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[原创]高等代数问题解答75

1. 高等代数问题解答75

Example 1. 设$A\in R^{n\times n}$是正定矩阵,$X_{0}\in R^{n}$为$AX=b$的解,则二次函数
$$P(X)=\dfrac{1}{2}X^{T}AX-X^{T}b$$
在$X_{0}$处达到最小值.

\textbf{解:} 由$A$正定知存在可逆矩阵$P$使得$P^{T}AP=E,$于是令$X=PY$,设
$$Y=(y_{1},y_{2},\cdots,y_{n})^{T},P^{T}AX_{0}=(c_{1},c_{2},\cdots,c_{n})^{T},$$

$$\begin{aligned}%
P(X)&=\dfrac{1}{2}X^{T}AX-X^{T}b\\
&=\dfrac{1}{2}Y^{T}P^{T}APY-Y^{T}P^{T}AX_{0}\\
&=\dfrac{1}{2}Y^{T}Y-Y^{T}P^{T}AX_{0}\\
&=\dfrac{1}{2}(y_{1}^{2}+y_{2}^{2}+\cdots+y_{n}^{2})-(c_{1}y_{1}+c_{2}y_{2}+\cdots+c_{n}y_{n})\\
&=\dfrac{1}{2}[(y_{1}^{2}-2c_{1}y_{1})+(y_{2}^{2}-2c_{2}y_{2})+\cdots+(y_{n}^{2}-2c_{n}y_{n})]\\
&=\dfrac{1}{2}[(y_{1}-c_{1})^{2}+(y_{2}-c_{2})^{2}+\cdots+(y_{n}-c_{n})^{2}]-\dfrac{1}{2}(c_{1}^{2}+c_{2}^{2}+\cdots+c_{n}^{2})\\
&\geq -\dfrac{1}{2}(c_{1}^{2}+c_{2}^{2}+\cdots+c_{n}^{2})\\
\end{aligned}$$
于是$P(X)$取得最小值的充要条件是
$$y_{1}=c_{1},y_{2}=c_{2},\cdots,y_{n}=c_{n},$$

$$Y=P^{T}AX_{0},$$
注意到$X=PY,P^{T}A=P^{-1}$,于是
$$X=PY=PP^{T}AX_{0}=X_{0}.$$
即结论成立.

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