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I have Euclidean vector space $(V,\langle -,-\rangle)$ and enter image description here

I have to show, that $s_x$ is orthogonal w.r.t inner product $\langle -,-\rangle$

Orthogonality means, that inner product is equal to 0. But in this case I cannot udnerstand which inner product should I take in order to prove the orthogonality.

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    $\begingroup$ What do you mean by the statement that $s_x$ is orthogonal? Orthogonality is generally between two objects, so what is the other object? $\endgroup$
    – Anurag A
    Commented Jun 13, 2017 at 6:51
  • $\begingroup$ As I understood : You have an application $s_x$ that takes some vector $y\in V$ as input, and outputs another vector $y -2\frac{<x,y>}{<x,x>} x$. You need to show that whatever vector $y$ you put in $s_x$, the resulting vector will be orthogonal to the first. Hence need to show that $\forall y \in V,\ <s_x(y),y> = 0$. That my interpretation of the question. I could be wrong. $\endgroup$
    – Zubzub
    Commented Jun 13, 2017 at 7:00
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    $\begingroup$ No, I think you are being asked to show that $s_x$ preserves orthogonality, that is, that if $u$ and $v$ are orthogonal, then so are $s_x(u)$ and $s_x(v)$. $\endgroup$
    – Gerry Myerson
    Commented Jun 13, 2017 at 7:08
  • $\begingroup$ Well, $s_x = -\sigma_x(y)$, where $\sigma_x$ is the axial symmetry with respect to the linear subspace spanned by $x$. So $s_x$ is the composition of an inversion and an axial symmetry, and now you can decide whether it is orthogonal or not, once you have understood what does orthogonal mean (IMHO a map is orthogonal if it preserves the inner product, as in the previous comment). $\endgroup$
    – SiD
    Commented Jun 13, 2017 at 9:37

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An orthogonal transformation on a real inner product space is, as mentioned in the comments, a linear transformation that preserves the inner product. So you have to show that for any $y,z \in V$ we have $$ \langle s_x(y), s_x(z) \rangle = \langle y, z \rangle. $$

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