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For W be a finite dimensional subspace of an inner product space V. Given V is the direct sum of W and its orthogonal complement W'.

For a map U defined on V as U (v + v') = v - v' , for all v in W , v' in W'.

I have to show that U is a self - adjoint and unitary operator.

The part that U is unitary operator is clear as U preserves length.

I am trying to show that U is self- adjoint. Please suggest.

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up vote 4 down vote accepted

You could also just check, for $v = v_1 + v_2$ and $w = w_1 + w_2$ where $v_1, w_1 \in W$ and $v_2, w_2 \in W'$, that \[ \langle Uv, w\rangle = \langle v_1 - v_2, w_1 + w_2\rangle \] is equal to $\langle v, Uw\rangle$. Use the linearity of the inner product and that, e.g., $\langle v_1, w_2\rangle = 0$.

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Thanks a lot for the help. –  Tav Mar 1 '12 at 17:19
    
I'm also trying to solve this, and I tried what you suggested, but it didn't work.I ended up with $<Uv,w> = ... = <v_1 , w_1> + <-w_2, v_2>$ how do you get to the wanted: $<v_1+v_2, w_1 - w_2> $ ? @Dylan Moreland –  Dor Shalom Dec 22 '13 at 9:33
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You could use the fact that an operator is self-adjoint iff it is represented by a Hermitian matrix. Form a basis of $V$ by combining a basis of $W$ and a basis of $W'$. In this basis, the operator is represented by a diagonal matrix with $1$s and $-1$s on the diagonal.

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