= Solution
For the <Kronecker quiver> representation with arrows $I$ and $N=\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)$, an endomorphism $(P,Q)$ satisfies $Q=P$ and $PN=NP$. Solving the second equation gives
$$
\operatorname{End}_Q(X)=\left\{\begin{pmatrix}a&b\\0&a\end{pmatrix}:a,b\in k\right\}\cong k[t]/(t^2).
$$
Thus \b[this representation is indecomposable but is not a brick]: its endomorphism ring is a <local endomorphism ring>, while the nonzero endomorphism $N$ is nilpotent.
For a general indecomposable non-brick, the <proof of Ringel lemma on bricks> finds a proper indecomposable submodule with nonzero self-extensions. Repetition in strictly decreasing dimension reaches a <brick module> $Y\subset X$ with $\operatorname{Ext}^1_Q(Y,Y)\ne0$. The linked proof supplies the minimal-rank, retraction and hereditary-extension steps.
Now assume the <Tits form of a quiver> is positive definite. If an indecomposable $X$ were not a brick, this $Y$ would give the contradiction
$$
0<q(\dim Y)=1-\dim\operatorname{Ext}^1_Q(Y,Y)\le0.
$$
Hence $X$ is a brick. For its nonzero dimension vector $\mathbf n$, positivity and integrality then imply
$$
0<q(\mathbf n)=1-\dim\operatorname{Ext}^1_Q(X,X)\le1,\qquad \boxed{q(\mathbf n)=1,\quad\operatorname{Ext}^1_Q(X,X)=0}.
$$
Thus every indecomposable in this case is a rigid brick. This deduction uses the <Ringel lemma on bricks> and the <Ringel form>, without requiring the full <Gabriel theorem>.
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