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Suppose $V$ is a vector space over a field not of characteristic $2$, and is equipped with an inner product. I want to show that, given vectors $v$ and $w$, there is some orthogonal (inner-product-preserving) transformation taking $v$ to $w$.

Any help would be appreciated. I was toying with the reflection $$R_z(x) := x - 2\frac{x \cdot z}{\left|\left|z\right|\right|^2}z,$$ which negates $z$ and fixes its orthogonal complement pointwise. I am not sure if that is helpful.

(Note: The set of notes I've been given lists this as Witt's Lemma, but upon researching that, it seems to be quite complicated and involve machinery out of my reach.)

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It's enough to know here that a transformation $T: V \to V$ is orthogonal iff it maps orthonormal bases to orthonormal bases. (If this isn't familiar to you, you might like to prove it as an exercise.)

By construction $$v_1 := \frac{v}{||v||}$$ has unit length (here $||v|| := \sqrt{\langle v, v\rangle}$), so we can extend it to an orthonormal basis $(v_a)$ of $V$, and similarly we can extend $$w_1 := \frac{w}{||w||}$$ to an orthogonal basis $(w_a)$ of $V$. Then, by the above fact, the transformation $T: V \to V$ characterized by $$T: v_a \mapsto w_a, \qquad a = 1, \ldots, n,$$ is orthogonal, and in particular, it maps $v$ to $w$ as desired: $$T(v) = T(||v|| v_1) = ||v|| T(v_1) = ||w|| w_1 = w.$$

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  • $\begingroup$ What if the norm is $v \cdot v$? Does that change things? $\endgroup$ – Johann Linus Oct 30 '14 at 2:29
  • $\begingroup$ Instead of the square root, that is. $\endgroup$ – Johann Linus Oct 30 '14 at 2:29
  • $\begingroup$ I ask as the problem is for a generic inner product. $\endgroup$ – Johann Linus Oct 30 '14 at 2:32
  • $\begingroup$ There's no choice in the norm here, it's simply the one the inner product defines. (Anyway, $v \mapsto v \cdot v$ does not satisfy the Triangle Inequality, so it is not a norm in the first place.) $\endgroup$ – Travis Oct 30 '14 at 3:58

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