[Notes] Reflection through hyperplane

2016-08-23_145553

(from Introduction to Linear Algebra [4th Edition] by Gilbert Strang, section 4.4, page 231)

I don’t know the official reflection definition in mathematics so it would be a lame to prove that Q is indeed a reflection matrix, but I will note several stuffs that I have observed. Also special thanks to anyone gave me a hint that in 3-D, u is a unit vector orthogonal to a plane on Wikipedia.

I guess that if vector t = Qv is mirror image of vector v through hyperplane H then: \frac{v + t}{2} \in H and H \times (v - t) = 0 (here H is also the name of a matrix). My observations will be based on this guess.

I have:  v + t = 2(I - uu^{T})v. So H = C(I - uu^{T}), for simplicity I will also call H = I - uu^{T}.

Now I check:

    \[ H \times (v - t) = (I - uu^{T})(v - v - 2uu^{T}v) = 2(u(u^{T}u)u^{T} - uu^{T}) = 2(uu^{T} - uu^{T}) = 0 \]

I also check:

    \[ H \times u = (I - uu^{T})u = (u - u(u^{T}u)) = u - u = 0 \]

So far so good, the only one concern left: Is H a hyperplane?

I have:

    \[ H \times x = 0 \Leftrightarrow (I - uu^{T})x = 0 \Leftrightarrow x = uu^{T}x \Rightarrow x \in C(uu^{T}) \]

and:

    \[ x \in C(uu^{T}) \Leftrightarrow  x = ku, k \in \mathbb{R} \Rightarrow H \times x = H \times ku = 0 \]

So if u \in R^{n} then:

 N(H) = C(uu^{T}) \Rightarrow dim N(H) = 1 \Leftrightarrow dim C(H) = dim H = n - dim N(H) = n - 1

So H is a hyperplane of the vector space R^{n}.

H also has a nice property: H^{2} = (I - uu^{T})^{2} = I - uu^{T} - uu^{T} + u(u^{T}u)u^{T} = I - uu^{T} = H.

 


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