中国科学院真题 第20067题
📝 题目
六、设 $A, B$ 为对称方阵，试证明
$\operatorname{Tr}(\mathrm{ABAB}) \leq \operatorname{Tr}(\mathrm{AABB})$ ，其中＂ $\operatorname{Tr}$＂表示方阵的追迹（即对角元素之和）。
证明：设 $A, B$ 为 $n$ 阶对称方阵
$$
\begin{gathered}
A=\left(\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n}\right)=\left(\begin{array}{c}
\alpha_{1}^{\prime} \\
\alpha_{2}^{\prime} \\
\vdots \\
\alpha_{n}^{\prime}
\end{array}\right), \\
B=\left(\beta_{1}, \beta_{2}, \cdots, \beta_{n}\right)=\left(\begin{array}{c}
\beta_{1}^{\prime} \\
\beta_{2}^{\prime} \\
\vdots \\
\beta_{n}^{\prime}
\end{array}\right) . \\
\text { 则 } A B=\left(\begin{array}{c}
\alpha_{1}^{\prime} \\
\alpha_{2}^{\prime} \\
\vdots \\
\alpha_{n}^{\prime}
\end{array}\right)\left(\beta_{1}, \beta_{2}, \cdots, \beta_{n}\right)=\left(\begin{array}{cccc}
\alpha_{1}^{\prime} \beta_{1} & \alpha_{1}^{\prime} \beta_{2} & \cdots & \alpha_{1}^{\prime} \beta_{n} \\
\alpha_{2}^{\prime} \beta_{1} & \alpha_{2}^{\prime} \beta_{2} & \cdots & \alpha_{2}^{\prime} \beta_{n} \\
\cdots & \cdots & \cdots & \cdots \\
\alpha_{n}^{\prime} \beta_{1} & \alpha_{n}^{\prime} \beta_{2} & \cdots & \alpha_{n}^{\prime} \beta_{n}
\end{array}\right)
\end{gathered}
$$
所以
$$
(A B)^{2}=\left(\begin{array}{cccc}
\alpha_{1}^{\prime} \beta_{1} & \alpha_{1}^{\prime} \beta_{2} & \cdots & \alpha_{1}^{\prime} \beta_{n} \\
\alpha_{2}^{\prime} \beta_{1} & \alpha_{2}^{\prime} \beta_{2} & \cdots & \alpha_{2}^{\prime} \beta_{n} \\
\cdots & \cdots & \cdots & \cdots \\
\alpha_{n}^{\prime} \beta_{1} & \alpha_{n}^{\prime} \beta_{2} & \cdots & \alpha_{n}^{\prime} \beta_{n}
\end{array}\right)\left(\begin{array}{cccc}
\alpha_{1}^{\prime} \beta_{1} & \alpha_{1}^{\prime} \beta_{2} & \cdots & \alpha_{1}^{\prime} \beta_{n} \\
\alpha_{2}^{\prime} \beta_{1} & \alpha_{2}^{\prime} \beta_{2} & \cdots & \alpha_{2}^{\prime} \beta_{n} \\
\cdots & \cdots & \cdots & \cdots \\
\alpha_{n}^{\prime} \beta_{1} & \alpha_{n}^{\prime} \beta_{2} & \cdots & \alpha_{n}^{\prime} \beta_{n}
\end{array}\right)
$$
由此得 $\operatorname{Tr}(\mathrm{AB})^{2}=\sum_{i=1}^{n} \sum_{j=1}^{n}\left(\alpha_{i}^{\prime} \beta_{j}\right) \cdot\left(\alpha_{j}^{\prime} \beta_{i}\right)$ 。
