How to prove that the reflection is an orthogonal tranformation

Question

From Humphreys: A reflection is a linear operator $s$ on $V$ which sends some nonzero vector $\alpha$ to its negative while fixing pointwise the hyperplane $H_{\alpha}$ orthogonal to $\alpha.$ We may write $s=s_{\alpha}.$ The formula for the reflection is

$s_{\alpha}\lambda=\lambda-\frac{(\lambda,\alpha)}{(\alpha,\alpha)}\alpha.$

How to explicitly show that $s_{\alpha}$ is an orthogonal transformation, i.e. $(s_{\alpha}\lambda, s_{\alpha}\mu)=(\lambda, \mu)$ for all $\lambda, \mu \in V$?

Thanks!

Edit: Sorry for the wrong formula given above, it should be

$s_{\alpha}\lambda=\lambda-2\frac{(\lambda,\alpha)}{(\alpha,\alpha)}\alpha.$

Here is my attempt.

$(s_{\alpha}\lambda, s_{\alpha}\mu)=(\lambda - 2\frac{(\lambda,\alpha)}{(\alpha,\alpha)}\alpha , \mu - 2\frac{(\mu,\alpha)}{(\alpha,\alpha)}\alpha )=(\lambda, \mu)+(\lambda, - 2\frac{(\mu,\alpha)}{(\alpha,\alpha)}\alpha)+ (- 2\frac{(\lambda,\alpha)}{(\alpha,\alpha)}\alpha, \mu)+ (- 2\frac{(\lambda,\alpha)}{(\alpha,\alpha)}\alpha, - 2\frac{(\mu,\alpha)}{(\alpha,\alpha)}\alpha)$

and I don't know if it's correct.

Answer 1

Maybe, your formula is wrong. I know $s_{\alpha}(\lambda)=\lambda-2\frac{(\lambda,\alpha)}{(\alpha, \alpha)}\alpha$

First, show that any reflection is linear.

Finally, show that any reflection preserves norm.

Then we get : (Norm preserving + linear $\Rightarrow$ Inner product preserving.)

$\textbf{Solution:}$ Clearly, $s_{\alpha}(c\lambda)=c\lambda-2\frac{(c\lambda,\alpha)}{(\alpha, \alpha)}\alpha=cs_{\alpha}(\lambda)$.

Likewise, you can show that $s_{\alpha}(\lambda+\mu)=s_{\alpha}(\lambda)+s_{\alpha}(\mu)$ easily. (Use linearity of inner product.)

$||s_{\alpha}(\lambda)||^2=||\lambda||^2 +4\frac{(\lambda,\alpha)^2}{(\alpha,\alpha)^2}||\alpha||^2-4\frac{(\lambda,\alpha)}{(\alpha,\alpha)}(\lambda,\alpha)=||\lambda||^2$. Hence, reflection is norm-preserving and linear.

$\textbf{Claim}:$ If $T$ is a linear operator on $V$ and norm-preserving, $T$ preserves inner product.

Suppose that $T$ is linear and $||T(v)||=||v||$ for every $v\in V$. Then

$||u-v||^2=||T(u-v)||^2=(T(u-v),T(u-v))=(Tu-Tv,Tu-Tv)$

= $(Tu,Tu)+(Tv,Tv)-2(Tu,Tv)=||Tu||^2+||Tv||^2-2(Tu,Tv)=||u||^2+||v||^2-2(Tu,Tv)$

Therefore we have $$||u-v||^2=||u||^2+||v||^2-2(Tu,Tv).$$

Thus,

$$(u,v)=(Tu,Tv)$$ for every $u,v \in V.$

Answer 2

Just keep going with your calculation: $$\begin{align} &(\lambda,\mu) + \left(\lambda, -2{(\mu,\alpha)\over(\alpha,\alpha)}\alpha\right) + \left(\mu,-2{(\lambda,\alpha)\over(\alpha,\alpha)}\alpha\right) + \left(-2{(\lambda,\alpha)\over(\alpha,\alpha)}\alpha,-2{(\mu,\alpha)\over(\alpha,\alpha)}\alpha\right) \\ =\; &(\lambda,\mu) - 2{(\lambda,\alpha)(\mu,\alpha)\over(\alpha,\alpha)} - 2{(\lambda,\alpha)(\mu,\alpha)\over(\alpha,\alpha)} + 4{(\lambda,\alpha)(\mu,\alpha)(\alpha,\alpha)\over(\alpha,\alpha)^2} \\ =\; &(\lambda,\mu). \end{align}$$

Answer 3

Choosing an orthonormal basis of $H_\alpha$, and completing with $\frac\alpha{\|\alpha\|}$, one has an orthonormal basis of the whole space. On this basis the matrix of the reflection is diagonal with diagonal entries $1$ except the last which is $-1$. That is clearly an orthogonal matrix. This suffices, as on any orthonormal basis, the orthogonal transformations are precisely those whose matrix is an orthogonal matrix.