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- Let $Q$ be a linear operator acting in an $n$-dimensional Euclidean space $\mathbf R_n (n \geq 3)$. Suppose $Q$ does not change the area of any parallelogram, so that $$V[x, y] = V[Qx, Qy].$$ Show that $Q$ is an isometric operator.
The hint for this problem is:
- Hint. It is sufficient to show that $Q$ is an isogonal operator (see Prob. 31). Assuming that there is a right angle which is not transformed into a right angle, construct a parallelogram whose area changes as a result of applying the operator $Q$.
Taking the approach suggested by the hint, we only need to prove $Q$ is an isogonal operator, that is, for any $x, y \in \mathbf R_n$, if $(x, y) = 0$, then $(Qx, Qy) = 0$. To prove it, the hint suggests proof by contradiction: assume that we have found a pair of vectors $x, y \in \mathbf R_n$ with $(x, y) = 0$, that is, being orthogonal, but $(Qx, Qy) \neq 0$. Then, construct a parallelogram (i.e., another pair of vectors $a, b \in \mathbf R_n$) whose area does not preserve across the transformation, that is, $V[a, b] \neq V[Qa, Qb]$. But, even after a whole week contemplating, I still don't know how to construct such a parallelogram. Am I missing something? Any help is appreciated!
I don't see this hint in Dover's reprint of Shilov's book. Anyway, I think the problem can be solved easily without that hint.
Since $Q^TQ$ is real symmetric, it has an orthonormal eigenbasis $u_1,u_2,\ldots,u_n$. Let $Q^TQu_i=\lambda_iu_i$. Then $\lambda_i=u_i^TQ^TQu_i=\|Qu_i\|^2\ge0$ and $$ 1=V[u_i,u_j]=V[Qu_i,Qu_j]=\sqrt{[u_i|u_j]^TQ^TQ[u_i|u_j]}=\sqrt{\lambda_i\lambda_j} $$ for every $i\ne j$. Hence $\lambda_i=1$ for every $i$, meaning that $Q^TQ=I$, i.e. $Q$ is an isometry.