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Given some upper Hessenberg matrix $H \in R^{n \text{x} n}$, i know how to find an orthogonal matrix which is a product of Givens rotations such that $P^THP$ is also upper Hessenberg, but I'm not sure how to find $P$ so that

$P^THP=\begin{bmatrix}\lambda & w^T \\ 0 & H_1\end{bmatrix}$

where $H_1 \in R^{(n-1) \text {x} (n-1)}$ and $\lambda$ is the eigenvalue of $H$ and its corresponding eigenvector is $x$.

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  • $\begingroup$ the first column of $P$ must be the eigenvector of $H$ corresponding to the eigenvalue $\lambda$. the rest of the columns are chosen to make $P$ orthogonal. $\endgroup$ – abel Apr 20 '15 at 0:18
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Since you know $\lambda$ you can apply standard Hessenberg QR algorithm with shift and deflate H.

Standard steps would result in

$P^THP = \begin{bmatrix}H_1 & w\\ 0 &\lambda\end{bmatrix}$

However I think if we apply the $QR$ step from the left we can get the desired form for the solution

(Another solution)

Say $(\lambda,x)$ is the eigenvalue,eigenvector pair for $H$.
I guess another way to solve this is to apply Givens rotations on the $(n-1,n),(n-2,n-1),\cdots,(1,2)\space$ entries of $x$ like so

$G_{(1,2)}\cdots G_{(n-2,n-1)}G_{(n-1,n)}Hx = \lambda G_{(1,2)}\cdots G_{(n-2,n-1)}G_{(n-1,n)}x$

$\implies GHx = \lambda Gx\quad\quad$ where $G = G_{(1,2)}\cdots G_{(n-2,n-1)}G_{(n-1,n)}$

$\implies GHG^TGx = \lambda Gx$

$GHG^T$ is a unitary transform and preserves form via the Implicit-Q theorem and $\lambda Gx = \lambda e_1$

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