题目

设A,P均为3阶方阵，P^T为P的转置矩阵，且P^TAP=(}1&0&00&1&00&0&2AQ=( ).

设A,P均为3阶方阵，$P^{T}$为P的转置矩阵，且$P^{T}AP=\left(\begin{matrix}1&0&0\\0&1&0\\0&0&2\end{matrix}\right)$，若$P=(\alpha_{1},\alpha_{2},\alpha_{3}),Q=(\alpha_{1}+\alpha_{2},\alpha_{2},\alpha_{3})$,则$Q^{T}AQ=( )$.

题目解答

答案

由题意，$P^TAP = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{pmatrix}$，其中 $P = (\alpha_1, \alpha_2, \alpha_3)$。
这表明 $\alpha_1, \alpha_2, \alpha_3$ 分别对应特征值 $1, 1, 2$。
矩阵 $Q = (\alpha_1 + \alpha_2, \alpha_2, \alpha_3)$，则
$Q^T = \begin{pmatrix} \alpha_1^T + \alpha_2^T \\ \alpha_2^T \\ \alpha_3^T \end{pmatrix}.$
计算 $Q^T A Q$：
$Q^T A Q = \begin{pmatrix} \alpha_1^T + \alpha_2^T \\ \alpha_2^T \\ \alpha_3^T \end{pmatrix} A \begin{pmatrix} \alpha_1 + \alpha_2 & \alpha_2 & \alpha_3 \end{pmatrix}.$
展开得：
$Q^T A Q = \begin{pmatrix} (\alpha_1^T + \alpha_2^T)A(\alpha_1 + \alpha_2) & (\alpha_1^T + \alpha_2^T)A\alpha_2 & (\alpha_1^T + \alpha_2^T)A\alpha_3 \\ \alpha_2^T A (\alpha_1 + \alpha_2) & \alpha_2^T A \alpha_2 & \alpha_2^T A \alpha_3 \\ \alpha_3^T A (\alpha_1 + \alpha_2) & \alpha_3^T A \alpha_2 & \alpha_3^T A \alpha_3 \end{pmatrix}.$
根据 $P^T A P$，有：
$\alpha_i^T A \alpha_j = \begin{cases} 1 & (i, j) = (1, 1) \text{ 或 } (2, 2), \\ 2 & (i, j) = (3, 3), \\ 0 & \text{其他}. \end{cases}$
代入计算：
$(\alpha_1^T + \alpha_2^T)A(\alpha_1 + \alpha_2) = \alpha_1^T A \alpha_1 + 2\alpha_1^T A \alpha_2 + \alpha_2^T A \alpha_2 = 1 + 0 + 1 = 2,$
$(\alpha_1^T + \alpha_2^T)A\alpha_2 = \alpha_1^T A \alpha_2 + \alpha_2^T A \alpha_2 = 0 + 1 = 1,$
$(\alpha_1^T + \alpha_2^T)A\alpha_3 = \alpha_1^T A \alpha_3 + \alpha_2^T A \alpha_3 = 0 + 0 = 0,$
$\alpha_2^T A (\alpha_1 + \alpha_2) = \alpha_2^T A \alpha_1 + \alpha_2^T A \alpha_2 = 0 + 1 = 1,$
$\alpha_2^T A \alpha_2 = 1, \quad \alpha_2^T A \alpha_3 = 0,$
$\alpha_3^T A (\alpha_1 + \alpha_2) = \alpha_3^T A \alpha_1 + \alpha_3^T A \alpha_2 = 0 + 0 = 0,$
$\alpha_3^T A \alpha_2 = 0, \quad \alpha_3^T A \alpha_3 = 2.$
综上，
$Q^T A Q = \begin{pmatrix} 2 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 2 \end{pmatrix}.$
因此，正确答案是 A。

答案：A. $\begin{pmatrix} 2 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 2 \end{pmatrix}$

解析

本题主要考察矩阵的转置、矩阵乘法以及特征值特征向量的性质，核心是利用矩阵$P$与$Q$的关系，通过$P^TAP$的已知对角矩阵反推$\alpha_i^TA\alpha_j$的取值，进而计算$Q^TAQ$。

步骤1：分析$P^TAP$的意义

已知$P=(\alpha_1,\alpha_2,\alpha_3)$，则$P^TAP$是对$A$进行合同变换（正交合同？不，仅合同变换），得到对角矩阵$\text{diag}(1,1,2)$。这表明：

  • 当$i=j$时：$\alpha_1^TA\alpha_1=1$，$\alpha_2^TA\alpha_2=1$，$\alpha_3^TA\alpha_3=2$（特征值对应）；
  • 当$i\neq j$时：$\alpha_i^TA\alpha_j=0$（因$P^TAP$非对角元为0，展开后交叉项消失）。

步骤2：表示$Q$并计算$Q^TAQ$

$Q=(\alpha_1+\alpha_2,\alpha_2,\alpha_3)$，则$Q^TAQ$可通过矩阵乘法展开为分块矩阵，每个元素为$\beta_i^TA\beta_j$（$\beta_1=\alpha_1+\alpha_2$，$\beta_2=\alpha_2$，$\beta_3=\alpha_3$）：

  • (1,1)元：$\beta_1^TA\beta_1=(\alpha_1^T+\alpha_2^T)A(\alpha_1+\alpha_2)=\alpha_1^TA\alpha_1+\alpha_1^TA\alpha_2+\alpha_2^TA\alpha_1+\alpha_2^TA\alpha_2=1+0+0+1=2$；
  • (1,2)元：$\beta_1^TA\beta_2=(\alpha_1^T+\alpha_2^T)A\alpha_2=\alpha_1^TA\alpha_2+\+\alpha_2^TA\alpha_2=0+1=1$；
  • (1,3)元：$\beta_1^TA\beta_3=(\alpha_1^T+\alpha_2^T)A\alpha_3=\alpha_1^TA\alpha_3+\alpha_2^TA\alpha_3=0+0=0$；
  • (2,1)元：$\beta_2^TA\beta_1=\alpha_2^TA(\alpha_1+\alpha_2)=\alpha_2^TA\alpha_1+\alpha_2^TA\alpha_2=0+1=1$（对称）；
  • (2,2)元：\(\beta_2^TA\beta_2=\alpha_2^TA\alpha_2=1\\)；
  • (2,3)元：$\beta_2^TA\beta_3=\alpha_2^TA\alpha_3=0$；
  • (3,1)元：$\beta_3^TA\beta_1=\alpha_3^TA(\alpha_1+\alpha_2)=\alpha_3^TA\alpha_1+\alpha_3^TA\alpha_2=0+0=0$；
  • (3,2)元：$\beta_3^TA\beta_2=\alpha_3^TA\alpha_2=0$；
  • (3,3)元：$\beta_3^TA\beta_3=\alpha_3^TA\alpha_3=2$。

步骤3：整理结果

综上，$Q^TAQ=\begin{pmatrix}2&1&0\\1&1&0\\0&0&2\end{pmatrix}$。