@liubenyuan, some remarks. Necessarily $S>0$. On the other hand, your result is correct up a signum (have a look at the case $n=1$); infortunately that makes a big difference ! The derivative is $0$ if $A=S^{-1}(qq^T-S)S^{-1}$ and you want that $A\geq 0$, that is $qq^T\geq S$. Let $(\lambda_i\geq 0)_i$ be the spectrum of $S$. According to the min-max theorem $||q||^2\geq \lambda_1$ and, for every $i>1$, $\lambda_i=0$ and $S$ cannot be invertible except if $n=1$. Finally, there are no critical points, except for $n=1$, $a=q^2/s^2-1/s$ with $q^2\geq s$.

EDIT: cf. the following post. Have you read my answer to this post ?

Hessian of function of covariance matrices

@liubenyuan, some remarks. Necessarily $S>0$. On the other hand, your result is correct up a signum (have a look at the case $n=1$); infortunately that makes a big difference ! The derivative is $0$ if $A=S^{-1}(qq^T-S)S^{-1}$ and you want that $A\geq 0$, that is $qq^T\geq S$. Let $(\lambda_i\geq 0)_i$ be the spectrum of $S$. According to the min-max theorem $||q||^2\geq \lambda_1$ and, for every $i>1$, $\lambda_i=0$ and $S$ cannot be invertible except if $n=1$. Finally, there are no critical points, except for $n=1$, $a=q^2/s^2-1/s$ with $q^2\geq s$.

EDIT: cf. the following post. Have you read my answer to this post ?

Hessian of function of covariance matrices

@liubenyuan, some remarks. Necessarily $S>0$. On the other hand, your result is correct up a signum (have a look at the case $n=1$); infortunately that makes a big difference ! The derivative is $0$ if $A=S^{-1}(qq^T-S)S^{-1}$ and you want that $A\geq 0$, that is $qq^T\geq S$. Let $(\lambda_i\geq 0)_i$ be the spectrum of $S$. According to the min-max theorem $||q||^2\geq \lambda_1$ and, for every $i>1$, $\lambda_i=0$ and $S$ cannot be invertible except if $n=1$. Finally, there are no critical points, except for $n=1$, $a=q^2/s^2-1/s$ with $q^2\geq s$.

@liubenyuan, some remarks. Necessarily $S>0$. On the other hand, your result is correct up a signum (have a look at the case $n=1$); infortunately that makes a big difference ! The derivative is $0$ if $A=S^{-1}(qq^T-S)S^{-1}$ and you want that $A\geq 0$, that is $qq^T\geq S$. Let $(\lambda_i\geq 0)_i$ be the spectrum of $S$. According to the min-max theorem $||q||^2\geq \lambda_1$ and, for every $i>1$, $\lambda_i=0$ and $S$ cannot be invertible except if $n=1$. Finally, there are no critical points, except for $n=1$, $a=q^2/s^2-1/s$ with $q^2\geq s$.

EDIT: cf. the following post. Have you read my answer to this post ?

Hessian of function of covariance matrices