system (\[sa:eveq.int.1\]) upon a substitution $$\label{qsa.1}
 v^\alpha = \varphi^\alpha(u), \quad \alpha = 1, \ldots, m,$$ such that $$\label{qsa.eq3}
 \varphi(u) \not= 0.$$ In other words, the equations (\[qsa.2\]) hold after the substitution (\[qsa.1\]), where not all $\varphi^\alpha(u)$ vanish simultaneously.

\[qsa0:rem2\] The condition (\[qsa.eq3\]), unlike (\[qsa0.eq2\]), does not guarantee the equivalence of Eqs. (\[sa:eveq.int.2\]) and (\[sa:eveq.int.1\]) because the matrix $\|\lambda^\beta_\alpha\|$ may be singular.

\[qsa:exa1\] It is well known that the linear heat equation (\[sa:eveq.exa1:eq1\]) is not self-adjoint (*not strictly self-adjoint* in the sense of Definition \[saeq:def\]). It is clear from Eqs. (\[sa:eveq.exa1:eq1\]) and (\[sa:eveq.exa1:eq4\]). Let us test Eq. (\[sa:eveq.exa1:eq1\]) for quasi self-adjointness. Letting $v = \varphi(u),$ we obtain $$v_t = \varphi' u_t, \ v_x = \varphi' u_x, \ v_{xx} = \varphi' u_{xx}
 + \varphi'' u_x^2,$$ and the condition (\[qsa.2\]) is written: $$\varphi'(u)[u_t + u_{xx}] + \varphi''(u) u_x^2 = \lambda [u_t - u_{xx}].$$ Whence, comparing the coefficients of $u_t$ in both sides, we obtain $\lambda = \varphi'(u).$ Then the above equation becomes $$\varphi'(u)[u_t + u_{xx}] + \varphi''(u) u_x^2 = \varphi'(u) [u_t - u_{xx}].$$ This equation yields that $\varphi'(u) = 0.$ Hence, Eq. (\[sa:eveq.exa1:eq1\]) is quasi self-adjoint with the substitution $v = C,$ where $C$ is any non-vanishing constant. This substitution does not satisfy 