# why is $Hx= x-2u^{H}xu$ ? why not $Hx= x(x-uu^H) + 2((u^Hx)u)^2?$

I have some confusion in this diagram

My confusion : why is $$Hx= x-2u^{H}xu$$ ? why not $$Hx=x(x-uu^H) + 2((u^Hx)u)^2?$$

My thinking is that by using pythogoras theorem blue line(vector) denotes $$(x-(u^Hx)u)^2 +(u^Hx)^2= x^2 + ((u^Hx)u)^2- 2xu^Hu +((u^Hx)u)^2 = x(x-uu^H) + 2((u^Hx)u)^2 \neq (I-2uu^H)x$$

• When you are squaring a vector I assume you are taking the squared 2-norm of that vector which would result in a scalar. Assuming that that is correct, then some of your calculations seem incorrect, namely you are adding vectors to scalars. Jan 2 at 3:58

You are splitting the vector $$x$$ into its components $$(u^Hx)u$$ that is parallel to $$u$$ and its complement $$x-(u^Hx)u$$ that is orthogonal to $$u$$. Then the reflected vector is composed by reducing the last one again by the parallel component. The idea being that the reflection on the plane that has $$u$$ as normal vector changes the sign of the component parallel to $$u$$. \begin{align} x &= [x-(u^Hx)u]+(u^Hx)u\\ Hx &= [x-(u^Hx)u]-(u^Hx)u=x-2u(u^Hx)=(I-2uu^H)x. \end{align}
By the application of Pythagoras you should have got that $$\|Hx\|=\|x\|$$. But you made several errors. The first is not to use the square of the vector (what is that?), but the square of the norm. Then you need to apply that the underlying scalar product is only semi-linear, so that the binomial expansion should read $$\|u+v\|^2=\|u\|^2+u^Hv+v^Hu+\|v\|^2.$$ Also, it should be suspicious if you start adding scalars to vectors or vectors to matrices as in $$(x−uu^H)$$. \begin{align} \|Hx\|^2 &= \|x-(u^Hx)u\|^2+\|(u^Hx)u\|^2\\ &=\|x\|^2-x^H((u^Hx)u)-((u^Hx)u)^Hx+2|u^Hx|^2\\ &=\|x\|^2-(x^Hu)(u^Hx)-\overline{(u^Hx)}(u^Hx)+2|u^Hx|^2\\ &=\|x\|^2 \end{align} using $$|z|^2=\bar zz$$ and $$x^Hu=\overline{u^Hx}$$.