$\text{A.}$ $y_1^2+y_2^2$. $\text{B.}$ $y_1^2-y_2^2$. $\text{C.}$ $y_1^2+y_2^2-y_3^2$. $\text{D.}$ $y_1^2-y_2^2-y_3^2$.
【答案】 A

【解析】 $\boldsymbol{A}^{\mathrm{T}}=\left(2 \boldsymbol{\alpha} \boldsymbol{\alpha}^{\mathrm{T}}+\boldsymbol{\beta} \boldsymbol{\beta}^{\mathrm{T}}\right)^{\mathrm{T}}=\left(2 \boldsymbol{\alpha} \boldsymbol{\alpha}^{\mathrm{T}}\right)^{\mathrm{T}}+\left(\boldsymbol{\beta} \boldsymbol{\beta}^{\mathrm{T}}\right)^{\mathrm{T}}=2 \boldsymbol{\alpha} \boldsymbol{\alpha}^{\mathrm{T}}+\boldsymbol{\beta} \boldsymbol{\beta}^{\mathrm{T}}=\boldsymbol{A}$, 即 $\boldsymbol{A}$ 为对称矩阵, 由 $\boldsymbol{\alpha}$, $\boldsymbol{\beta}$ 均为单位向量, $\boldsymbol{\alpha}^{\mathrm{T}} \boldsymbol{\beta}=0$, 知 $\boldsymbol{\alpha}^{\mathrm{T}} \boldsymbol{\alpha}=1, \boldsymbol{\beta}^{\mathrm{T}} \boldsymbol{\beta}=1, \boldsymbol{\beta}^{\mathrm{T}} \boldsymbol{\alpha}=0$, 于是
$$\begin{gathered} \boldsymbol{A} \boldsymbol{\alpha}=\left(2 \boldsymbol{\alpha} \boldsymbol{\alpha}^{\mathrm{T}}+\boldsymbol{\beta} \boldsymbol{\beta}^{\mathrm{T}}\right) \boldsymbol{\alpha}=2 \boldsymbol{\alpha}\left(\boldsymbol{\alpha}^{\mathrm{T}} \boldsymbol{\alpha}\right)+\boldsymbol{\beta}\left(\boldsymbol{\beta}^{\mathrm{T}} \boldsymbol{\alpha}\right)=2 \boldsymbol{\alpha}, \\ \boldsymbol{A} \boldsymbol{\beta}=\left(2 \boldsymbol{\alpha} \boldsymbol{\alpha}^{\mathrm{T}}+\boldsymbol{\beta} \boldsymbol{\beta}^{\mathrm{T}}\right) \boldsymbol{\beta}=2 \boldsymbol{\alpha}\left(\boldsymbol{\alpha}^{\mathrm{T}} \boldsymbol{\beta}\right)+\boldsymbol{\beta}\left(\boldsymbol{\beta}^{\mathrm{T}} \boldsymbol{\beta}\right)=\boldsymbol{\beta}, \end{gathered}$$

$$r(\boldsymbol{A})=r\left(2 \boldsymbol{\alpha} \boldsymbol{\alpha}^{\mathrm{T}}+\boldsymbol{\beta} \boldsymbol{\beta}^{\mathrm{T}}\right) \leqslant r\left(2 \boldsymbol{\alpha} \boldsymbol{\alpha}^{\mathrm{T}}\right)+r\left(\boldsymbol{\beta} \boldsymbol{\beta}^{\mathrm{T}}\right)=2 < 3,$$