$. If $c^{\mathsf{T}}Sc=0$, the Poincaré inequality and zero Dirichlet boundary condition force $v_c=0$, contradicting $c\ne 0$. Thus both $S$ and $M$ are positive-definite matrices.

For $X=(p,q)\in {\mathbb{R}}^N\times {\mathbb{R}}^N$, symmetry of $M$ cancels the cross terms:ThereforeThis is the positivity of a quadratic form intended by the last sentence of the question. For $N>0$, $A$ is a nonsymmetric matrix, since $M\ne 0$. Under the convention that “positive-definite matrix” includes symmetry, that last sentence needs this qualification: the symmetric part of a matrix is positive definite. The quadratic lower bound for a nonsymmetric matrix still implies that $A$ is an invertible matrix: $AX=0$ would give $X^{\mathsf{T}}AX=0$, hence $X=0$.