 Solution

For the analogous positive off-diagonal entries in size $n$, takeThe same entrywise computation gives $(D^{-1}AD{)}_{ii}=a_i$ and $(D^{-1}AD{)}_{i,i+1}=(D^{-1}AD{)}_{i+1,i}=\sqrt{b_i{c}_i}$, with every other entry zero. Thus the result is a symmetric matrix with a tridiagonal pattern. Positivity of the diagonal entries $a_i$ is not needed for this positive diagonal symmetrization of a tridiagonal matrix; positivity of $b_i,c_i$ ensures the displayed positive, invertible $D$. In particular, the assertion is not an arbitrary-real-coefficient symmetrization claim when the off-diagonal signs differ.