Minimum of a matrix function Let $\boldsymbol F_s$ be a $3\times3$ real matrix, and
$$
Q_3^s\left(\boldsymbol F_s\right)=\frac{1}{2}\left[\operatorname{tr}^2\boldsymbol F_s-\operatorname{tr}\boldsymbol F_s^2\right].
$$
How to find
$$
Q_2^s\left(\hat{\boldsymbol F_s}\right)=\min_{\boldsymbol c\in\mathbb{R}^3}Q_3^s\left(\hat{\boldsymbol F_s}+\boldsymbol c\otimes\boldsymbol e_3\right),
$$
where $\hat{\boldsymbol F_s}=\sum_{i,j=1}^2F_{ij}\boldsymbol e_i\otimes\boldsymbol e_j$? Here $F_{ij}$ are the elements of $\boldsymbol F_s$.
My try
Since
$$
Q_3^s\left(\hat{\boldsymbol F_s}+\boldsymbol c\otimes\boldsymbol e_3\right)=\frac{1}{2}\left[\left(F_{11}+F_{22}+c_3\right)^2-\left(F_{11}^2+2F_{12}F_{21}+F_{22}^2+c_3^2\right)\right],
$$
therefore
$$
\frac{\partial Q_3^s\left(\hat{\boldsymbol F_s}+\boldsymbol c\otimes\boldsymbol e_3\right)}{\partial c_3}=F_{11}+F_{22},
$$
which is independent of $c_3$. I don`t know how to proceed.
 A: Why make things so complicated? Your $\hat{F}_s$ is just a $3\times3$ matrix whose last column and last row are zero. Let
$$
F=\pmatrix{a&b&0\\ c&d&0\\ 0&0&0}+\pmatrix{x\\ y\\ z}e_3^T=\pmatrix{a&b&x\\ c&d&y\\ 0&0&z}.
$$
Then $\frac12(\operatorname{tr}^2F-\operatorname{tr}F^2)=2(a+d)z+\text{constant}$. Hence $\min_{x,y,z\in\mathbb R}\frac12(\operatorname{tr}^2F-\operatorname{tr}F^2)$ does not exist when $a+d\ne0$, and the objective function is constant when $a+d=0$. (By the way, $\frac12(\operatorname{tr}^2F-\operatorname{tr}F^2)=\operatorname{tr}(\operatorname{adj}F)$, but that probably doesn't matter here.) 
A: Let 
$$\eqalign{
 b &= e_3 \cr
 X &= F+cb^T \cr
\operatorname{tr}(X) &= \operatorname{tr}(F) + c^Tb \cr
}$$
Now write the function in terms of this new variable and find its differential and gradient
$$\eqalign{
 Q &= (X:I)^2 - X^T:X \cr
\cr
dQ &= 2(X:I)I:dX - 2X^T:dX \cr
   &= 2\Big((\operatorname{tr}F+c^Tb)I-F^T-bc^T\Big):dX \cr
   &= 2\Big((\operatorname{tr}F+c^Tb)I-F^T-bc^T\Big):dc\,b^T \cr
   &= 2\Big((\operatorname{tr}F+c^Tb)I-F^T-bc^T\Big)b:dc \cr
   &= 2(I\operatorname{tr}F-F^T)\,b:dc \cr
\cr
\frac{\partial Q}{\partial c} &= 2(I\operatorname{tr}F-F^T)\,b \cr
\cr
}$$
Since the gradient is independent of $c$, there is no minimum or maximum. We can increase or decrease the function value as much as we please by varying $c$.
