We have now found $\hat{Q}$ and $\hat{U}$ and this allows us to find the matrix $\hat{B}$ which is, due to Schur's theorem, unitarily similar to a triangular matrix (upper in our case) since we have that $$\hat{Q}^{-1}\hat{A}\hat{Q} = \hat{B},$$ from Schur's theorem the matrix $\hat{B}$ is triangular and the eigenvalues the same as those of $\hat{A}$ and are given by the diagonal matrix elements of $\hat{B}$. Why?

Our matrix $\hat{B}=\hat{U}\hat{Q}$.