而 $A^{2}=\left(\begin{array}{c}\alpha_{1}^{\prime} \\ \alpha_{2}^{\prime} \\ \vdots \\ \alpha_{n}^{\prime}\end{array}\right)\left(\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n}\right)=\left(\begin{array}{cccc}\alpha_{1}^{\prime} \alpha_{1} & \alpha_{1}^{\prime} \alpha_{2} & \cdots & \alpha_{1}^{\prime} \alpha_{n} \\ \alpha_{2}^{\prime} \alpha_{1} & \alpha_{2}^{\prime} \alpha_{2} & \cdots & \alpha_{2}^{\prime} \alpha_{n} \\ \cdots & \cdots & \cdots & \cdots \\ \alpha_{n}^{\prime} \alpha_{1} & \alpha_{n}^{\prime} \alpha_{2} & \cdots & \alpha_{n}^{\prime} \alpha_{n}\end{array}\right)$
$$
B^{2}=\left(\begin{array}{c}
\beta_{1}^{\prime} \\
\beta_{2}^{\prime} \\
\vdots \\
\beta_{n}^{\prime}
\end{array}\right)\left(\beta_{1}, \beta_{2}, \cdots, \beta_{n}\right)=\left(\begin{array}{cccc}
\beta_{1}^{\prime} \beta_{1} & \beta_{1}^{\prime} \beta_{2} & \cdots & \beta_{1}^{\prime} \beta_{n} \\
\beta_{2}^{\prime} \beta_{1} & \beta_{2}^{\prime} \beta_{2} & \cdots & \beta_{2}^{\prime} \beta_{n} \\
\cdots & \cdots & \cdots & \cdots \\
\beta_{n}^{\prime} \beta_{1} & \beta_{n}^{\prime} \beta_{2} & \cdots & \beta_{n}^{\prime} \beta_{n}
\end{array}\right)
$$
所以
$$
A^{2} B^{2}=\left(\begin{array}{cccc}
\alpha_{1}^{\prime} \alpha_{1} & \alpha_{1}^{\prime} \alpha_{2} & \cdots & \alpha_{1}^{\prime} \alpha_{n} \\
\alpha_{2}^{\prime} \alpha_{1} & \alpha_{2}^{\prime} \alpha_{2} & \cdots & \alpha_{2}^{\prime} \alpha_{n} \\
\cdots & \cdots & \cdots & \cdots \\
\alpha_{n}^{\prime} \alpha_{1} & \alpha_{n}^{\prime} \alpha_{2} & \cdots & \alpha_{n}^{\prime} \alpha_{n}
\end{array}\right)\left(\begin{array}{cccc}
\beta_{1}^{\prime} \beta_{1} & \beta_{1}^{\prime} \beta_{2} & \cdots & \beta_{1}^{\prime} \beta_{n} \\
\beta_{2}^{\prime} \beta_{1} & \beta_{2}^{\prime} \beta_{2} & \cdots & \beta_{2}^{\prime} \beta_{n} \\
\cdots & \cdots & \cdots & \cdots \\
\beta_{n}^{\prime} \beta_{1} & \beta_{n}^{\prime} \beta_{2} & \cdots & \beta_{n}^{\prime} \beta_{n}
\end{array}\right)
$$
由此得 $\operatorname{Tr}\left(\mathrm{A}^{2} \mathrm{~B}^{2}\right)=\sum_{i=1}^{n} \sum_{j=1}^{n}\left(\alpha_{i}^{\prime} \alpha_{j}\right) \cdot\left(\beta_{j}^{\prime} \beta_{i}\right)$ 。
最后由柯西－布涅柯夫斯基不等式易知
$$
\begin{aligned}
& \left(\alpha_{i}^{\prime} \beta_{j}\right) \cdot\left(\alpha_{j}^{\prime} \beta_{i}\right) \leq\left(\alpha_{i}^{\prime} \alpha_{j}\right) \cdot\left(\beta_{j}^{\prime} \beta_{i}\right), \quad 1 \leq i, \quad j \leq n . \\
& \text { 从而得 } \operatorname{Tr}(\mathrm{AB})^{2} \leq \operatorname{Tr}\left(\mathrm{A}^{2} \mathrm{~B}^{2}\right) .
