I am working on the following problem:
Let $x \in \mathbb{R}^n$ and let $H$ be a Householder matrix such that $Hx = \|x\|_2e_1$. Let $G_{1,2}, \ldots, G_{1,n}$ be Givens rotation matrices such that $Gx = G_{1,n} \cdots G_{1,2}x = \|x\|_2e_1$. Is $G$ equal to $H$?
My idea: Since $H$ is a Householder matrix, it is unitary, so $H^TH = I$. If we multiply both sides of $Hx = Gx$ by $H^T$ then we obtain $H^TGx = x$. Based on this, I guess $H$ does not have to be equal to $G$; the only thing we need is that $1$ is an eigenvalue of $H^TG$. Can you guys give me a hint? I am having troubles to come up with a counterexample.
If you want a counterexample, here is probably the simplest one. Let $\phi\in\mathbb{R}$ and consider $c:=\cos\phi$ and $s:=\sin\phi$. Set $$ H:=\begin{bmatrix}c&s\\s&-c\end{bmatrix}, \quad G:=\begin{bmatrix}c&s\\-s&c\end{bmatrix}. $$ The matrix $H$ is a reflection and $G$ is a rotation. Clearly $H\neq G$, even though $$ H\begin{bmatrix}c\\s\end{bmatrix} =G\begin{bmatrix}c\\s\end{bmatrix} =\begin{bmatrix}1\\0\end{bmatrix}. $$
The rotation matrix $G$ is clearly a Givens rotation. We just need to show that the reflection $H$ is a Householder reflection, that is, it is of the form $H=I-2vv^T$ with $v:=[v_1,v_2]^T$, such that $\|v\|_2^2=v_1^2+v_2^2=1$.
This can be verified by comparing the equation $I-2vv^T$ with the defining expression for $H$: $$ 1-2v_1^2=c,\quad 1-2v_2^2=c, \quad s=-2v_1v_2. $$ You can guess the solution from the first two equations and then adjust the signs of $v_1$ and $v_2$ in order to satisfy the last one. For example, one solution is $$ v_1=-\mathrm{sgn}(s)\sqrt{\frac{1-c}{2}}, \quad v_2=\sqrt{\frac{1+c}{2}}. $$