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Richard Kaye, Robert Wilson - Linear algebra (1998)
| $α,β∈\mathscr L(V,V)$为自伴算子,$V$是实/复内积空间,则$\abs{⟨v∣αβ\left(v\right)⟩}^2≥\frac{1}{4}\abs{⟨v∣\left(αβ-βα\right)\left(v\right)⟩}^2$. |
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将$αβ$写成$\frac12(αβ-βα)+\frac12(αβ+βα)$.
$⟨v∣αβ v⟩=\frac12⟨v∣(αβ-βα)v⟩+\frac12⟨v∣(αβ+βα)v⟩$
只要证明$⟨(αβ-βα)v∣(αβ+βα)v⟩=0$就能由勾股定理得\begin{align*}\abs{⟨v∣αβ\left(v\right)⟩}^2&=\frac{1}{4}\abs{⟨v∣\left(αβ-βα\right)\left(v\right)⟩}^2+\frac{1}{4}\abs{⟨v∣\left(αβ+βα\right)\left(v\right)⟩}^2\\&\ge\frac{1}{4}\abs{⟨v∣\left(αβ-βα\right)\left(v\right)⟩}^2\end{align*}红色的正交性如何证明
消去$⟨(αβ)v∣(βα)v⟩-⟨(βα)v∣(αβ)v⟩$得$$⟨αβ v∣αβ v⟩=⟨βα v∣βα v⟩$$以上还没有用到自伴算子这个条件... |
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