How to check for local extrema or saddle point given an semidefinite matrix I've computed the Hessian of a given function $f(a,b,c) = y-a\sin(bx-c)$ and got the following result:
$\begin{pmatrix}
0 & -x\cdot\cos(bx - c) & \cos(bx - c)  \\
-x\cdot\cos(bx - c) & ax^2\cdot\sin(bx - c) & -ax\sin(bx - c) \\
\cos(bx - c) & -ax\cdot\sin(bx - c) & a\cdot\sin(bx - c)
\end{pmatrix}$
This matrix is positive semi-definite and thus one can not state for a given point $P=(a_i,b_i,c_i)$ if it is a local min, max or a saddle point. Is there any other way to explicitly determine if we have a loc. min, max or saddle point?

With reference to the given matrix, we have that
$$\det(a\sin(bx - c))=a\sin(bx - c)$$
$$\begin{vmatrix}
    0 & -x\cos(bx - c)  \\
-x\cos(bx - c) & ax^2\sin(bx - c)  \\
\end{vmatrix}=-x^2\cos^2(bx - c)$$
$$\begin{vmatrix} 
0 & -x\cdot\cos(bx - c) & \cos(bx - c)  \\
   -x\cdot\cos(bx - c) & ax^2\cdot\sin(bx - c) & -ax\sin(bx - c) \\ 
\cos(bx - c) & -ax\cdot\sin(bx - c) & a\cdot\sin(bx - c)
   \end{vmatrix}=$$
$$=x\cdot\cos(bx - c)(-ax\cos(bx - c)\sin(bx - c)+ax\cos(bx - c)\sin(bx - c))+\cos(bx - c)(ax^2\cos(bx - c)\sin(bx - c)-ax^2\cos(bx - c)\sin(bx - c))=0$$
therefore for $x^2\cos^2(bx - c)\neq 0$


*

*$a\sin(bx - c)\le 0$ the matrix is negative semidefinite

*$a\sin(bx - c)>0$ the matrix is indefinite
and for $x^2\cos^2(bx - c)= 0$


*

*$a\sin(bx - c)< 0$ the matrix is negative semidefinite

*$a\sin(bx - c)>0$ the matrix is positive semidefinite
Note that the condition for critical points implies


*

*$f_a=-\sin(bx - c)=0\implies \sin(bx - c)=0$

*$f_b=-ax\sin(bx - c)=0\implies \sin(bx - c)=0 \lor x=0$

*$f_c=a\sin(bx - c)=0\implies \sin(bx - c)=0$
therefore 


*

*$\cos(bx - c)=\pm 1$


and the expression for the Hessian simplifies.
