[问题2014S11] 解答
我们先引用一下复旦高代书 P310 的习题 6, 其证明可参考白皮书 P257 的例 8.33:
习题6 设实二次型 f(x1,x2,?,xn)=y21+?+y2k?y2k+1???y2k+s<script id="MathJax-Element-1" type="math/tex">f(x_1,x_2,\cdots,x_n)=y_1^2+\cdots+y_k^2-y_{k+1}^2-\cdots-y_{k+s}^2</script>, 其中 yi=ai1x1+ai2x2+?+ainxn(i=1,2,?,k+s)<script id="MathJax-Element-2" type="math/tex">y_i=a_{i1}x_1+a_{i2}x_2+\cdots+a_{in}x_n\,(i=1,2,\cdots,k+s)</script>, 求证: f<script id="MathJax-Element-3" type="math/tex">f</script> 的正惯性指数 p≤k<script id="MathJax-Element-4" type="math/tex">p\leq k</script>, 负惯性指数 q≤s<script id="MathJax-Element-5" type="math/tex">q\leq s</script>.
把上述结论转化为实对称阵的语言, 马上可以得到如下引理:
引理 设 A<script id="MathJax-Element-6" type="math/tex">A</script> 为 m<script id="MathJax-Element-7" type="math/tex">m</script> 阶实对称阵, C<script id="MathJax-Element-8" type="math/tex">C</script> 为 m×n<script id="MathJax-Element-9" type="math/tex">m\times n</script> 阶实矩阵, 则 p(A)≥p(C′AC)<script id="MathJax-Element-10" type="math/tex">p(A)\geq p(C‘AC)</script>, q(A)≥q(C′AC)<script id="MathJax-Element-11" type="math/tex">q(A)\geq q(C‘AC)</script>, 其中 p(?),q(?)<script id="MathJax-Element-12" type="math/tex">p(\,\cdot\,),q(\,\cdot\,)</script> 分别表示正负惯性指数.
回到本题的证明.
考虑 2n<script id="MathJax-Element-13" type="math/tex">2n</script> 阶实对称阵 [A00B]<script id="MathJax-Element-14" type="math/tex">\begin{bmatrix} A & 0 \\ 0 & B \end{bmatrix}</script> 以及 2n×n<script id="MathJax-Element-15" type="math/tex">2n\times n</script> 阶实矩阵 [ In In]<script id="MathJax-Element-16" type="math/tex">\begin{bmatrix} I_n \\ I_n \end{bmatrix}</script>, 则有
[In In ][A00B][InIn]=A+B. <script id="MathJax-Element-17" type="math/tex; mode=display">\begin{bmatrix} I_n & I_n \end{bmatrix}\begin{bmatrix} A & 0 \\ 0 & B \end{bmatrix}\begin{bmatrix} I_n \\ I_n \end{bmatrix}=A+B.</script> 由上述引理即得 p(A)+p(B)=p([A00B])≥p(A+B); <script id="MathJax-Element-18" type="math/tex; mode=display">p(A)+p(B)=p(\begin{bmatrix} A & 0 \\ 0 & B \end{bmatrix})\geq p(A+B);</script> q(A)+q(B)=q([A00B])≥q(A+B), <script id="MathJax-Element-19" type="math/tex; mode=display">q(A)+q(B)=q(\begin{bmatrix} A & 0 \\ 0 & B \end{bmatrix})\geq q(A+B),</script> 故结论得证. □<script id="MathJax-Element-20" type="math/tex">\Box</script>