\end{aligned}
$$
💡 答案解析
六、设 $A, B$ 为对称方阵，试证明
$\operatorname{Tr}(\mathrm{ABAB}) \leq \operatorname{Tr}(\mathrm{AABB})$ ，其中＂ $\operatorname{Tr}$＂表示方阵的追迹（即对角元素之和）。
证明：设 $A, B$ 为 $n$ 阶对称方阵
$$
\begin{gathered}
A=\left(\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n}\right)=\left(\begin{array}{c}
\alpha_{1}^{\prime} \\
\alpha_{2}^{\prime} \\
\vdots \\
\alpha_{n}^{\prime}
\end{array}\right), \\
B=\left(\beta_{1}, \beta_{2}, \cdots, \beta_{n}\right)=\left(\begin{array}{c}
\beta_{1}^{\prime} \\
\beta_{2}^{\prime} \\
\vdots \\
\beta_{n}^{\prime}
\end{array}\right) . \\
\text { 则 } A B=\left(\begin{array}{c}
\alpha_{1}^{\prime} \\
\alpha_{2}^{\prime} \\
\vdots \\
\alpha_{n}^{\prime}
\end{array}\right)\left(\beta_{1}, \beta_{2}, \cdots, \beta_{n}\right)=\left(\begin{array}{cccc}
\alpha_{1}^{\prime} \beta_{1} & \alpha_{1}^{\prime} \beta_{2} & \cdots & \alpha_{1}^{\prime} \beta_{n} \\
\alpha_{2}^{\prime} \beta_{1} & \alpha_{2}^{\prime} \beta_{2} & \cdots & \alpha_{2}^{\prime} \beta_{n} \\
\cdots & \cdots & \cdots & \cdots \\
\alpha_{n}^{\prime} \beta_{1} & \alpha_{n}^{\prime} \beta_{2} & \cdots & \alpha_{n}^{\prime} \beta_{n}
\end{array}\right)
\end{gathered}
$$
所以
$$
(A B)^{2}=\left(\begin{array}{cccc}
\alpha_{1}^{\prime} \beta_{1} & \alpha_{1}^{\prime} \beta_{2} & \cdots & \alpha_{1}^{\prime} \beta_{n} \\
\alpha_{2}^{\prime} \beta_{1} & \alpha_{2}^{\prime} \beta_{2} & \cdots & \alpha_{2}^{\prime} \beta_{n} \\
\cdots & \cdots & \cdots & \cdots \\
\alpha_{n}^{\prime} \beta_{1} & \alpha_{n}^{\prime} \beta_{2} & \cdots & \alpha_{n}^{\prime} \beta_{n}
\end{array}\right)\left(\begin{array}{cccc}
\alpha_{1}^{\prime} \beta_{1} & \alpha_{1}^{\prime} \beta_{2} & \cdots & \alpha_{1}^{\prime} \beta_{n} \\
\alpha_{2}^{\prime} \beta_{1} & \alpha_{2}^{\prime} \beta_{2} & \cdots & \alpha_{2}^{\prime} \beta_{n} \\
\cdots & \cdots & \cdots & \cdots \\
\alpha_{n}^{\prime} \beta_{1} & \alpha_{n}^{\prime} \beta_{2} & \cdots & \alpha_{n}^{\prime} \beta_{n}
\end{array}\right)
$$
由此得 $\operatorname{Tr}(\mathrm{AB})^{2}=\sum_{i=1}^{n} \sum_{j=1}^{n}\left(\alpha_{i}^{\prime} \beta_{j}\right) \cdot\left(\alpha_{j}^{\prime} \beta_{i}\right)$ 。
