Proving $A$ is an isometry on $\mathbb{R}^3$ if $A = I - 2vv^T$ 
Let $v$ be a vector of unit length in $\mathbb{R}^3$, and $A = I - 2vv^T$. Prove that $A$ is an isometry.

I need to prove that $A$ preserves dot products, i.e., $A x \cdot Ay = x \cdot y$ for any $x,y \in \mathbb{R}^3$, but I have no clue how to get started and how the assumption that $v$ has unit length plays a role.
 A: Hint: One way to do this (assuming you are using the standard euclidean dot product) is to write $Ax \cdot Ay = (Ay)^TAx = y^TA^TAx$. Then you can show that $A^TA = I$ using the fact that $v$ is a vector of unit length (use that $v^Tv = 1$).
Then $Ax \cdot Ay = y^Tx = x \cdot y$.
Proof that $A^TA = I$:

 We have $A^TA = (I - 2vv^T)^T(I - 2vv^T)$, using the properties of the transpose and expanding we have $A^TA = I - 4vv^T + 4vv^Tvv^T$. Since matrix multiplication is associative and $v^Tv = 1$ we have $$A^TA = I - 4vv^T + 4v(v^Tv)v^T = I - 4vv^T + 4vv^T = I.$$

A: Let $P$ be the plane orthogonal to $v$.
$$A = I - 2vv^T$$ is the matrix of the orthogonal symmetry with respect to $P$ (proof below) ; as such, it preserves lengths (or dot-products, which is equivalent), therefore, it is an isometry.
Proof :

*

*For any $X \in P$ (i.e., such that $X \perp v$ , otherwise said  such that $v^TX=0$.), we have  $AX=(I-vv^T)X=IX-v(v^TX)=X-v0=X$.


*For any $X \perp P$, (i.e., such that $X=av$ for some $a \in \mathbb R$), we have $AX=(I-vv^T)X=av-v(v^Tav)=av-a \|v\|^2 v=av-av=0$.
Remark: in the same vein $B=I - vv^T$ is the matrix of orhtogonal projection onto plane $P$. Not an isometry, this one.
