[问题2014S14]  解答

首先, 满足条件的 \varphiφ<script id="MathJax-Element-1" type="math/tex">\varphi</script> 的全体特征值都为零. 事实上, 任取 \varphiφ<script id="MathJax-Element-2" type="math/tex">\varphi</script> 的特征值 \lambdaλ<script id="MathJax-Element-3" type="math/tex">\lambda</script>, 对应的特征向量为 0\neq\xi\in V0≠ξ∈V<script id="MathJax-Element-4" type="math/tex">0\neq\xi\in V</script>, 即 \varphi(\xi)=\lambda\xiφ(ξ)=λξ<script id="MathJax-Element-5" type="math/tex">\varphi(\xi)=\lambda\xi</script>, 则由假设可得 0=(\varphi(\xi),\xi)=(\lambda\xi,\xi)=\lambda(\xi,\xi),

0=(φ(ξ),ξ)=(λξ,ξ)=λ(ξ,ξ),

ξ≠0

(ξ,ξ)>0

λ=0

我们用反证法来证明结论. 若 \varphi\neq 0φ≠0<script id="MathJax-Element-10" type="math/tex">\varphi\neq 0</script>, 则 \varphiφ<script id="MathJax-Element-11" type="math/tex">\varphi</script> 的 Jordan 标准型中至少有一个 Jordan 块的阶数大于 1, 不妨设为 J_m(0),\,m\geq 2Jm(0),m≥2<script id="MathJax-Element-12" type="math/tex">J_m(0),\,m\geq 2</script>. 设这个 Jordan 块对应的基向量为 e_1,e_2,\cdots,e_me1,e2,?,em<script id="MathJax-Element-13" type="math/tex">e_1,e_2,\cdots,e_m</script>, 则有 \varphi(e_1)=0,\,\,\varphi(e_2)=e_1,\,\,\cdots.

φ(e1)=0,φ(e2)=e1,?.

(φ(e2),e2)=0

(e1,e2)=0

0=(φ(e1+e2),e1+e2)=(e1,e1+e2)=(e1,e1)>0,