# Hessenberg Matrices

$$A=\begin{bmatrix} 2&3&4\\ 3&-5&5\\ 4&5&0\end{bmatrix}$$ Find a unitary matrix $Q$ such that $A = QHQ^{H}$, where $H$ is Hessenberg.

I am having a little trouble finding my $Q$. I know that the first row and column in $Q$ look like $[1\quad 0 \quad 0]$, and the rest of the matrix (what's left is a $2\times 2$ matrix) is what my teacher called $F$. I know that $$F = I - \frac{2vv^{T}}{v^{T}v},$$ but I'm not sure how to come up with the $v$-vector in order to find $F$, so that I can find my $Q$. Then to find $H$, I know $H = Q^{H}AQ$, which once I find $Q$, seems to be easy to find, especially since all the eigenvalues of $A$ are real and don't have any imaginary parts, since $A$ is symmetric. And since $A$ is symmetric, then $H$ will be tridiagonal.

Any help would be greatly appreciated.

-