I know that $AB+BA$ is not necessarily positive definite, as this question has been asked before on here. What I don't understand is how one would go about constructing counter-examples. Previous answers to this question just state the counter-examples, such as $A = \begin{bmatrix} 6 & 0 \\ 0 & 1 \\ \end{bmatrix}$, $B = \begin{bmatrix} 2 & 1 \\ 1 & 1 \\ \end{bmatrix}$. Is there some intuition behind how these matrices are chosen, or is it more a matter of just fiddling with $2 \times 2$ matrices until we find a counter-example?
Thanks in advance.
Here is one way to think about the problem. If $AB+BA$ is always positive definite, by passing $A$ to the limit, $AB+BA$ should be positive semidefinite too when $A$ is a rank-one positive semidefinite matrix.
So, with such an $A$, if you can construct a counterexample such that $B\succ0$ but $AB+BA$ is indefinite, then by a continuity argument, you can get a counterexample with a positive definite $A$.
E.g. $AB+BA$ is indefinite when $A=\pmatrix{1&0\\ 0&0}$ and $B=\pmatrix{2&1\\ 1&1}$. It follows that $AB+BA$ is indefinite when $A=\pmatrix{1\\ &t}$ and $t>0$ is sufficiently small (the value of $B$ here is actually unimportant; as long as $B$ is not diagonal, we get a counterexample). Multiply $A$ by $\frac1t$, we may pick $A=\pmatrix{\frac1t\\ &1}$ as well. Now we love integer matrices. So, let's set $\frac1t=n$ for some positive integer $n$. It remains to find an $n$ that really works. Since $$ \det(AB+BA)=\det\pmatrix{4n&n+1\\ n+1&2}=8n-(n+1)^2=8-(n-3)^2, $$ any $n>3+\sqrt{8}$ with do the job. The least integer such $n$ is $6$, which is the value used in the answer you have mentioned.