而 $A^{2}=\left(\begin{array}{c}\alpha_{1}^{\prime} \\ \alpha_{2}^{\prime} \\ \vdots \\ \alpha_{n}^{\prime}\end{array}\right)\left(\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n}\right)=\left(\begin{array}{cccc}\alpha_{1}^{\prime} \alpha_{1} & \alpha_{1}^{\prime} \alpha_{2} & \cdots & \alpha_{1}^{\prime} \alpha_{n} \\ \alpha_{2}^{\prime} \alpha_{1} & \alpha_{2}^{\prime} \alpha_{2} & \cdots & \alpha_{2}^{\prime} \alpha_{n} \\ \cdots & \cdots & \cdots & \cdots \\ \alpha_{n}^{\prime} \alpha_{1} & \alpha_{n}^{\prime} \alpha_{2} & \cdots & \alpha_{n}^{\prime} \alpha_{n}\end{array}\right)$
$$
B^{2}=\left(\begin{array}{c}
\beta_{1}^{\prime} \\
\beta_{2}^{\prime} \\
\vdots \\
\beta_{n}^{\prime}
\end{array}\right)\left(\beta_{1}, \beta_{2}, \cdots, \beta_{n}\right)=\left(\begin{array}{cccc}
\beta_{1}^{\prime} \beta_{1} & \beta_{1}^{\prime} \beta_{2} & \cdots & \beta_{1}^{\prime} \beta_{n} \\
\beta_{2}^{\prime} \beta_{1} & \beta_{2}^{\prime} \beta_{2} & \cdots & \beta_{2}^{\prime} \beta_{n} \\
\cdots & \cdots & \cdots & \cdots \\
\beta_{n}^{\prime} \beta_{1} & \beta_{n}^{\prime} \beta_{2} & \cdots & \beta_{n}^{\prime} \beta_{n}
\end{array}\right)
$$
所以
$$
A^{2} B^{2}=\left(\begin{array}{cccc}
\alpha_{1}^{\prime} \alpha_{1} & \alpha_{1}^{\prime} \alpha_{2} & \cdots & \alpha_{1}^{\prime} \alpha_{n} \\
\alpha_{2}^{\prime} \alpha_{1} & \alpha_{2}^{\prime} \alpha_{2} & \cdots & \alpha_{2}^{\prime} \alpha_{n} \\
\cdots & \cdots & \cdots & \cdots \\
\alpha_{n}^{\prime} \alpha_{1} & \alpha_{n}^{\prime} \alpha_{2} & \cdots & \alpha_{n}^{\prime} \alpha_{n}
\end{array}\right)\left(\begin{array}{cccc}
\beta_{1}^{\prime} \beta_{1} & \beta_{1}^{\prime} \beta_{2} & \cdots & \beta_{1}^{\prime} \beta_{n} \\
\beta_{2}^{\prime} \beta_{1} & \beta_{2}^{\prime} \beta_{2} & \cdots & \beta_{2}^{\prime} \beta_{n} \\
\cdots & \cdots & \cdots & \cdots \\
\beta_{n}^{\prime} \beta_{1} & \beta_{n}^{\prime} \beta_{2} & \cdots & \beta_{n}^{\prime} \beta_{n}
\end{array}\right)
$$
由此得 $\operatorname{Tr}\left(\mathrm{A}^{2} \mathrm{~B}^{2}\right)=\sum_{i=1}^{n} \sum_{j=1}^{n}\left(\alpha_{i}^{\prime} \alpha_{j}\right) \cdot\left(\beta_{j}^{\prime} \beta_{i}\right)$ 。
最后由柯西－布涅柯夫斯基不等式易知
$$
\begin{aligned}
& \left(\alpha_{i}^{\prime} \beta_{j}\right) \cdot\left(\alpha_{j}^{\prime} \beta_{i}\right) \leq\left(\alpha_{i}^{\prime} \alpha_{j}\right) \cdot\left(\beta_{j}^{\prime} \beta_{i}\right), \quad 1 \leq i, \quad j \leq n . \\
& \text { 从而得 } \operatorname{Tr}(\mathrm{AB})^{2} \leq \operatorname{Tr}\left(\mathrm{A}^{2} \mathrm{~B}^{2}\right) .
\end{aligned}
$$
📋 详细解题步骤
步骤 1/1
目标：计算tr((AB-BA)²)=2tr(A²B²)-2tr(ABAB)≥0，利用迹的循环性，得tr(ABAB)≤tr(A²B²)。
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