八元多项式⩾0 取等时有4个自由变量

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来自$\det(A\overline{A}+I) \geqslant 0$的$A\inC^{2\times2}$的情况：WolframAlpha
如何证明$\forall a_1,\dots,a_8\inR,$
\begin{multline*}
(a_1^2 + a_2^2+1) (a_7^2 + a_8^2 + 1)\\
+ a_1 (a_3 (-2 a_5 a_7 - 2 a_6 a_8) + a_4 (2 a_6 a_7 - 2 a_5 a_8))\\
+ a_2 (a_3 (2 a_5 a_8 - 2 a_6 a_7) + a_4 (-2 a_5 a_7 - 2 a_6 a_8))\\
+ (a_3^2 + a_4^2) (a_5^2 + a_6^2)+ 2 a_3 a_5 + 2 a_4 a_6\geqslant0\end{multline*}
验证：

1. Minimize[Det[IdentityMatrix[2]+{{a1+a2 I,a3+a4 I},{a5+a6 I,a7+a8 I}}.{{a1-a2 I,a3-a4 I},{a5-a6 I,a7-a8 I}}],{a1,a2,a3,a4,a5,a6,a7,a8}]

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输出：$$\left\{0,\left\{\text{a1}\to 0,\text{a2}\to 0,\text{a3}\to -1,\text{a4}\to -1,\text{a5}\to \frac{1}{2},\text{a6}\to \frac{1}{2},\text{a7}\to 0,\text{a8}\to 0\right\}\right\}$$

hbghlyj

在 mathoverflow.net/a/384916 这个证明中包含了这个不等式的证明：（又见MSE）

$\det(1+\bar{X}X)$
apply the consimilarity lemma, to write $X=SR\bar{S}^{-1}$ with $R$ a real matrix. This gives
$$\det(1+\bar{S}R^2\bar{S}^{-1})=\det(1+R^2)=\det(1+iR)\det(1-iR)=|\det(1+iR)|^2\geq 0.$$

取等条件为$$\det(1+iR)=0\tag1\label1$$

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\begin{multline*}
(a_1^2 + a_2^2) (a_7^2 + a_8^2 + 1) + (a_3^2 + a_4^2) (a_5^2 + a_6^2)\\
+ a_1 (a_3 (-2 a_5 a_7 - 2 a_6 a_8) + a_4 (2 a_6 a_7 - 2 a_5 a_8))\\
+ a_2 (a_3 (2 a_5 a_8 - 2 a_6 a_7) + a_4 (-2 a_5 a_7 - 2 a_6 a_8))\\
+ 2 a_3 a_5 + 2 a_4 a_6 + a_7^2 + a_8^2+1\end{multline*}\begin{multline*}
=\frac{\left(a_3 \left(a_1^2+a_2^2+1\right)+a_5 \left(a_3^2+a_4^2\right)\right)^2+\left(a_4 \left(a_1^2+a_2^2+1\right)+a_6 \left(a_3^2+a_4^2\right)\right)^2}{\left(a_1^2+a_2^2+1\right) \left(a_3^2+a_4^2\right)}\\+\frac{\left(-a_7 \left(a_1^2+a_2^2+1\right)+a_1 a_3 a_5-a_1 a_4 a_6+a_2 a_3 a_6+a_2 a_4 a_5\right)^2}{a_1^2+a_2^2+1}\\+\frac{\left(-a_8 \left(a_1^2+a_2^2+1\right)+a_1 a_3 a_6+a_1 a_4 a_5-a_2 a_3 a_5+a_2 a_4 a_6\right)^2}{a_1^2+a_2^2+1}\ge0\end{multline*}

1. Det[IdentityMatrix[2]+{{a1+a2 I,a3+a4 I},{a5+a6 I,a7+a8 I}}.{{a1-a2 I,a3-a4 I},{a5-a6 I,a7-a8 I}}]==
2. ((((1+a1^2+a2^2) a3+(a3^2+a4^2) a5)^2+(a3^2 a6+a4 (1+a1^2+a2^2+a4 a6))^2)/((1+a1^2+a2^2) (a3^2+a4^2))+1/(1+a1^2+a2^2) ((a1 a3 a5+a2 a4 a5+a2 a3 a6-a1 a4 a6-(1+a1^2+a2^2) a7)^2+(-a2 a3 a5+a1 a4 a5+a1 a3 a6+a2 a4 a6-(1+a1^2+a2^2) a8)^2))//FullSimplify

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True

取等条件有$4$个自由变量：
$a_1,a_2,a_3,a_4$取任意实数，$a_3^2+a_4^2\ne0$
$a_5=-\frac{a_3 \left(a_1^2+a_2^2+1\right)}{a_3^2+a_4^2}$，$a_6=-\frac{a_4 \left(a_1^2+a_2^2+1\right)}{a_3^2+a_4^2}$
$a_7=\frac{a_1 a_3 a_5-a_1 a_4 a_6+a_2 a_3 a_6+a_2 a_4 a_5}{a_1^2+a_2^2+1}$，$a_8=\frac{a_1 a_3 a_6+a_1 a_4 a_5-a_2 a_3 a_5+a_2 a_4 a_6}{a_1^2+a_2^2+1}$

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验证以上“取等条件”与\eqref{1}相同：
将以上用复数$a_1+a_2i,a_3+a_4i,a_5+a_6i,a_7+a_8i$写出：
$a_1+a_2i,a_3+a_4i$取任意实数，$a_3+a_4i\ne0$
$a_5+a_6i=-\frac{|a_1+a_2i|^2+1}{a_3-i a_4}$
$a_7+a_8i=-\frac{(a_1-i a_2) (a_3+ia_4)}{a_3-i a_4}$
$A\overline{A}=-I$
$A=\pmatrix{a_1+a_2i&a_3+a_4i\\-\frac{|a_1+a_2i|^2+1}{a_3-i a_4}&-\frac{(a_1-i a_2) (a_3+ia_4)}{a_3-i a_4}}$
取$S=\left(
\begin{array}{cc}
-\frac{1}{a_4-i a_3} & 0 \\
\frac{a_2+i a_1}{a_3-i a_4} & i \\
\end{array}
\right)$，$SA\overline{S}^{-1}=R=\pmatrix{0&1\\-1&0}$是实矩阵，且$R$满足\eqref{1}.

1. A={{a1+a2 I,a3+a4 I},{-((1+a1^2+a2^2)/(a3-I a4)),-(a1-I a2) (a3+I a4)/(a3-I a4)}};
2. S={{-(1/(-I a3+a4)),0},{(I a1+a2)/(a3-I a4),I}};
3. S.A.Inverse[S/.{I->-I,-I->I}]//Factor

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{{0,1},{-1,0}}

hbghlyj

对于$A\inC^{n\times n}$有两条不等式：$$\det(A\overline{A}+I) \geqslant 0$$
$$\det(AA^H+I) \geqslant 1$$
是否成立$\det(AA^H+I)\geqslant\det(A\overline{A}+I)$？

当$A\inC^{2\times2}$时的验证：

1. Det[IdentityMatrix[2]+{{a1+a2 I,a3+a4 I},{a5+a6 I,a7+a8 I}}.{{a1-a2 I,a5-a6 I},{a3-a4 I,a7-a8 I}}]-Det[IdentityMatrix[2]+{{a1+a2 I,a3+a4 I},{a5+a6 I,a7+a8 I}}.{{a1-a2 I,a3-a4 I},{a5-a6 I,a7-a8 I}}]//FullSimplify

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$(a_3-a_5)^2+(a_4-a_6)^2$

math.stackexchange.com/questions/5003482