= Solution
{c}
Let $x_1,\ldots,x_N$ be the interior nodes, and let $\phi_1,\ldots,\phi_N$ be the nodal <basis> of the common <conforming finite element space>, with $\phi_j(x_i)=\delta_{ij}$ and zero boundary values. These <piecewise-linear hat functions> give the expansions
$$
u_h=\sum_{j=1}^N U_j\phi_j,\qquad w_h=\sum_{j=1}^N W_j\phi_j.
$$
Testing the first equation with $\phi_i$ and then the second equation with $\phi_i$ gives the <linear system>
$$
\boxed{\begin{pmatrix}S&-M\\M&S\end{pmatrix}\begin{pmatrix}U\\W\end{pmatrix}=\begin{pmatrix}0\\F\end{pmatrix},\qquad
S_{ij}=\int_\Omega\nabla\phi_j\cdot\nabla\phi_i,\quad
M_{ij}=\int_\Omega\phi_j\phi_i,\quad
F_i=\int_\Omega f\phi_i.}
$$
Here $S$ 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 $v_c=\sum_jc_j\phi_j$. The two <quadratic forms> are
$$
c^{\mathsf T}Mc=\|v_c\|_0^2,\qquad c^{\mathsf T}Sc=\|\nabla v_c\|_0^2.
$$
If $c\ne0$, the <linear independence> of the <basis> ensures $v_c\ne0$, 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 $c\ne0$. 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:
$$
\begin{aligned}
X^{\mathsf T}AX
&=p^{\mathsf T}Sp-p^{\mathsf T}Mq+q^{\mathsf T}Mp+q^{\mathsf T}Sq\\
&=p^{\mathsf T}Sp+q^{\mathsf T}Sq.
\end{aligned}
$$
Therefore
$$
\boxed{X^{\mathsf T}AX>0\quad(X\ne0),\qquad \frac{A+A^{\mathsf T}}2=\begin{pmatrix}S&0\\0&S\end{pmatrix}.}
$$
This is the positivity of a <quadratic form> intended by the last sentence of the question. For $N>0$, $A$ is a <nonsymmetric matrix>, since $M\ne0$. 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 $X=0$.
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