Past exam of the year 1 of the computer science course of the University of Oxford / 2022 / Digital systems, linear algebra, introduction to formal proof / 5 / b

For $D=\mathrm{diag}(d_1,d_2,d_3)$, the upper entry of $D^{-1}AD$ in positions $i,i+1$ is $b_i{d}_{i+1}/d_i$, and the corresponding lower entry is $c_i{d}_i/d_{i+1}$. They are equal when $(d_{i+1}/d_i{)}^2=c_i/b_i$. All coefficients used here are positive, so chooseDirect matrix multiplication gives the symmetric matrixIts pattern remains tridiagonal. This is positive diagonal symmetrization of a tridiagonal matrix.

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.

They are similar matrices, so their eigenvalues, with algebraic multiplicities, are identical. Explicitly,Moreover their eigenvectors correspond by $Av=\lambda v⟺(D^{-1}AD)(D^{-1}v)=\lambda D^{-1}v$. Since the transformed matrix is a real symmetric matrix, the spectral theorem for real symmetric matrices also shows that $A$ has real eigenvalues and is a diagonalizable matrix.