How to show: Matrix $H_v = I - 2vv^{\top}$ with $||v|| =1$ to the reflection on hyperplane $v^{\bot}$ is symetrical and orthogonal and $det H_v = -1$?

Question

I have several questions

1. How can I show that a Matrix $H_v = I - 2vv^{\top}$ with $||v|| =1$ to the reflection on the hyperplane $v^{\bot}$ is symetrical and orthogonal and the determinant is $det H_v = -1$?\

2. And is every symetrical orthogonal matrix $A$ with $A = -1$ formed $A=H_v$ for a $v$?

3. How to show that every orthogonal matrix $A$ is a product of reflection matrices $H_v$ on the hyperplane $v^{\bot}$

4. How can one describe every quadratic upper triangular matrices $A$ which are orthogonal?

Answer 1

I had to work a bit on it, and with the help of other answers on this board I came up with some answers:

1. $I$ is symetric. The dot product §vv^{\top}§ gives a symmetric matrix as result. The $2$ in front is irrelevant. It's a mutpile so that still counts. It is $(AB)^{\top} = B^{\top} A^{\top} $, so $I^{\top} = I$ and $(vv^{\top})^{\top} = vv^{\top}$. And with that $H^{\top} = (I- vv^{\top})^{\top} = I^{\top} - (vv^{\top})^{\top} = I - vv^{\top} = H$. $H$ is therefor symmetrical.

2. counter example matrix [-1 0 0; 0 -1 0; 0 0 -1]

3. Let $u,v \in \mathbb{R^n}$ be colmn vectors, their dot product is then $\langle u,v \rangle = u^{\top} \cdot v$. Let $A$ be a orthogonal Matrix, then:

$\langle A \cdot u,A \cdot v \rangle = (A \cdot u)^{\top} \cdot (A \cdot v) = u^{\top} \cdot A^{\top} \cdot A \cdot v = u^{\top} \cdot A^{-1} \cdot A \cdot v = u^{\top} \cdot v$ This shows that the transformation keeps the dot product. no change of angles is happening. This is in general the product of reflections. Bonus: Every reflection can be made with a series of householder reflections. So it is about householder reflections and matrices, too.

4. There are no upper triangular matrices which are orthogonal. Under the main diagonal there are only zeros, and multiplied with another upper tirangular matrix, the result will again be an upper triangular matrix. Thus the Inverse of the upper triangular matrix $R$ is an upper triangular matrix. The transpose on the other hand is an lower triangular matrix $R^{\top} = L$ But from that follows $R^{\top} \neq R^{-1}$