Hessenberg reduction Given $A \in \mathbb{R}^{nxn}$ and $z \in \mathbb{R}^n$, find orthogonal $Q$ such that $Q^TAQ$ is upper Hessenberg and $Q^Tz = \beta e_1$.
My attempt so far,
Individually I can find the Householder matrix $H_1 z= \beta e_1$ and $Q_1^TAQ_1 = R$ such that $R$ is upper Heisenberg but combining $Q_1$ and $H_1$ makes them lose one of the two properties.
Any hint on the property of $A$ or $z$ that I need to look out for would be appreciated.
(Edit)
The solution $Q=Q_1H_1$ is sufficient for the problem. It will preserve the $\beta e_1$ property of $H_1$ if we apply appropriate Householder transformations in the standard manner to compute $Q_1$.
 A: First, determine the HH transformation $Q_1$ such that $Q_1^Tz=\beta e_1$ (where $\beta=\pm\|z\|_2$ depending on the implementation of the HH transformation; you can multiply $Q_1$ by $-1$ if you are not happy with the sign of $\beta$) and set $B:=Q_1^TAQ_1$. Then apply the usual "Hessenbergization" to $B$ instead of $A$: $Q_{n-1}^T\cdots Q_2^TBQ_2\cdots Q_{n-1}=H$. The matrix $Q:=Q_1\cdots Q_{n-1}$ then satisfies $Q^Tz=\beta e_1$ and $Q^TAQ=H$.
Here is a (very simplified but working) MATLAB code:
function [Q, H] = hess2(A, z)

assert(ismatrix(A) && size(A,1) == size(A,2),...
       'A must be a square matrix.');

n = size(A, 1);

assert(all(size(z) == [n, 1]),...
       'Z must be a %d x 1 vector.', n);

Q = eye(n);
H = A;
for k = 1 : n - 1
    if k == 1
        v = hhmat(z);
    else
        v = hhmat(H(k:end,k-1));
    end
    V = eye(n - k + 1) - 2 * v * v';
    W = blkdiag(eye(k - 1), V);
    H = W * H * W;
    Q = Q * W;
end

function v = hhmat(x)
% Compute the vector V so that H := I - 2 * V * V' is a Householder
% transformation satisfying H * V = s * e_1, where s is + or - 1.

v = x;
alpha = -norm(v);
if (v(1) < 0) alpha = -alpha; end
v(1) = v(1) - alpha; 
v = v / norm(v);

Alternatively (as a better option), you can first compute $B:=Q_1^TAQ_1$ and then use the MATLAB's hess function to compute the Hessenberg form of $B$ (and compute the product of the two orthogonal matrices $Q_1$ and that obtained from hess). 
The reason why this works is that the HH matrices involved in the "Hessenbegization" have the form
$$
\pmatrix{I_k&0\\0&\tilde{Q}},
$$
where $I_k$ is a $k\times k$ identity and $k\geq 1$. So in
$$
Q^Tz=Q_{n-1}^T\cdots Q_2^TQ_1^T=Q_{n-1}^T\cdots Q_2^T(\beta e_1),
$$
they act only on the zero components of the vector $Q_1^Tz=\beta e_1$.
