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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.

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    Does $e_i \otimes e_j$ mean $e_ie_j^T$? If so, then what is the difference between $F_s$ and $\hat F_s$? In the case that $[e_3 \quad F_s^T e_3 \quad (F_s^T)^2 e_3]$ has full rank, we should get the same minimum since we can completely steer the characteristic polynomial of $F_s + ce_3^T$. – 2017-02-16
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    what does $\text{tr}^2$ mean? – 2017-02-16

2 Answers 2

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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.)

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    That's basically what I derived, isn't it? – 2017-02-17
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    @AsaturKhurshudyan Yes, but without those scary symbols. – 2017-02-18
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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 $c$.

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    Because you used the pseduo-inverse, you can't guarantee that the gradient will be zero at your $c$. – 2017-02-16
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    @Omnomnomnom The pseudo-inverse has been removed, but now I'm stuck. – 2017-02-16
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    $X:II-X^T$ should really be $(X:I)I - X^T$, no? – 2017-02-16
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    In fact, without more assumptions on $F$, we can't guarantee that the function has any lower bound, so it makes sense that the gradient might never be zero. – 2017-02-16