Reflection matrix in $ \mathbb{R}^{3} $.

Question

I need help in understanding how they got the transformation matrix $ Q_{L} $ from Theorem 2 and $ P_{M} $ at the bottom of the page. They skipped some steps and I find it confusing. Any help would be greatly appreciated. :)

Cheers.

Answer 1

You have one vector, $d,$ written as a column vector. The transpose $d^T$ is a row vector. First, the squared length of $d,$ which is $a^2 + b^2 + c^2,$ is just $d^T d.$ When regarded as matrices, the result of $d^T d$ is officially a $1$ by $1$ matrix; the tradition is to regard this matrix, with a single entry, as that entry itself, that is $d^T d = d \cdot d.$

Next, $d d^T$ really is a $3$ by $3$ matrix. The first item in the book is $$ P = \frac{1}{d \cdot d} d d^T. $$ I suggest you write this out, confirm.

Suppose we make an orthogonal basis, with $d$ and then two new ones, $e$ and $f.$ We are demanding $d \cdot e = 0,$ $d \cdot f = 0,$ $e \cdot f = 0.$ Using the description above, we find $$ P e = \frac{1}{d \cdot d} d d^T e = \frac{1}{d \cdot d} d (d^T e) = \frac{1}{d \cdot d} d (0) = 0,$$ similar for $f.$

In comparison $$ P d = \frac{1}{d \cdot d} d d^T d = \frac{1}{d \cdot d} d (d^T d) = \frac{1}{d \cdot d} d ( d \cdot d) = d.$$ So it sends $d$ to itself.

Very similar, $$ Q = I - 2P, $$ so $Qe=e,$ also $Qf = f,$ but $Qd = d - 2d = -d.$

There is more to be said but this is a good start.