Induction step of proof of 'Every element of $O(n)$ is product of hyperplane reflections' The following came up in induction step of proof of 'Every element of $O(n)$ is product of hyperplane reflections'.
An element $A$ in $O(n)$ is called hyperplane reflection if  $$A=Pdiag(1,\cdots , 1,-1)P^T$$
where $P\in O(n)$.
If $$A'=P \begin{bmatrix}
A_{n-1} & \\
 & \pm 1 
\end{bmatrix} P^T
$$
 and $A_{n-1}$ is in $O(n-1)$ then $A'$ is product of hyperplane reflections in $O(n)$. Why is this true?
All we know is $A_{n-1}$ is product of hyperplane reflections. How does this give result?   
 A: I think this is the lemma you need:
Lemma:  Suppose $B$ has the block diagonal form $diag(A,\pm 1)$, where $A\in O(n-1)$ is a hyperplane reflection.  Then $B$ is a product of at most 2 hyperplane reflections.
Proof:  We first handle the case when $B = diag(A,1)$.  Write $A = P \, diag(1,1..,1,-1)\, P^t$, where all of these matrices are in $O(n-1)$.
Consider the homomorphism $\hat{}:O(n-1)\rightarrow O(n)$ with $\hat{C}$ the block diagonal matrix $diag(C,1)$.
Then $B = \hat{A} = \hat{P}\,diag(1,....,1,-1,1)\, \hat{P}^t$.  This is almost a hyperplane reflection.  The issue is that that $-1$ appears in the wrong slot.  So, let's fix it.
Let $Q\in O(n)$ be obtained from the $n\times n$ identity matrix by swapping the last two rows.  Then $diag(1,...,1,-1,1) =Q\, diag(1,....,1,-1)\, Q^t$.  Substituting this into our formula for $B$ above, we get $B = \hat{A} = \hat{P}Q\,diag(1,...,1,-1),^t \hat{P}^t = (\hat{P}Q)diag(1,...,1,-1) (\hat{P}Q)^t.$
Since $\hat{P},Q\in O(n)$, it follows that $\hat{P}Q\in O(n)$, so $B$ is a hyperplane reflection in this case.
Now, we may assume $B = diag(A,-1) = diag(A,1)diag(1,...,1,-1)$, where, as above, $A = P\,diag(1,...,1,-1)\,P^t$.  From the first case, $B$ is of the form $C\,diag(1,...,1-1)$ where $C$ is a hyperplane reflection.  Since $diag(1,...,1,-1)$ is obviously a hyperplane reflection, $B$ is a product of hyperplane reflections in this case.
