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This problem is from Kunze Hoffman book. I think I go in the right direction to solve this but I miss some point to finish it. Can anyone help me?

Suppose $U$ is a self-adjoint unitary linear operator on a finite-dimensional inner product space $V$. Prove that we can find a subspace $W$ such that, if we express every $\alpha$ in $V\,$ in the form $\alpha = \beta + \gamma$, here $\beta$ in $W$ and $\gamma$ in $W^{\bot}$, then we have $U(\alpha) = \beta - \gamma$

If we denote $W = range(U+I)$, $W' = range(U - I)$,then must have $W \subset null(U-I)$ and $W \subset W^{'\bot}$. We can finish the proof if we can prove that $W' = null(U+I)$ or $W' = W^{\bot}$. But I can't prove this...

Thanks

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Hint: Geometrically this all wants to be the reflection through the subspace $W$. For any $\alpha$, define $$\beta:=\frac{\alpha+U\alpha}2\,, \quad \gamma:=\frac{\alpha-U\alpha}2$$ (Then the $\beta$'s will span $W={\rm ran}(U+I)$ and $\gamma$'s span all $W'={\rm ran}(U-I)$.) Prove that $\beta\perp\gamma$ using the hypothesis on $U$.

Ah, you also need that the whole $W$ is perpendicular to $W'$. But it's the same proof: $$\langle \alpha+U\alpha,\,\theta-U\theta\rangle=\langle \alpha,\theta\rangle+\langle U\alpha,\theta\rangle-\langle \alpha,U\theta\rangle- \langle U\alpha,U\theta\rangle = \dots$$

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    $\begingroup$ Thanks so much. It's unbelievably simple and clear. Great!!! $\endgroup$ – le duc quang Mar 11 '13 at 1:11

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