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 \( \hat{B}=\hat{U}\hat{Q} \).