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.)


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: $$T(v) = T(||v|| v_1) = ||v|| T(v_1) = ||w|| w_1 = w.$$

  • $\begingroup$ 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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